Let's assume that the DM (decision maker) is considering several possible solutions: i = 1,...,m. The situation in which the decision maker operates is uncertain. It is only known that one of the options is present: j = 1,…, n. If the i -e decision is made, and the situation is j -th, then the firm headed by the decision maker will receive income q ij . The matrix Q = (q ij) is called the matrix of consequences (possible solutions). What decision does the decision maker need to make? In this situation of complete uncertainty, only some preliminary recommendations can be made. They will not necessarily be accepted by the decision maker. Much will depend, for example, on his appetite for risk. But how to assess the risk in this scheme?
Let's say we want to estimate the risk posed by the i -e decision. We don't know the real situation. But if they knew it, they would choose the best solution, i.e. generating the most income. Those. if the situation is j, then a decision would be made that would yield income q ij.
This means that by making the i -e decision we risk getting not q j , but only q ij , which means that making the i -th decision carries the risk of not getting r ij = q j - q ij . The matrix R = (r ij) is called the risk matrix.

Example No. 1. Let there be a matrix of consequences
Let's create a risk matrix. We have q 1 = max(q i 1) = 8, q 2 = 5, q 3 = 8, q 4 = 12.. Therefore, the risk matrix is

Decision making under conditions of complete uncertainty

Not everything random can be “measured” by probability. Uncertainty is a broader concept. The uncertainty of which number the die will land on is different from the uncertainty of what the state of the Russian economy will be in 15 years. Briefly speaking, unique individual random phenomena are associated with uncertainty, while massive random phenomena necessarily allow for some patterns of a probabilistic nature.
A situation of complete uncertainty is characterized by the absence of any additional information. What rules and recommendations exist for making decisions in this situation?

Wald's rule(rule of extreme pessimism). Considering the i -e solution, we will assume that in fact the situation is the worst, i.e. bringing the smallest income a i But now let’s choose the solution i 0 with the largest a i0 . So, Wald's rule recommends making a decision i0 such that
So, in the above example, we have a 1 = 2, a 2 = 2, a 3 = 3, a 4 = 1. Of these numbers, the maximum is number 3. This means that Wald’s rule recommends making the 3rd decision.

Savage Rule(minimum risk rule). When applying this rule, the risk matrix R = (rij) is analyzed. Considering the i -e solution, we will assume that in fact there is a situation of maximum risk b i = max
But now let’s choose the solution i 0 with the smallest b i0 . So, Savage's rule recommends making a decision i 0 such that
In the example under consideration we have b 1 = 8, b 2 = 6, b 3 = 5, b 4 = 7. The minimum of these numbers is the number 5. I.e. Savage's rule recommends making the 3rd decision.

Hurwitz rule(weighing pessimistic and optimistic approaches to a situation). Decision i is made, at which the maximum is achieved
, where 0 ≤ λ ≤ 1.
The value of λ is chosen for subjective reasons. If λ approaches 1, then the Hurwitz rule approaches the Wald rule; as λ approaches 0, the Hurwitz rule approaches the “pink optimism” rule (guess for yourself what this means). In the above example, with λ = 1/2, Hurwitz's rule recommends the 2nd solution.

Decision making under conditions of partial uncertainty

Let us assume that in the scheme under consideration the probabilities pj are known that the real situation develops according to option j. This situation is called partial uncertainty. How to make a decision here? You can select one of the following rules.
Rule for maximizing average expected income. The income received by the company when implementing the i-th solution is a random variable Qi with a distribution series

qi1	qi2	…	qin
p1	p2	…	pn

The mathematical expectation M is the average expected income, denoted by . The rule recommends making the decision that yields the maximum average expected return.
Suppose that in the circuit from the previous example the probabilities are (1/2, 1/6, 1/6, 1/6). Then Q 1 =29/6, Q 2 =25/6, Q 3 =7, Q 4 =17/6. The maximum average expected return is 7, corresponding to the third solution.
Rule for minimizing average expected risk. The risk of the company when implementing the i-th decision is a random variable R i with a distribution series

ri1	ri2	…	rin
p1	p2	…	pn

The mathematical expectation M is the average expected risk, also denoted R i. The rule recommends making a decision that entails the minimum average expected risk.
Let us calculate the average expected risks for the above probabilities. We get R 1 =20/6, R 2 =4, R 3 =7/6, R 4 =32/5. The minimum average expected risk is 7/6, corresponding to the third solution.
Analysis of decisions made according to two criteria: average expected income and average expected risk and finding Pareto optimal solutions is similar to the analysis of profitability and risk of financial transactions. In the example, the set of solutions that are Pareto optimal operations consists of only one 3rd solution.
If the number of Pareto-optimal solutions is more than one, then the weighting formula f(Q)=2Q -R is used to determine the best solution.

Laplace's rule

Sometimes, in conditions of complete uncertainty, Laplace's rule is used, according to which all probabilities p j are considered equal. After this, you can choose one of the two decision-making rules-recommendations given above.

Example No. 2. Let's consider an example of solving a statistical game in an economic problem.
An agricultural enterprise can sell some products:
A1) immediately after cleaning;
A2) in the winter months;
A3) in the spring months.
Profit depends on the selling price in a given period of time, storage costs and possible losses. The amount of profit calculated for different states-ratios of income and costs (S1, S2 and S3), during the entire period of implementation, is presented in the form of a matrix (million rubles)

	S1	S2	S3
A1	2	-3	7
A2	-1	5	4
A3	-7	13	-3

Determine the most profitable strategy according to all criteria (Bayes criterion, Laplace criterion, Wald maximin criterion, Hurwitz pessimism-optimism criterion, Hodge-Lehman criterion, Savage minimax risk criterion) if the probabilities of demand states: 0.2; 0.5; 0.3; pessimism coefficient C = 0.4; coefficient of reliability of information about demand conditions u = 0.6.
Solution
The calculation results will be entered into the table:

	S1	S2	S3	B	BUT	MM	BY	H-L
A1	2	-3	7	1	2	-3	3	-0,6
A2	-1	5	4	3,5	2,7	-1	2,6	1,7
A3	-7	13	-3	4,2	1	-7	5	-0,28
p j	0,2	0,5	0,3	A3	A2	A2	A3	A2

1. Bayes criterion (maximum mathematical expectation)

The calculation is carried out according to the formula:
;
W 1 = 2∙0.2 + (-3) ∙0.5 + 7∙0.3 = 0.4 – 1.5 + 2.1 = 1
W 2 = -1∙0.2 + 5 ∙0.5 + 4∙0.3 = -0.2 + 2.5 + 1.2 = 3.5
W 3 = -7∙0.2 + 13∙0.5 + (-3)∙0.3 = -1.2 + 6.5 - 0.9 = 4.2
We enter the found values ​​in the first column (B) and select the maximum
W = max(1;3.5;4.2) = 4.2,

This means that strategy A3 is optimal according to this criterion – sell in the spring months.

2. Laplace's insufficient base criterion (LCR)

Find the average value of the elements of each row:
.
;
;
.
We enter the found values ​​in the second column (BUT) and select the maximum W = max(2; 2.7; 1) = 2.7, which means that strategy A2 is optimal according to this criterion - sell in the winter months.

3. Maximin Wald criterion (MM)

In each line we find the minimum element: .
W 1 = min(2; -3; 7) = -3
W 2 = min(-1; 5; 4) = -1
W 3 = min(-7; 13; -3) = -7
We enter the found values ​​in the third column (MM) and select the maximum W = max(-3; -1; 7) = -1, which means that strategy A2 is optimal according to this criterion - sell in the winter months.

4. Hurwitz criterion of pessimism-optimism (P-O)

For each line, we calculate the value of the criterion using the formula: . According to the condition, C = 0.4, which means:
W 1 = 0.4∙min(2; -3; 7) + (1-0.4) ∙ max(2; -3; 7) = 0.4∙(-3) + 0.6∙7 = -1.2 + 4.2 = 3
W 2 = 0.4∙min(-1; 5; 4) + (1-0.4) ∙ max(-1; 5; 4) = 0.4∙(-1) + 0.6∙5 = -0.4 + 3 = 2.6
W 3 = 0.4∙min(-7; 13; -3) + (1-0.4) ∙ max(-7; 13; -3) = 0.4∙(-7) + 0.6∙ 13 = -2.8 + 7.2 = 5
We enter the found values ​​in the fourth column (P-O) and select the maximum W = max(3; 2.6 5) = 5, which means that strategy A3 is optimal according to this criterion - sell in the spring months.

5. Hodge-Lehman criterion (HL)

For each line, we calculate the criterion value using the formula: . According to the condition u = 0.6 and the factors in each term have already been calculated, they can be taken from the first column (B) and from the third column (MM), which means:
W 1 = 0.6∙1 + (1-0.6) ∙(-3) = 0.6 – 1.2 = -0.6
W 2 = 0.6∙3.5 + (1-0.6) ∙(-1) = 2.1 – 0.4 = 1.7
W 3 = 0.6∙4.2 + (1-0.6) ∙(-7) = 2.52 – 2.8 = -0.28
We enter the found values ​​in the fifth column (Х-Л) and select the maximum W = max(-0.6; 1.7; -0.28) = 1.7, which means that strategy A2 is optimal according to this criterion - sell in the winter months.

5. Savage's minimax risk criterion

Let's calculate the risk matrix. It is better to fill it in columns. In each column we find the maximum element and you read from it all the other elements of the column, writing the results in the appropriate places.
Here's how the first column is calculated. The maximum element in the first column: a 11 = 2, which means according to the formula :
r 11 = 2 – a 11 = 2 -2 = 0
r 21 = 2 – a 21 = 2 –(-1) = 3
r 31 = 2 – a 31 = 2 –(-7) = 9
Let's calculate the second column of the risk matrix. The maximum element in the second column is: a 32 = 13, which means:
r 12 = 13 – a 12 = 13 –(-3) = 16
r 22 = 13 – a 22 = 13 –5 = 8
r 32 = 13 – a 32 = 13 –13 = 0
Let's calculate the third column of the risk matrix. The maximum element in the third column is: a 13 = 7, which means:
r 13 = 7 – a 13 = 7 –7 = 0
r 23 = 7 – a 23 = 7 –4 = 3
r 33 = 7 – a 33 = 7 –(-3) = 10
Thus, the risk matrix has the form (in each column, in place of the maximum element of the payment matrix there should be a zero):

			W i
0	16	0	16
3	8	3	8
9	0	10	10

Let's supplement the risk matrix with the calculated values ​​of the Wi criterion - in each row we select the maximum element ():
W 1 = max(0; 16; 0) = 16
W 2 = max(3; 8; 3) = 8
W 3 = max(9; 0; 10) = 10
We enter the found values ​​in the column (W i) and select the minimum W = min(16,8,10) = 8, which means that strategy A2 is optimal according to this criterion - sell in the winter months.

Conclusion:

1. Strategy A1 (sell immediately after harvesting) is not optimal according to any of the criteria.
2. Strategy A2 (sell in the winter months) is optimal according to the Laplace insufficient base criterion, the Wald maximin criterion and the Savage minimax criterion.
3. Strategy A3 (sell in the spring months) is optimal according to the Bayesian, Hurwitz, Hodge-Lehman pessimism-optimism criteria.

Example No. 2. In a regular strategic game, each player takes exactly those actions that are most beneficial to him and less beneficial to his opponent. This assumes that the players are rational and antagonistic opponents. However, very often there is uncertainty, which is not associated with the conscious opposition of the enemy, but depends on some objective reality.
The agricultural enterprise has three plots of land: wet, medium moisture and dry. One of these plots is supposed to be used for growing potatoes, the rest - for sowing green mass. To obtain a good potato harvest, a certain amount of moisture in the soil is required during the growing season. If there is excessive moisture, planted potatoes may rot in some areas, and if there is insufficient rainfall, they will develop poorly, which leads to a decrease in yield. Determine in which area to sow potatoes in order to get a good harvest, if the average potato yield in each area is known, depending on weather conditions. Location on A 1 the yield is 200, 100 and 250 centners per 1 ha when the normal amount of precipitation falls, more and less than the norm, respectively. Similarly on the site A 2– 230, 120 and 200 cwt, and on the site A 3– 240, 260 and 100 c.
We use a game approach. Agricultural enterprise – player A, which has three strategies: A 1– sow potatoes in a damp area, A 2– in an area of ​​average humidity, A 3- on a dry area. Player P– nature, which has three strategies: P 1 corresponds to the amount of precipitation below normal, P 2– normal, P 3- more than normal. The gain of the agricultural enterprise for each pair of strategies ( A i, P j) is determined by the potato yield per hectare.

P A	P 1	P 2	P 3
A 1	250	200	100
A 2	200	230	120
A 3	100	240	260

Let's consider a general situation where some party needs to perform an operation in a poorly known environment. About the state of this situation we can make n assumptions: P 1, P 2,…, P n. For example, consumer demand. By analogy with example 8, these states are considered as strategies of nature. In statistical game theory, nature is not an intelligent player; it is viewed as a kind of disinterested entity that does not choose optimal strategies for itself. Its possible states are realized randomly. Such situations are usually called games with nature. Operating party A has at its disposal m possible strategies: A 1, A 2,…, Am. Player winnings A for each pair of strategies A i And P j assumed to be known a ij.
It may seem that playing with nature is easier than playing strategy because nature does not oppose the player A. In reality, this is not the case, since in an uncertain situation it is more difficult to make an informed decision. Although he will win A, most likely, more than in a game against a conscious opponent.

Example 9. The company produces popular children's dresses and suits, the sale of which depends on weather conditions. The company's costs during August-September per unit of production were: dresses - 7 den. units, suits – 28 den. units The selling price is 15 and 50 den. units respectively. According to observations over several previous years, the company can sell 1,950 dresses and 610 suits in warm weather, and 630 dresses and 1,050 suits in cool weather.
Create a payment matrix.
Solution. The company has two strategies: A 1: release products, believing that the weather will be warm; A 2: release products believing that the weather will be cool.
Nature has two strategies: B 1: the weather is warm; B 2: The weather is cool.
Let's find the elements of the payment matrix:
1) a 11 – the company’s income when choosing a strategy A 1 given that B 1:
a 11 =(15-7) 1950+(50-28) 610=29020.
2) a 12 – the company’s income when choosing A 1 given that B 2. The company will produce 1,950 dresses and sell 630, income from the sale of dresses
(15-7) 630-7 (1950-630)=5040-9240
a 12 =5040-9240+22·610=9220.
3) similarly for the strategy A 2 in conditions B 1 the company will produce 1,050 suits and sell 610;
a 21 =8 630+22 610-28 (1050-610)=6140
4) a 22 =8 630+22 1050=28140
Payment matrix:

20 020	9 220
6 140	28 140

Example 2. The association carries out mineral exploration in three deposits. The association's fund is 30 den. units Money to the first deposit M 1 can be invested in multiples of 9 den. units, second M 2– 6 days units, in the third M 3– 15 den. units Prices for mineral resources at the end of the planning period may be in two states: C 1 And C 2. Experts have found that in the situation C 1 profit from the field M 1 will be 20% of the amount of money invested. units for development, for M 2– 12% and on M 3- 15 %. In a situation C 1 at the end of the planning period the profit will be 17%, 15%, 23% in the fields M 1, M 3, M 3 respectively.
Player A- Union. Player P(nature) – a set of external circumstances that determine a particular profit in the fields. The player has A There are four possibilities that make full use of the available facilities. The first strategy A 1 is that A will invest in M 1 9 days units, in M 2 – 6 days units, in M 3 – 15 days units Second strategy A 2: in M 1 – 18 days units, in M 2 – 12 days units, in M 3 don’t invest money. Third strategy A 3: 30 days units invest in M 3. The fourth strategy A 4:. 30 den. units invest in M 2. Briefly we can write A 1 (9, 6, 15), A 2 (18, 12, 0), A 3 (0, 0, 30), A 4 (0, 30, 0).
Nature can realize one of its two states, characterized by different prices for minerals at the end of the planning period. Let us denote the states of nature P 1 (20 %, 12 %, 15 %), P 2 (17 %, 15 %, 23 %).
Elements a ij of the payment matrix have the meaning of the total profit received by the association in various situations ( A i, P j) (i=1, 2, 3, 4, j= 1, 2). For example, let's calculate a 12, corresponding to the situation ( A 1, P 2), i.e. the case when the association invests in deposits M 1 , M 2 , M 3, respectively 9 days. units, 6 days units, 15 days units, and at the end of the planning period prices were in a state C 2:
a 12= 9·0.17+6·0.15+15·0.23 = 5.88 den. units

Example 3. Flooding is expected and may range from category one to five. Amount of flood damage:

Flood category	1	2	3	4	5
Damage, den. units	5	10	13	16	20

As a preventive action, a dam can be built; There are five options for choosing the dam height: h 1 < h 2 < h 3 < h 4 < h 5, and the dam height h 1 protects only from floods of the first category, height h 2– from floods of the first and second categories, etc., dam height h 5 protects against floods of any category.
Dam construction costs:

Dam height	h 1	h 2	h 3	h 4	h 5
Costs, den. units	2	4	6	8	10

The decision maker has six strategies (not to build a dam at all ( A 0) or build a height dam h i (A i), i= 1, 2, 3, 4, 5). Nature also has six strategies (not to flood ( P 0) or cause a flood j th category ( P j), 1≤j≤5).
We get loss matrix:

P/A	P 0	P 1	P 2	P 3	P 4	P 5
A 0	0	5	10	13	16	20
A 1	2	2	12	15	18	22
A 2	4	4	4	17	20	24
A 3	6	6	6	6	22	26
A 4	8	8	8	8	8	28
A 5	10	10	10	10	10	10

For example, if you build a dam height h 2, and the flood will be of the third category, then construction costs will be 4 den. units, and damage from flooding is 13 den. units Thus, the total loss will be 4 + 13 = 17 den. units If the flood is of the second category, then there will be no damage from the flood, and losses are associated only with the construction of the dam, i.e. 4 days units
So that from the loss matrix ( b ij) to obtain the winning matrix, it is enough to change the sign of all elements and add any constant C(in this case C can be interpreted as the amount allocated for the construction of the dam, then the gain a ij =C-b ij represents the amount saved). For example, with C =30 the payoff matrix is:

P / A	P 0	P 1	P 2	P 3	P 4	P 5
A 0	30	25	20	17	14	10
A 1	28	28	18	15	12	8
A 2	26	26	26	13	10	6
A 3	24	24	24	24	8	4
A 4	22	22	22	22	22	2
A 5	20	20	20	20	20	20

Games with nature

Term "nature" in game theory is understood in a broad sense. These can be actual natural physical (climatic), biological, chemical, social, etc. processes that accompany economic activity. “Nature” can also mean a market opposing the entrepreneur, a competitive environment, a monopoly, etc. “Nature” can act as an antagonistic side, or maybe as a cooperative environment. “Nature” in the form of natural processes, as part of the economy, does not seek to “specifically” harm the entrepreneur, but it incurs certain damage from his economic activity and this the “loss” for her should be minimal, if, in general, it is impossible to do without it for the environment. Player A in such games are economic entities, and player B is “nature”. Where does physical “nature” come from? The loss of player B, physical “nature,” must be compensated from outside, for example, by government subsidies or funds included in investment projects for the renewal of natural resources. Knowledge of the optimal strategies of “nature” allows us to determine the most unfavorable conditions for player A (entrepreneur) that await him (“hope for the best, but prepare for the worst”), and estimate the necessary resources for the restoration of natural resources, giving him the opportunity to receive a guaranteed income.
If “nature” implies a competitive environment, then the loss of the second player is the price of fighting competitors in the market.
Let's move on to examples of meaningful formulations of problems for playing with “nature”.
1. Antagonistic games
Example 1. (Crop planning). A farmer who has a limited plot of land can plant it with three different crops A 1, A 2, A 3. The harvest of these crops depends mainly on the weather (“nature”), which can be in three different states: B 1, B 2, B 3. The farmer has information (statistical data) about the average yield of these crops (the number of centners of crop obtained per hectare of land) under three different weather conditions, which is reflected in the table: Then the income matrix (payment matrix) of farmer A has the form:

Matrix element A - ( a ij) shows how much income a farmer can receive from one hectare of land if he sows a crop i ( i =1, 2, 3), and the weather will be in the state j (j = 1, 2, 3).
It is necessary to determine the proportions in which the farmer must sow the available plot of land in order to obtain the maximum guaranteed income, regardless of what weather conditions occur.
This problem can be reduced to an antagonistic game. In this case, the farmer is the first player, and nature is the second player. We will assume that nature, as a player, can behave in such a way as to cause maximum harm to the farmer, thereby pursuing opposing interests (these assumptions allow us to estimate the income that he can receive if the weather conditions are as unfavorable as possible for him) . In this case, the farmer has three pure strategies at his disposal:

• the first pure strategy assumes that the entire plot of land will be sown with crop A 1;
• the second pure strategy assumes that the entire plot of land will be sown with crop A 2 ;
• the third pure strategy assumes that the entire plot will be sown with crop A 3 .

As a player, nature can also use three possible strategies:

• dry weather, which corresponds to the first pure strategy B 1;
• normal weather, which corresponds to the second pure strategy B 2;
• rainy weather, which corresponds to the third pure strategy B 3.

Solution



2. Let's check whether this game has a saddle point.

V * =max i min j a ij = 50.
V * =min j max i a ij = 100.

3. The solution to the game should be sought in mixed strategies. Let's reduce the game problem to a linear programming problem. If first player - farmer- applies its optimal mixed strategy P *, and second player - nature- consistently applies his pure strategies, then the mathematical expectation of the income that a farmer can receive from his plot will not be less than the game price V.


.


Let's divide the equality:
p* 1 + p* 2 + p* 3 = 1
on V, we find that the new variables y 1, y 2, y 3 satisfy the condition:
y 1 + y 2 + y 3 = 1/V
Because the The first player's goal is to maximize his winnings, A the mathematical expectation of his winnings is not less than the price of the game, then the first player will strive to maximize the cost of the game, which is equivalent to minimizing the value of 1/V.
So, for the first player (farmer), the problem of determining the optimal behavior strategy has been reduced to a linear programming problem:
find the minimum of the function F = y 1 + y 2 + y 3


and direct restrictions:
y 1 ≥ 0, y 2 ≥ 0, y 3 ≥ 0
Let's move on to the second player, nature. If second player - nature - will apply its optimal mixed strategy Q * , and the first player - the farmer - will consistently apply his pure strategies, then the mathematical expectation of the second player's loss will not be greater than the cost of the game. Therefore, the following system of inequalities must be satisfied:

Let us divide each of the inequalities included in the system by V and introduce new variables:
.
As a result, we obtain a new system of inequalities:

Let's divide the equality:
q* 1 + q* 2 + q* 3 = 1
on V, we find that the new variables q 1, q 2, q 3 satisfy the condition:
q 1 + q 2 + q 3 = 1/V
Because the target second player - nature- minimizing his loss, A the mathematical expectation of his loss is no more than the price of the game, then the second player will strive to minimize the cost of the game, which is equivalent to maximizing the value 1/V.
So, for the second player (nature), the problem of determining the optimal behavior strategy has been reduced to a linear programming problem:
find the maximum of the function F / = x 1 + x 2 + x 3
with the following functional limitations:

and direct restrictions:
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0
Thus, in order to find the optimal mixed strategy of the second player, it is also necessary to solve the linear programming problem.
The problems of both players were reduced to a pair of dual linear programming problems:

Second Player's Problem minimizing loss V	First Player's Problem maximizing payoff V
Objective function
F / = x 1 +x 2 +x 3 = → max	F = y 1 +y 2 +y 3 = → min
Functional limitations
Direct restrictions
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0	y 1 ≥ 0, y 2 ≥ 0, y 3 ≥ 0

The first player's problem is solved using the simplex method. Score results:
conclusions. According to the results obtained the farmer is guaranteed an average income of 66.67 units from every hectare of land used for crops under the most unfavorable conditions. Optimal strategy for him - growing two crops, A 1 and A 3, and, under first culture he should be given 0,67 part of the whole land, and under third crop 0.33 part of the total land.
Nature threatens the farmer with heat for 0.33 of the growing season and rain for 0.67 of the season.

Example. Planning of production under different states of nature - demand market.
An enterprise can produce 4 types of products: A 1, A 2, A 3, A 4, while making a profit. Its value is determined by the state of demand (the nature of the market), which can be in one of four possible states: B 1, B 2, B 3, B 4. The dependence of the amount of profit on the type of product and market conditions is presented in the table:

Types of products	Possible states of the demand market
	B 1	B 2	B 3	B 4
A 1	4	3	5	6
A 2	2	6	1	5
A 3	3	0	7	2
A 4	3	5	1	3

The payment matrix looks like:

Matrix element A - ( a ij) characterizes how much profit an enterprise can receive if it produces i-th type of product( i=1, 2, 3, 4) at jth demand( j = 1, 2, 3, 4).
It is necessary to determine the optimal proportions of the types of products produced by the enterprise, the sale of which would provide it with the maximum possible revenue, regardless of what state of demand will be realized
This task can be reduced to an antagonistic game.
In this case, as first player stands company, and as second player - nature, which affects the state of demand and can make it as unfavorable as possible for the enterprise. We will assume that nature, as a player, will behave in such a way as to cause maximum harm to the enterprise, thereby pursuing opposing interests.
In this case, the conflict between the two parties can be characterized as antagonistic, and the use of a model of this conflict allows the enterprise. estimate the revenue that it can receive regardless of what state of demand is realized.
Acting as first player, company can use four strategies:
· the first pure strategy corresponding to the production of only products A 1 by the enterprise
· the second pure strategy, corresponding to the production of only products A 2 by the enterprise
· third pure strategy, corresponding to the production of only products A 3 by the enterprise
· the fourth pure strategy, corresponding to the production of only products A 4 by the enterprise
Acting as second player, nature can also use four strategies:
· the first pure strategy, in which the state of demand B 1 is realized;
· the second pure strategy, in which the state of demand B 2 is realized;
· the third pure strategy, in which the state of demand B 3 is realized;
· the fourth pure strategy, in which the state of demand B 4 is realized.
Solution
1. Let's analyze the payment matrix A.

Matrix A has no dominated strategies and cannot be simplified.
2. Let's check whether this game has a saddle point.
Let's find the lower and upper price of the game:
V * =max i min j a ij = 3.
V * =min j max i a ij = 4.
Since V * ≠V * , then this antagonistic game does not have a saddle point and a solution in pure strategies.
The solution to the game should be sought in mixed strategies. Let us reduce the antagonistic conflict under consideration to a direct and dual linear programming problem.
If first player - company - applies my optimal mixed strategy P*, a second player - nature - applies consistently their pure strategies, That mathematical expectation of income, which the enterprise can receive will be no less than the price of the gameV.
And vice versa, if second player - nature - will apply your optimal mixed strategyQ*, A first player - enterprise will be consistentapply your pure strategies, That mathematical expectation of loss the second player will be no more than the price of the game. Therefore, the following system of inequalities must be satisfied:

Second Player's Problem minimizing lossesV	First Player's Problem maximizing winningsV
Objective function
F / = x 1 +x 2 +x 3 +x 4 =→ max	F = y 1 +y 2 +y 3 +y 4 =→ min
Functional limitations
Direct restrictions
x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, x 4 ≥ 0	y 1 ≥ 0, y 2 ≥ 0, y 3 ≥ 0, y 4 ≥ 0

Using the simplex method for solving the first player problem, we get:
Y * = (y 1 * = 0.182; y 2 ​​* = 0; y 3 * = 0; y 4 * = 0.091)
F= y 1 * + y 2 * + y 3 * +y 4 * = 0.273
From the relation y 1 * + y 2 * + y 3 * +y 4 * =1/V we find V:

From the relations:

Let's find:
p* 1 = y* 1 V = 0.67, p* 2 = y* 2 V = 0, p* 3 = y* 3 V = 0, p* 4 = y* 4 V =0.33

Finally we have:
P * = (p * 1 =0.67; p * 2 = 0; p * 3 =0; p * 4 = 0.33), V = 3.67
Based on the solution found for the dual linear programming problem, we find solution the original task - Second player's tasks:
X * = (x 1 * = 0.121; x 2 * = 0.121; x 3 * = 0.03; x 4 * = 0)
F / = x 1 * + x 2 * + x 3 * +x 4 * = 0.273
From the relation x 1 * + x 2 * + x 3 * +x 4 * = 1/V we find V:

From the relations:

Let's find:
q* 1 = x* 1 V = 0.445, q* 2 = x* 2 V = 0.444, q* 3 = x* 3 V = 0.111, q* 4 = x* 4 V = 0.
Finally we have:
Q * = (q * 1 = 0.445; q * 2 =0.444; q * 3 = 0.111; q * 4 = 0), V = 3.67

Example. The company plans to sell its products in the markets, taking into account possible options for consumer demand P j , j = 1.4 (low, medium, high, very high). The company has developed three sales strategies for goods A 1, A 2, A 3. The volume of turnover (money units), depending on the strategy and consumer demand, is presented in the table.

A j	P j
	P 1	P 2	P 3	P 4
A 1	30+N	10	20	25 + N/2
A 2	50	70 - N	10 + N/2	25
A 3	25 – N/2	35	40	60 - N/2

where N=3

Solution find using a calculator.
Bayes criterion.
According to the Bayes criterion, the strategy (pure) A i that maximizes the average gain a or minimizes the average risk r is accepted as optimal.
We count the values ​​of ∑(a ij p j)
∑(a 1,j p j) = 33 0.3 + 10 0.2 + 20 0.4 + 26.5 0.1 = 22.55
∑(a 2,j p j) = 50 0.3 + 67 0.2 + 11.5 0.4 + 25 0.1 = 35.5
∑(a 3,j p j) = 23.5 0.3 + 35 0.2 + 40 0.4 + 58.5 0.1 = 35.9

A i	P 1	P 2	P 3	P 4	∑(a ij p j)
A 1	9.9	2	8	2.65	22.55
A 2	15	13.4	4.6	2.5	35.5
A 3	7.05	7	16	5.85	35.9
p j	0.3	0.2	0.4	0.1

Laplace criterion.
If the probabilities of states of nature are plausible, Laplace’s principle of insufficient reason is used to assess them, according to which all states of nature are assumed to be equally probable, i.e.:
q 1 = q 2 = ... = q n = 1/n.
q i = 1/4

A i	P 1	P 2	P 3	P 4	∑(a ij)
A 1	8.25	2.5	5	6.63	22.38
A 2	12.5	16.75	2.88	6.25	38.38
A 3	5.88	8.75	10	14.63	39.25
p j	0.25	0.25	0.25	0.25

Conclusion: choose strategy N=3.
Wald criterion.
According to the Wald criterion, a pure strategy is taken as optimal, which under the worst conditions guarantees the maximum gain, i.e.
a = max(min a ij)
The Wald criterion focuses statistics on the most unfavorable states of nature, i.e. this criterion expresses a pessimistic assessment of the situation.

A i	P 1	P 2	P 3	P 4	min(a ij)
A 1	33	10	20	26.5	10
A 2	50	67	11.5	25	11.5
A 3	23.5	35	40	58.5	23.5

Conclusion: choose strategy N=3.
Savage criterion.
Savage's minimum risk criterion recommends choosing as the optimal strategy the one in which the magnitude of the maximum risk is minimized under the worst conditions, i.e. provided:
a = min(max r ij)
Savage's criterion focuses statistics on the most unfavorable states of nature, i.e. this criterion expresses a pessimistic assessment of the situation.
We find the risk matrix.
Risk– a measure of the discrepancy between different possible outcomes of adopting certain strategies. The maximum gain in the jth column b j = max(a ij) characterizes the favorable state of nature.
1. Calculate the 1st column of the risk matrix.
r 11 = 50 - 33 = 17; r 21 = 50 - 50 = 0; r 31 = 50 - 23.5 = 26.5;
2. Calculate the 2nd column of the risk matrix.
r 12 = 67 - 10 = 57; r 22 = 67 - 67 = 0; r 32 = 67 - 35 = 32;
3. Calculate the 3rd column of the risk matrix.
r 13 = 40 - 20 = 20; r 23 = 40 - 11.5 = 28.5; r 33 = 40 - 40 = 0;
4. Calculate the 4th column of the risk matrix.
r 14 = 58.5 - 26.5 = 32; r 24 = 58.5 - 25 = 33.5; r 34 = 58.5 - 58.5 = 0;

A i	P 1	P 2	P 3	P 4
A 1	17	57	20	32
A 2	0	0	28.5	33.5
A 3	26.5	32	0	0

A i	P 1	P 2	P 3	P 4	max(a ij)
A 1	17	57	20	32	57
A 2	0	0	28.5	33.5	33.5
A 3	26.5	32	0	0	32

Conclusion: choose strategy N=3.
Hurwitz criterion.
The Hurwitz criterion is a criterion of pessimism - optimism. The optimal strategy is taken to be one for which the following relation holds:
max(s i)
where s i = y min(a ij) + (1-y)max(a ij)
For y = 1 we obtain the Walde criterion, for y = 0 we obtain the optimistic criterion (maximax).
The Hurwitz criterion takes into account the possibility of both the worst and the best behavior of nature for humans. How is y chosen? The worse the consequences of erroneous decisions, the greater the desire to insure against errors, the closer y is to 1.
We calculate s i.
s 1 = 0.5 10+(1-0.5) 33 = 21.5
s 2 = 0.5 11.5+(1-0.5) 67 = 39.25
s 3 = 0.5 23.5+(1-0.5) 58.5 = 41

A i	P 1	P 2	P 3	P 4	min(a ij)	max(a ij)	y min(a ij) + (1-y)max(a ij)
A 1	33	10	20	26.5	10	33	21.5
A 2	50	67	11.5	25	11.5	67	39.25
A 3	23.5	35	40	58.5	23.5	58.5	41

Conclusion: choose strategy N=3.
Thus, as a result of solving the statistical game according to various criteria, strategy A 3 was recommended more often than others.

The company's management decides to locate the production of a new product in a certain location. To form an idea of ​​the situation on the market of a new product at the time of mastering production, it is necessary to take into account the costs of delivering finished products to the consumer, the development of the transport and social infrastructure of the region, competition in the market, the relationship between supply and demand, exchange rates and much more. Possible solutions, the investment attractiveness of which is defined as the percentage of income growth in relation to the amount of capital investment, are presented in the table.
Choose:
1) a place to locate production, if the head of the enterprise is confident that situation 4 will develop on the market;
2) a place to locate production if management estimates the probability of situation 1 to be 0.2; situations 2 in 0.1; situation 3 at 0.25;
3) select an option under conditions of uncertainty according to the criterion: maximax, maximin, Laplace criterion, Savage criterion, Hurwitz criterion (y = 0.3);
4) will the best solution according to the Hurwitz criterion change if the value of a is increased to 0.5?
5) assuming that the table data represents the costs of the enterprise, determine the choice that the enterprise will make when using each of the following criteria: maximin; maximax; Hurwitz criterion(? = 0.3); Savage criterion; Laplace criterion

Typical tasks

1. Select the optimal project for construction using the Laplace, Wald, maximum optimism, Savage and Hurwitz criteria with a=0.58. The cost matrix looks like:

   0.07	0.26	0.11	0.25	0.1	0.21
   68	45	54	79	47	99
   56	89	42	56	74	81
   72	87	56	40	62	42
   65	48	75	89	52	80
   69	93	93	56	45	43
   73	94	79	68	67	46
   66	100	64	89	94	49
   70	42	97	42	42	50

2. A retail trade enterprise has developed several options for a plan for selling goods at the upcoming fair, taking into account changing market conditions and customer demand, the resulting profit amounts from their possible combinations are presented in the form of a winning matrix. Determine the optimal plan for selling goods.
   x=0.7
3. The company plans to sell its products in the markets, taking into account possible options for consumer demand Pj, j=1͞,4͞ (low, medium, high, very high). The company has developed three sales strategies for goods A 1, A 2, A 3. The volume of turnover (money units), depending on the strategy and consumer demand, is presented in the table.

   A j	P j
   	P 1	P 2	P 3	P 4
   A 1	30+N	10	20	25 + N/2
   A 2	50	70 - N	10 + N/2	25
   A 3	25 – N/2	35	40	60 - N

   
   Where N=3
   The possible states of consumer demand are known, which are, respectively, q 1 =0.3, q 2 =0.2, q 3 =0.4, q 4 =0.1. It is necessary to find a sales strategy that maximizes the average turnover of the company. In this case, use the criteria of Wald, Hurwitz, Savage, and Bayes.
   Solution
4. The factory's costs per unit of production during April - May were: dresses - 8 monetary units, suits - 27, and the selling price is 16 and 48, respectively. According to past observations, the factory can sell during these months in warm weather conditions 600 suits and 1975 dresses, and in cool weather - 625 dresses and 1000 suits.

The minimum risk method is used to determine the boundary value of the determining parameter for making a decision on the condition of an object, based on the condition of minimum average costs.

Let the state of some object be determined by the value of some parameter X. you must select this value for this parameter X 0 , to:

The serviceable state is characterized by the distribution density of the parameter X,f(x/ D1) and the faulty one is f(x/ D2) (Figure 2.8). Curves f(x/ D1) And f(x/ D2) intersect and therefore impossible to choose X 0 so that rule (2.16) would not give erroneous solutions.

Errors that arise when making decisions are divided into errors of the first and second kind.

Error of the first kind– making a decision about the malfunction (presence of a defect) of an object, when in fact the object is in good condition.

Error of the second type– making a decision about the good condition of an object, when in reality the object is in a faulty state (the object contains a defect).

The probability of a type I error is equal to the product of the probability of two events:

the likelihood that the object is in good condition;

the probability that the value of the defining parameter x will exceed the boundary value X 0 .

The expression for determining the probability of a type I error has the form:

Where p(D 1 ) – a priori probability of the object being in good condition (considered known based on preliminary statistical data).

The probability of a type II error is determined similarly:

Rice. 2.8. Probability densities of states of the diagnostic object

Elements of information collection systems: unifying measuring transducers.

To coordinate the primary transducer with the devices of the information acquisition system, its output signal must be unified, i.e. meet certain requirements for level, power, type of storage medium, etc., which are determined by the relevant GOSTs.

To convert the output signals of primary converters into unified ones, a number of normalizing converters are used. Natural signals from primary converters of various physical quantities can be supplied to the input of normalizing converters, and corresponding unified signals are generated at the output.

The group of means that ensure unification of the signal between its source or the output of the primary transducer and the input of the secondary device belongs to the class of unifying measuring transducers (UMT).

The following types of UIP are distinguished:

individual;

group;

multichannel.

Individual UIP(Fig. 3.36a)) serve one PP and are connected between the PP and the switch or subsequent measuring transducer. Individual UIPs are placed together with the PP directly at the research site.

They are used to unify signals with a relatively small number of measured parameters and with limited measurement time, which does not allow the use of group UPS.

Individual UIPs allow you to produce:

converting one unified signal into another;

galvanic isolation of input circuits;

multiplication of the input signal over several outputs.

However, the use of its own UIP in each IMS measuring complex complicates the system and reduces its reliability and economic efficiency.

Group UIP(Fig. 3.36b)) are more efficient from this point of view; they serve a certain group of primary converters, the output signals of which are homogeneous physical quantities. They are located in Iis after the switch and are controlled together with the last control unit.

When constructing multi-channel IMS of heterogeneous physical quantities, the latter are grouped according to the type of physical quantity, and each group is connected to the corresponding group UIP.

Multichannel UIP.(Fig. 3.36c)) If the measured physical quantities are mostly heterogeneous, then the IIS can use multi-channel UIPs, which are several individual UIPs combined in one case or one board. Information conversion is carried out according to n entrances and n exits. The main design feature of a multi-channel UPS is the use of a common power source and control system for all individual UPSs.

Rice. 3.36. main types of unifying

measuring transducers

The main functions performed by the UIP:

linear (scaling, zeroing, temperature compensation);

nonlinear (linearization) signal transformations.

With a linear characteristic of the primary converter, the UIP performs linear operations, which are called scaling. The essence of scaling is as follows. Let the input signal vary from y 1 before y 2 , and the dynamic range of the output signal of the UIP should be in the range from 0 before z. Then, to match the beginning of the dynamic ranges of the UIP and the primary converter, a signal must be added to the PP signal, and then the total signal must be amplified at the same time.

It is also possible that the output signal of the PP is first amplified, and then the beginnings of the dynamic ranges are combined.

The first option for bringing the output signal to a unified form is usually used in individual UIPs, and the second in group ones.

Because The relationship between the output signal yPP and the measured parameter is most often nonlinear (for example, with thermocouples, platinum resistance thermal converters, etc.) The UIP must perform the operation linearization. Linearization consists of straightening the PP transformation function. In this case, the linearizing function should have the form of an inverse PP transformation function.

To linearize the transformation function in the UIP, special nonlinear links are used. They can be turned on up to linear

a unifying converter, after it or into the feedback circuit of an amplifier used to change the scale of the measured value.

U input

U OS

U out

R 1

R 2

R 3

R 4

R 5

D 1

D 2

D 3

Most often, linearization is achieved by piecewise linear approximation and is performed using a chain of series-connected resistors shunted by zener diodes or diodes D 1  D 3

Rice. 3.37.block diagram of UIP

As the voltage at the amplifier output increases, the divider current and the voltage drop across each resistor increase. R 1  R 5 .as soon as the voltage drop across any of the resistors reaches the breakdown voltage of the corresponding zener diode, the zener diode begins to bypass this resistor. The resistor resistances are selected in such a way as to obtain the required feedback voltage dependence U OS inverting amplifier U, removed from the resistor R 5 , from the output voltage of the amplifier.

A typical analog UIP contains:

output amplifier;

galvanic isolation device;

functional converter that linearizes the PP signal;

output amplifier;

stabilized power supply.

Some primary converters have an alternating current signal as an output signal; this signal is modulated either in amplitude (for example, differential transformer converters) or in frequency (for example, piezoresonators).

As an example, consider the block diagram of a UIS designed to convert alternating voltage from pressure, differential pressure, flow, level, and steam content sensors into a unified direct current signal 0...5 mA (Fig. 3.38.).

Rice. 3.38. Block diagram of UIP

The alternating voltage from the differential transformer primary converter is converted by the demodulator into a proportional direct current voltage, which is amplified by a magnetic MU and electronic U DC amplifiers covered by deep negative feedback through a feedback device OS, which allows, if necessary, to linearize the characteristic of the primary converter.

Unifying measuring converters working with frequency PPs must perform the same functions as amplitude PPs.

Example 2.5. For the matrix of consequences given in example 2.1, select the best solution based on the Hurwitz criterion with λ =1/2.

Solution. Considering the matrix of consequences Q row by row, for each i we calculate the values ​​ci= 1/2minqij + 1/2maxqij. For example, c1=1/2*2+1/2*8=5; similarly found c2=7; c3=6.5; c4= 4.5. The largest is c2=7. Consequently, the Hurwitz criterion for a given λ =1/2 recommends choosing the second option ( i=2).

2.3. Analysis of a related group of solutions under conditions of partial

uncertainty

If, when making a decision, the decision maker knows the probabilities pj If the real situation can develop according to option j, then they say that the decision maker is in conditions of partial uncertainty. In this case, you can be guided by one of the following criteria (rules).

Criterion (rule) for maximizing average expected income. This criterion is also called criterion for maximum average winnings. If the probabilities are known pj options for the development of the real situation, then the income received from the i-th solution is a random variable Qi with a distribution series

Expected value M[Qi] of the random variable Qi is the average expected income, also denoted by:

= M[Qi ] = .

For each i-th solution option, the values ​​are calculated, and in accordance with the criterion under consideration, an option is selected for which

Example 2.6. For the initial data of Example 2.1, let the probabilities of the development of a real situation be known for each of the four options that form a complete group of events:

p1 =1/2, p2=1/6, p3=1/6, p4=1/6. Find out which solution option achieves the highest average income and what is the amount of this income.

Solution. Let us find for each i-th solution option the average expected income: =1/2*5+1/6*2+1/6*8+1/6*4= 29/6, = 25/6, = 7, = 17/6. The maximum average expected return is 7 and corresponds to the third solution.

Rule for minimizing average expected risk (other name - minimum average loss criterion).

Under the same conditions as in the previous case, the decision maker’s risk when choosing the i-th solution is a random variable Ri with a distribution series

Expected value M and is the average expected risk, also denoted by: = M = . . The rule recommends making a decision that entails the minimum average expected risk: .

Example 2.7 . The initial data are the same as in example 2.6. Determine which solution option achieves the lowest average expected risk and find the value of the minimum average expected risk (loss).

Solution. For each i-th solution option, we find the value of the average expected risk. Based on the given risk matrix R, we find: = 1/2*3+1/6*3+1/6*0+1/6*8=20/6, = 4, = 7/6, = 32/6.

Therefore, the minimum average expected risk is 7/6 and corresponds to the third solution: = 7/6.

Comment. When they talk about the average expected income (gain) or the average expected risk (loss), they mean the possibility of repeated repetition of the decision-making process according to the described scheme or the actual repeated repetition of such a process in the past. The conditionality of this assumption is that the actually required number of such repetitions may not exist.

Laplpas criterion (rule) of equal opportunity (indifference). This criterion does not directly relate to the case of partial uncertainty, and it is applied under conditions of complete uncertainty. However, here it is assumed that all states of the environment (all variants of the real situation) are equally probable - hence the name of the criterion. Then the calculation schemes described above can be applied, considering the probabilities pj identical for all variants of the real situation and equal to 1/n. Thus, when using the criterion of maximizing the average expected income, a solution is selected that achieves . And in accordance with the criterion of minimizing the average expected risk, a solution option is selected for which .

Example 2.8. Using the Laplace criterion of equal opportunity for the initial data of Example 2.1, select the best solution based on: a) the rule for maximizing the average expected income; b) rules for minimizing the average expected risk.

Solution. a) Taking into account the equiprobability of the options in the real situation, the average expected income for each of the solution options is = (5+2+8+4)/4=19/4, = 21/4, = 26/4, = 15/4. Therefore, the best solution would be the third one, and the maximum average expected return would be 26/4.

b) For each solution option, we calculate the average expected risk based on the risk matrix, taking into account the equiprobability of the situation options: = (3+3+0+8)/4 = 14/4, = 3, = 7/4, = 18/4 . It follows that the third option will be the best, and the minimum average expected risk will be 7/4.

2.4. Pareto optimality of two-criteria financial

operations under conditions of uncertainty

From what was discussed above, it follows that each decision (financial transaction) has two characteristics that need to be optimized: average expected income and average expected risk. Thus, choosing the best solution is a two-criteria optimization problem. In multicriteria optimization problems, the main concept is the concept Pareto optimality. Let's consider this concept for financial transactions with the two indicated characteristics.

Let each operation A has two numerical characteristics E(a),r(A)(e.g. effectiveness and risk); during optimization E strive to increase and r decrease.

There are several ways to formulate such optimization problems. Let us consider this problem in general form. Let A - a certain set of operations, and different operations necessarily differ in at least one characteristic. When choosing the best operation, it is advisable that E was more and r was less.

We will say that the operation A dominates surgery b, and designate a > b, If E(a) ≥ E(b) And r(a) ≤ r(b) and at least one of these inequalities is strict. In this case, the operation A called dominant, and the operation b –dominated. It is obvious that no dominated operation can be recognized the best. Consequently, the best operation must be sought among non-dominated operations. The set of non-dominated operations is called Pareto set (region) or Pareto optimality set.

For the Pareto set, the following statement is true: each of the characteristics E,r is an unambiguous function of another, i.e., on the Pareto set, one characteristic of an operation can be used to unambiguously determine another.

Let's return to the analysis of financial decisions under conditions of partial uncertainty. As shown in Section 2.3, each operation has an average expected risk and average expected income. If you introduce a rectangular coordinate system, on the abscissa axis of which you plot the values , and on the ordinate axis there are values ​​, then each operation will correspond to a point ( , ) on the coordinate plane. The higher this point on the plane, the more profitable the operation; the further to the right the dot, the more risky the operation. Therefore, when searching for non-dominated operations (Pareto sets), you need to choose points above and to the left. Thus, the Pareto set for the initial data of examples 2.6 and 2.7 consists of only one third operation.

To determine the best operation in some cases, you can use some weighing formula in which the characteristics and enter with certain weights, and which gives one number specifying the best operation. Let, for example, for the operation i with characteristics ( , ) the weighing formula has the form f(i) = 3 - 2, and the best operation is selected based on the maximum value f(i). This weighting formula means that the decision maker agrees to increase the risk by three units if the income of the operation increases by at least two units. Thus, the weighting formula expresses the relationship of the decision maker to the indicators of income and risk.

Example 2.9. Let the initial data be the same as in examples 2.6 and 2.7, i.e. for the consequences and risk matrices of example 2.1, the probabilities of options for the development of the real situation are known: p1 = 1/2, p2 = 1/6, p3 = 1/6, p4=1/6. Under these conditions, the decision maker agrees to increase the risk by two units if the income of the operation increases by at least one unit. Determine the best operation for this case.

Solution. The weighing formula has the form f(i) = 2 - . Using the calculation results in examples 2.6 and 2.7, we find:

f(1) = 2*29/6 – 20/6 = 6,33; f(2) = 2*25/6 – 4 = 4,33;

f(3) = 2*7 – 7/6 = 12,83; f(4) = 2*17/6 – 32/6 = 0,33

Therefore, the third operation is the best, and the fourth is the worst.

Topic 3. Measurements and indicators of financial risks

Quantitative risk assessment. Risk of a separate operation. General risk measures.

This topic discusses criteria and methods for decision-making in cases where it is assumed that the probability distributions of possible outcomes are either known or they can be found, and in the latter case it is not always necessary to explicitly specify the distribution density.

3.1. General methodological approaches to quantitative risk assessment

Risk is a probabilistic category, therefore methods for its quantitative assessment are based on a number of the most important concepts of probability theory and mathematical statistics. Thus, the main tools of the statistical method of risk calculation are:

1) expected value m, for example, such a random variable as the result of a financial transaction k: m = E{k};

2) dispersion as a characteristic of the degree of variation of the values ​​of a random variable k around the grouping center m(recall that variance is the mathematical expectation of the squared deviation of a random variable from its mathematical expectation );

3) standard deviation ;

4) the coefficient of variation , which has the meaning of risk per unit of average income.

Comment. For a small set n values ​​– small sample! – discrete random variable Strictly speaking, we are talking only about estimates listed risk measures .

So, average (expected) sample value, or selective analogue of mathematical expectation , is the quantity where Ri – probability of realizing the value of a random variable k. If all values ​​are equally likely, then the expected value of a random sample is calculated using the formula.

Likewise, sample variance (sample variance ) is defined as the standard deviation in the sample: or

. In the latter case, the sample variance is biased estimate of theoretical variance . Therefore, it is preferable to use an unbiased estimate of the variance, which is given by the formula .

Obviously, the assessment can be calculated as follows or .

It is clear that the assessment coefficient of variation now takes the form .

In economic systems under risk conditions, decision making is most often based on one of the following criteria.

1. Expected value (profitability, profit or expenses).

2. Sample variance or standard (mean square) deviation .

3. Expected value combinations And variances or sample standard deviation .

Comment . Under the random variable k in each specific situation, the indicator corresponding to this situation is understood, which is usually written in the accepted notation: mp – portfolio return securities, IRR – (Internal Rate of Return) internal (rate) of return etc.

Let's look at the idea presented using specific examples.

3.2. Probability distributions and expected returns

As has been said more than once, risk is associated with the likelihood that the actual return will be lower than its expected value. Therefore, probability distributions are the basis for measuring the risk of an operation. However, we must remember that the estimates obtained are probabilistic in nature.

Example 1. Let's say, for example, that you intend to invest $100,000. for a period of one year. Alternative investment options are given in table. 3.1.

Firstly, these are GKO-OFZ with a maturity of one year and an income rate of 8%, which can be purchased at a discount, i.e. at a price below par, and at the time of redemption their par value will be paid.

Table 3.1

Profitability assessment for four investment alternatives

State economy	Probability Ri		Return on investment in a given state of the economy, %
corporate securities
Deep recession
Slight decline
Stagnation
Slight rise
Strong rise
Expected return

Note. Profitability corresponding to different states of the economy should be considered as an interval of values, and its individual values ​​as points within this interval. For example, a 10% yield on a corporate bond with a slight decline represents most likely return value for a given state of the economy, and the point value is used for convenience of calculations.

Secondly, corporate securities (blue chips), which are sold at par with a coupon rate of 9% (i.e., for $100,000 of invested capital you can receive $9,000 per annum) and a maturity of 10 years. However, you intend to sell these securities at the end of the first year. Consequently, the actual yield will depend on the level of interest rates at the end of the year. This level in turn depends on the state of the economy at the end of the year: rapid economic development is likely to cause interest rates to rise, which will reduce the market value of blue chips; In the event of an economic downturn, the opposite situation is possible.

Third, capital investment project 1, whose net cost is $100,000. Cash flow during the year is zero, all payments are made at the end of the year. The amount of these payments depends on the state of the economy.

And finally, alternative investment project 2, identical in all respects to project 1 and differing from it only probability distribution of payments expected at the end of the year .

Under probability distribution , we will understand the set of probabilities of possible outcomes (in the case of a continuous random variable, this would be the probability distribution density). It is in this sense that the data presented in Table 1 should be interpreted. 3.1 four probability distributions corresponding to four alternative investment options. The yield on GKO-OFZ is precisely known. It is 8% and does not depend on the state of the economy.

Question 1 . Can the risk on GKO-OFZ be unconditionally considered equal to zero?

Answer: a) yes; b) I think that not everything is so simple, but I find it difficult to give a more complete answer; c) no.

The correct answer is c).

For any answer, see reference 1.

Help 1 . Investments in GKO-OFZ are risk-free only in the sense that they nominal profitability does not change during a given period of time. At the same time they real the yield contains a certain amount of risk, since it depends on the actual growth rate of inflation during the period of holding this security. Moreover, GKOs can pose a problem for an investor who holds a portfolio of securities with the goal of generating continuous income: when a GKO-OFZ payment matures, the funds must be reinvested, and if interest rates decline, the portfolio's income will also decrease. This type of risk, which is called reinvestment rate risk , is not taken into account in our example, since the period during which the investor owns GKO-OFZ corresponds to their maturity date. Finally, we note that relevant yield of any investment is the after-tax return, so the return values ​​used to make a decision must reflect the after-tax return.

For the other three investment options, real or actual returns will not be known until the end of the respective holding periods. Since return values ​​are not known with certainty, these three types of investments are risky .

There are probability distributions discrete or continuous . Discrete distribution has a finite number of outcomes; so, in table. Table 3.1 shows discrete probability distributions of returns for various investment options. The GKO-OFZ yield takes only one possible value, while each of the three remaining alternatives has five possible outcomes. Each outcome is associated with the probability of its occurrence. For example, the probability that GKO-OFZ will have a yield of 8% is 1.00, and the probability that the yield of corporate securities will be 9% is 0.50.

If we multiply each outcome by the probability of its occurrence, and then add the results, we get a weighted average of the outcomes. The weights are the corresponding probabilities, and the weighted average is expected value . Since the outcomes are internal rates of return (Internal Rate of Return, abbreviated as IRR), the expected value is expected rate of return (Expected Rate of Return, abbreviation ERR), which can be represented as follows:

ERR = IRRi, (3.1)

where IRRi , - i-th possible outcome; pi- probability of occurrence of the i-th outcome; P - number of possible outcomes.

TECHNICAL DIAGNOSTICS OF ELECTRONIC MEANS

UDC 678.029.983

Compiled by: V.A. Pikkiev.

Reviewer

Candidate of Technical Sciences, Associate Professor O.G. Cooper

Technical diagnostics of electronic equipment: methodological recommendations for conducting practical classes in the discipline “Technical diagnostics of electronic equipment” / South-West. state University; comp.: V.A. Pikkiev, Kursk, 2016. 8 p.: ill. 4, table 2, appendix 1. Bibliography: p. 9 .

Methodological instructions for conducting practical classes are intended for students of the training direction 11.03.03 “Design and technology of electronic means”.

Signed for printing. Format 60x84 1\16.

Conditional oven l. Academician-ed.l. Circulation 30 copies. Order. For free

Southwestern State University.

INTRODUCTION. PURPOSE AND OBJECTIVES OF STUDYING THE DISCIPLINE.
1. Practical lesson No. 1. Method of the minimum number of erroneous decisions
2. Practical lesson No. 2. Minimum risk method
3. Practical lesson No. 3. Bayes method
4. Practical lesson No. 4. Maximum likelihood method
5. Practical lesson No. 5. Minimax method
6. Practical lesson No. 6. Neyman-Pearson method
7. Practical lesson No. 7. Linear separating functions
8. Practical lesson No. 8. Generalized algorithm for finding the separating hyperplane

INTRODUCTION. PURPOSE AND OBJECTIVES OF STUDYING THE DISCIPLINE.

Technical diagnostics considers diagnostic tasks, principles of organizing test and functional diagnostic systems, methods and procedures of diagnostic algorithms for checking malfunctions, operability and correct functioning, as well as for troubleshooting various technical objects. The main attention is paid to the logical aspects of technical diagnostics with deterministic mathematical models of diagnosis.

The purpose of the discipline is to master the methods and algorithms of technical diagnostics.

The objective of the course is to train technical specialists who have mastered:

Modern methods and algorithms for technical diagnostics;

Models of diagnostic objects and faults;

Diagnostic algorithms and tests;

Object modeling;

Equipment for element-by-element diagnostic systems;

Signature analysis;

Automation systems for diagnosing REA and EVS;

Skills in developing and constructing element models.

The practical classes provided for in the curriculum allow students to develop professional competencies of analytical and creative thinking by acquiring practical skills in diagnosing electronic equipment.

Practical classes involve working with applied problems of developing algorithms for troubleshooting electronic devices and constructing control tests for the purpose of their further use in modeling the functioning of these devices.

PRACTICAL LESSON No. 1

METHOD OF MINIMUM NUMBER OF ERROR DECISIONS.

In reliability problems, the method under consideration often gives “careless decisions”, since the consequences of erroneous decisions differ significantly from each other. Typically, the cost of missing a defect is significantly higher than the cost of a false alarm. If the indicated costs are approximately the same (for defects with limited consequences, for some control tasks, etc.), then the use of the method is completely justified.

The probability of an erroneous decision is determined as follows

D 1 - diagnosis of good condition;

D 2 - diagnosis of a defective condition;

P 1 - probability of 1 diagnosis;

P 2 - probability of the 2nd diagnosis;

x 0 - limit value of the diagnostic parameter.

From the condition for the extremum of this probability we obtain

The minimum condition gives

For unimodal (i.e., contain no more than one maximum point) distributions, inequality (4) is satisfied, and the minimum probability of an erroneous decision is obtained from relation (2)

The condition for choosing the boundary value (5) is called the Siegert–Kotelnikov condition (ideal observer condition). The Bayesian method also leads to this condition.

The solution x ∈ D1 is taken when

which coincides with equality (6).

The dispersion of the parameter (the value of the standard deviation) is assumed to be the same.

In the case under consideration, the distribution densities will be equal to:

Thus, the resulting mathematical models (8-9) can be used to diagnose ES.

Example

Diagnosis of the performance of hard drives is carried out by the number of bad sectors (Reallocated sectors). When producing the “My Passport” HDD model, Western Digital uses the following tolerances: Disks with an average value of x 1 = 5 per unit volume and standard deviation σ 1 = 2. In the presence of a magnetic deposition defect (faulty state), these values ​​are equal to x 2 = 12, σ 2 = 3. The distributions are assumed to be normal.

It is necessary to determine the maximum number of bad sectors, above which the hard drive must be removed from service and disassembled (to avoid dangerous consequences). According to statistics, a faulty state of magnetic sputtering is observed in 10% of hard drives.

Distribution densities:

1. Distribution density for good condition:

2. Distribution density for the defective state:

3. Let us divide the densities of states and equate them to the probabilities of states:

4. Let’s take the logarithm of this equality and find the maximum number of faulty sectors:

This equation has a positive root x 0 =9.79

The critical number of bad sectors is 9 per unit volume.

Task options

No.	x 1	σ 1	x 2	σ 2

Conclusion: Using this method allows you to make a decision without assessing the consequences of errors, based on the conditions of the problem.

The downside is that the listed costs are approximately the same.

The use of this method is widespread in instrument making and mechanical engineering.

Practical lesson No. 2

MINIMUM RISK METHOD

Purpose of the work: to study the minimal risk method for diagnosing the technical condition of the electrical system.

Job Objectives:

Study the theoretical foundations of the minimum risk method;

Carry out practical calculations;

Draw conclusions on the use of the minimum risk ES method.

Theoretical explanations.

The probability of making an erroneous decision consists of the probabilities of a false alarm and missing a defect. If we assign “prices” to these errors, we obtain an expression for the average risk.

Where D1 is the diagnosis of good condition; D2- diagnosis of defective condition; P1-probability of 1 diagnosis; P2 - probability of 2nd diagnosis; x0 - limit value of the diagnostic parameter; C12 - cost of false alarm.

Of course, the cost of an error is relative, but it must take into account the expected consequences of a false alarm and missing a defect. In reliability problems, the cost of missing a defect is usually significantly greater than the cost of a false alarm (C12 >> C21). Sometimes the cost of correct decisions C11 and C22 is introduced, which is taken negative for comparison with the cost of losses (errors). In general, the average risk (expected loss) is expressed by the equality

Where C11, C22 are the price of correct decisions.

The value x presented for recognition is random and therefore equalities (1) and (2) represent the average value (mathematical expectation) of risk.

Let us find the boundary value x0 from the condition of minimum average risk. Differentiating (2) with respect to x0 and equating the derivative to zero, we first obtain the extremum condition

This condition often determines two values ​​of x0, one of which corresponds to the minimum, and the second to the maximum of risk (Fig. 1). Relation (4) is a necessary but not sufficient condition for a minimum. For a minimum of R to exist at the point x = x0, the second derivative must be positive (4.1.), which leads to the following condition

(4.1.)

with respect to derivative distribution densities:

If the distributions f (x, D1) and f(x, D2) are, as usual, unimodal (i.e., contain no more than one maximum point), then when

Condition (5) is satisfied. Indeed, on the right side of the equality there is a positive quantity, and for x>x1 the derivative f "(x/D1), while for x

In what follows, by x0 we will understand the boundary value of the diagnostic parameter, which, according to rule (5), provides a minimum of average risk. We will also consider the distributions f (x / D1) and f (x / D2) to be unimodal (“one-humped”).

From condition (4) it follows that the decision to assign object x to state D1 or D2 can be associated with the value of the likelihood ratio. Recall that the ratio of the probability densities of the distribution of x under two states is called the likelihood ratio.

Using the minimum risk method, the following decision is made about the state of an object having a given value of parameter x:

(8.1.)

These conditions follow from relations (5) and (4). Condition (7) corresponds to x< x0, условие (8) x >x0. Quantity (8.1.) represents the threshold value for the likelihood ratio. Let us recall that diagnosis D1 corresponds to a serviceable state, D2 – to a defective state of the object; C21 – cost of false alarm; C12 – cost of missing the goal (the first index is the accepted state, the second is the valid one); C11< 0, C22 – цены правильных решений (условные выигрыши). В большинстве практических задач условные выигрыши (поощрения) для правильных решений не вводятся и тогда

It is often convenient to consider not the likelihood ratio, but the logarithm of this ratio. This does not change the result, since the logarithmic function increases monotonically along with its argument. The calculation for normal and some other distributions when using the logarithm of the likelihood ratio turns out to be somewhat simpler. Let us consider the case when the parameter x has a normal distribution under good D1 and faulty D2 states. The dispersion of the parameter (the value of the standard deviation) is assumed to be the same. In the case under consideration, the distribution density

Introducing these relations into equality (4), we obtain after logarithm

Diagnostics of the health of flash drives is carried out by the number of bad sectors (Reallocated sectors). When producing the “UD-01G-T-03” model, Toshiba TransMemory uses the following tolerances: Drives with an average value of x1 = 5 per unit volume are considered serviceable. Let us take the standard deviation equal to ϭ1 = 2.

If there is a NAND memory defect, these values ​​are x2 = 12, ϭ2 = 3. The distributions are assumed to be normal. It is necessary to determine the maximum number of bad sectors above which the hard drive must be removed from service. According to statistics, a faulty condition is observed in 10% of flash drives.

Let us accept that the ratio of the costs of missing a target and a false alarm is , and refuse to “reward” correct decisions (C11=C22=0). From condition (4) we obtain

Task options:

Var.

X 1 mm.

X 2 mm.

b1

b2

Conclusion

The method allows you to estimate the probability of making an erroneous decision, defined as minimizing the extremum point of the average risk of erroneous decisions at maximum likelihood, i.e. The minimum risk of an event occurring is calculated if information about the most similar events is available.

PRACTICAL WORK No. 3

BAYES METHOD

Among technical diagnostic methods, the method based on the generalized Bayes formula occupies a special place due to its simplicity and efficiency. Of course, the Bayes method has disadvantages: a large amount of preliminary information, “suppression” of rare diagnoses, etc. However, in cases where the volume of statistical data allows the Bayes method to be used, it is advisable to use it as one of the most reliable and effective.

Let there be a diagnosis D i and a simple sign k j occurring with this diagnosis, then the probability of the joint occurrence of events (the presence of the state D i and the sign k j in the object)

From this equality follows Bayes' formula

It is very important to determine the exact meaning of all quantities included in this formula:

P(D i) – probability of diagnosis D i, determined from statistical data (a priori probability of diagnosis). So, if N objects were previously examined and N i objects had state D i , then

P(k j/D i) – probability of appearance of feature k j in objects with state D i . If among N i objects with a diagnosis D i , N ij exhibited sign k j , then

P(k j) – the probability of the appearance of feature k j in all objects, regardless of the state (diagnosis) of the object. Let out of the total number of N objects the feature k j was found in N j objects, then

To establish a diagnosis, a special calculation of P(k j) is not required. As will be clear from what follows, the values ​​of P(D i) and P(k j /D v), known for all possible states, determine the value of P(k j).

In equality (2) P(D i / k j) is the probability of diagnosis D i after it has become known that the object in question has attribute k j (posterior probability of diagnosis).

The generalized Bayes formula refers to the case when the survey is carried out using a set of characteristics K, including characteristics k 1, k 2, ..., k ν. Each of the features k j has m j digits (k j1, k j2, …, k js, …, k jm). As a result of the examination, the implementation of the characteristic becomes known

and the entire complex of characteristics K *. The index *, as before, means the specific value (implementation) of the attribute. The Bayes formula for a set of features has the form

where P(D i / K *) is the probability of diagnosis D i after the results of the examination for a set of signs K are known; P(D i) – preliminary probability of diagnosis D i (according to previous statistics).

Formula (7) applies to any of n possible states (diagnoses) of the system. It is assumed that the system is in only one of the indicated states and therefore

In practical problems, the possibility of the existence of several states A 1, ..., Ar is often allowed, and some of them can occur in combination with each other. Then, as different diagnoses D i, one should consider individual states D 1 = A 1, ..., D r = A r and their combinations D r+1 = A 1 /\ A 2.

Let's move on to the definition P (K * / D i) . If a complex of features consists of n features, then

Where k * j = k js– the category of a sign revealed as a result of the examination. For diagnostically independent signs;

In most practical problems, especially with a large number of features, it is possible to accept the condition of independence of features even in the presence of significant correlations between them.

Probability of appearance of a complex of traits K *

The generalized Bayes formula can be written

where P(K * / D i) is determined by equality (9) or (10). From relation (12) it follows

which, of course, should be the case, since one of the diagnoses is necessarily realized, and the realization of two diagnoses at the same time is impossible.

It should be noted that the denominator of the Bayes formula is the same for all diagnoses. This allows us to first determine the probabilities of the joint occurrence of the i-th diagnosis and a given implementation of a set of characteristics

and then the posterior probability of diagnosis

To determine the probability of diagnoses using the Bayes method, it is necessary to create a diagnostic matrix (Table 1), which is formed on the basis of preliminary statistical material. This table contains the probabilities of character categories for various diagnoses.

Table 1

If the signs are two-digit (simple signs “yes - no”), then in the table it is enough to indicate the probability of occurrence of the sign P(k j / D i).

Probability of missing feature P (k j / D i) = 1 − P (k j / D i) .

However, it is more convenient to use a uniform form, assuming, for example, for a two-digit sign P(kj/D) = P(kj 1/D) ; P(k j/D) = P(kj 2/D).

Note that ∑ P (k js / D i) =1 , where m j is the number of digits of the sign k j .

The sum of the probabilities of all possible implementations of a feature is equal to one.

The diagnostic matrix includes a priori probabilities of diagnoses. The learning process in the Bayes method consists of forming a diagnostic matrix. It is important to provide for the possibility of clarifying the table during the diagnostic process. To do this, not only the values ​​of P(k js / D i) should be stored in the computer memory, but also the following quantities: N – the total number of objects used to compile the diagnostic matrix; N i - number of objects with diagnosis D i ; N ij – number of objects with diagnosis D i, examined according to characteristic k j. If a new object arrives with a diagnosis D μ, then the previous a priori probabilities of diagnoses are adjusted as follows:

Next, corrections are introduced to the probabilities of the features. Let a new object with a diagnosis D μ have a rank r of sign k j identified. Then, for further diagnostics, new values ​​of the probability of intervals of the feature k j are accepted for diagnosis D μ:

Conditional probabilities of signs for other diagnoses do not require adjustment.

Practical part

1.Study the guidelines and receive the assignment.

PRACTICAL WORK No. 4

State Committee of the Russian Federation for Fisheries

Federal State Educational

Institution of higher professional education

Kamchatka State Technical University

Department of Math

Coursework in the discipline

"Mathematical Economics"

On the topic: “Risk and insurance.”

Introduction………………………………………………………..……………….....3

1. CLASSICAL SCHEME FOR ASSESSING FINANCIAL OPERATIONS UNDER CONDITIONS OF UNCERTAINTY …………………..................................... ........................................4 1.1. Definition and essence of risk…………………………………..……………..…...4

1.2. Matrices of consequences and risks…………………………………….……..……6

1.3.Analysis of a related group of decisions under conditions of complete uncertainty……………………………………………………………...………………...7

1.4. Analysis of a related group of decisions under conditions of partial uncertainty………………………………………………………………..8

1.5. Pareto optimality…………………………………………………….9

2. CHARACTERISTICS OF PROBABILISTIC FINANCIAL OPERATIONS……..…..…...12

2.1. Quantitative risk assessment……………………………………………..12

2.2. Risk of a separate operation……………………………………………………..13 2.3. Some common risk measures…………………………………….15

2.4. Risk of ruin……………………………………………………………..…16

2.5. Risk indicators in the form of ratios……………………………………..17

2.6. Credit risk…………………………………………………………….17

3. GENERAL RISK REDUCTION METHODS………………………………………………………….…….18

3.1. Diversification……………………………………………………………18

3.2. Hedging…………………………………………………………………………………21

3.3. Insurance………………………………………………………………………………...22

3.4. Quality risk management………………………………….……….24

Practical part……………………………………………………………...….27

Conclusion………………………………………………………..………….…. ..29

References…………………………………………………………….……….……..….30

Applications……………………………………………………….…………..…...31

INTRODUCTION

The development of world financial markets, characterized by the intensification of the processes of globalization, internationalization, and liberalization, has a direct impact on all participants in the global economic space, the main members of which are large financial institutions, manufacturing and trading corporations. All participants in the global market directly feel the impact of all of the above processes and in their activities must take into account new trends in the development of financial markets. The number of risks arising in the activities of such companies has increased significantly in recent years. This is due to the emergence of new financial instruments actively used by market participants. The use of new instruments, although it makes it possible to reduce the risks assumed, is also associated with certain risks for the activities of financial market participants. Therefore, awareness of the role of risk in the company’s activities and the ability of the risk manager to adequately and timely respond to the current situation and make the right decision regarding risk are becoming increasingly important for the successful operation of the company. To do this, it is necessary to use various insurance and hedging instruments against possible losses, the range of which has expanded significantly in recent years and includes both traditional insurance methods and hedging methods using financial instruments.

The efficiency of the company as a whole will ultimately depend on how correctly one or another tool is chosen.

The relevance of the research topic is also predetermined by the incompleteness of the development of the theoretical basis and classification of financial risk insurance and the identification of its features in Russia.

Chapter 1. CLASSIC SCHEME FOR FINANCIAL ASSESSMENT

OPERATIONS UNDER UNCERTAINTY

Risk – one of the most important concepts accompanying any active human activity. At the same time, this is one of the most unclear, ambiguous and confusing concepts. However, despite its ambiguity, ambiguity and complexity, in many situations the essence of risk is very well understood and perceived. These same qualities of risk are a serious obstacle to its quantitative assessment, which in many cases is necessary both for the development of theory and in practice.

Let's consider the classic decision-making scheme under conditions of uncertainty.

1.1. Definition and essence of risk

Let us remind you that financial is an operation whose initial and final states have a monetary value and the purpose of which is to maximize income – difference between final and initial

grades (or some other similar indicator).

Almost always, financial transactions are carried out under conditions of uncertainty and therefore their results cannot be predicted in advance. Therefore, financial transactions risky : when they are carried out, both profit and loss are possible (or not a very large profit compared to what those who carried out this operation hoped for).

The person conducting the operation (making the decision) is called the decision maker – Face ,

decision maker . Naturally, the decision maker is interested in the success of the operation and is responsible for it (sometimes only to himself). In many cases, the decision maker – is an investor who invests money in a bank, in which – then a financial transaction, buying securities, etc.

Definition. The operation is called risky , if it can have several outcomes that are not equivalent for the decision maker.

Example 1 .

Consider three operations with the same set of two outcomes –

alternatives A , IN, which characterize the income received by the decision maker. All three

operations are risky. It is clear that the first and second are risky

operations, since each operation may result in losses.

But why should a third operation be considered risky? After all, it promises only positive income for decision makers? Considering the possible outcomes of the third operation, we see that we can receive an income of 20 units, so the possibility of receiving an income of 15 units is considered as a failure, as a risk of not getting 5 units of income. So, the concept of risk necessarily presupposes taking risks – the one to whom this risk applies, who is concerned about the result of the operation. The risk itself arises only if the operation may end in outcomes that are not equivalent for him, despite, perhaps, all his efforts to manage this operation.

So, in conditions of uncertainty, the operation acquires another characteristic – risk. How to evaluate an operation in terms of its profitability and risk? This question is so easy to answer, mainly because the concept of risk is multifaceted. There are several different ways to do this assessment. Let's consider one of these approaches.

1.2. Consequence and Risk Matrices

Let's say the issue of conducting a financial transaction is being considered. It is unclear how it might end. In this regard, several possible solutions and their consequences are analyzed. So we come to the following general scheme for making decisions (including financial ones) under conditions of uncertainty.

Let's assume that the decision maker is considering several possible solutions

i =1, …,n. The situation is uncertain, it is only clear that there is some – then from the options j =1,….,n. If accepted i– This is not a solution, but there is a situation j– I, then the company headed by the decision maker will receive income q ij . Matrix Q =(q ij) is called matrix of consequences(possible solutions). Let's say we want to estimate the risk posed by i-th solution. We don't know the real situation. But if we knew it, we would choose the best solution, i.e. generating the most income. If the situation j-i, then a decision would be made that would generate income q i =max q ij. So, taking i-th decision, we risk getting q j , but only q ij , those. Adoption i- decision carries the risk of not being reached r ij = q j –q ij is called risk matrix .

Example 2.

Let there be a matrix of consequences

Let's create a risk matrix. We have q 1 =max q i1 =8, q 2 =5, q 3 =8, q 4 =12. Therefore, the risk matrix is

1.3. Analysis of a coupled group of decisions under conditions of complete uncertainty

A situation of complete uncertainty is characterized by the absence of any additional information (for example, about the probabilities of certain options for the real situation). What are the rules? – recommendations for making decisions in this situation?

Wald's rule (rule of extreme pessimism).

Considering i-th decision, we will assume that in fact the situation is the worst, i.e. bringing the least income: a i =min q a 0 with the greatest a i0. So, Wald's rule recommends making a decision i 0 such that a i0 =max a i =max(min q ij).So, in example 2 we have a 1 =2, a 2 =2, a 3 =3, a 4 = 1. Now from the numbers 2, 2, 3, 1 we find the maximum - 3. This means that Wald’s rule recommends making the 3rd decision.

Savage's rule (minimum risk rule).

When applying this rule, the risk matrix is ​​analyzed R =(r ij). Considering i decision, we will assume that in fact a situation of maximum risk is emerging b i =max r ij. But now let's choose a solution i 0 with the smallest b i0. So, Savage's rule recommends making a decision i 0 such that b i0 =min b i =min(max r ij).So, in example 2 we have b 1 =8, b 2 =6, b 3 =5, b 4 =7. Now from the numbers 8, 6 , 5, 7 we find the minimum - 5.

Hurwitz's rule (weighing pessimistic and optimistic approaches to a situation).

A decision is made i, which reaches the maximum

{λ min q ij +(1 – λ max q ij)),

where 0≤ λ ≤1. Meaning λ selected for subjective reasons. If λ approaches 1 , then Hurwitz’s rule approaches Wald’s rule, as we approach λ to 0, Hurwitz’s rule approaches the rule of “pink optimism” (guess for yourself what this means). In example 2, with λ=1/2, the Hurwitz rule recommends the second solution.

1.4. Analysis of a coupled group of decisions under conditions of partial uncertainty

Let us assume that in the scheme under consideration the probabilities are known R j that the real situation is developing according to the variant j. This situation is called partial uncertainty. How to make a decision here? You can select one of the following rules.

Rule for maximizing average expected income.

Income received by the company from sales i-th solution is a random variable Q i with a distribution series. Expected value M [Q i ] is the average expected income, also denoted Q i . So, the rule recommends making the decision that yields the maximum average expected return. Suppose that in the scheme of example 2 the probabilities are – 1/2, 1/6, 1/6, 1/6.

Then Q 1 =29/6, Q 2 =25/6, Q 3 =7, Q 4 =17/6. The maximum average expected return is 7 and corresponds to the third solution.

Rule for minimizing average expected risk.

The company's risk during implementation i-th solution is a random variable R i with distribution series

Expected value M [R i ] and is the average expected risk, also denoted R i. The rule recommends making a decision that entails the minimum average expected risk. Let us calculate the average expected risks for the above probabilities. We get R 1 =20/6, R 2 =4, R 3 =7/6, R 4 =32/6. The minimum average expected risk is 7/6 and corresponds to the third solution.

Comment. The difference between partial (probabilistic) uncertainty and complete uncertainty is very significant. Of course, no one considers decision-making according to the rules of Wald, Savage, and Hurwitz to be final or the best. But when we begin to assess the probability of an option, this already presupposes the repeatability of the decision-making pattern in question: it has already happened in the past, or it will happen in the future, or it is repeated somewhere in space, for example, in the branches of the company.

1.5. Pareto optimality

So, when trying to choose the best solution, we were faced in the previous paragraph with the fact that each solution has two characteristics – average expected return and average expected risk. Now we have a two-criteria optimization problem of choosing the best solution.

There are several ways to formulate such optimization problems.

Let us consider this problem in general form. Let A - some set of operations, each operation A has two numerical characteristics E (A), r (A) (efficiency and risk, for example) and different operations necessarily differ in at least one characteristic. When choosing the best operation, it is advisable that E there was more and r less.

We will say that the operation A dominates the operation b, and designate A >b, If E (A)≥E (b) And r (A)≤r (b) and at least one of these inequalities is strict. In this case, the operation A called dominant , and the operation b- dominated . It is clear that under no reasonable choice of the best operation, a dominated operation cannot be recognized as such. Consequently, the best operation must be sought among non-dominated operations. The set of these operations is called Pareto set or Pareto optimality set .

This is an extremely important statement.

Statement.

On the Pareto set, each of the characteristics E , r-(unambiguous) function is different. In other words, if an operation belongs to the Pareto set, then one of its characteristics can be used to uniquely determine another.

Proof. Let A ,b - two operations from the Pareto set, then r (A) And r (b) – numbers. Let's pretend that r (A)≤r (b), Then E (A) cannot be equal E (b), since both points A , b belong to the Pareto set. It has been proven that according to the characteristics r E. It is also simply proved that, according to the characteristic E characteristic can be determined r .

Let us continue the analysis of the example given in § 10.2. Let's look at a graphic illustration. Each operation (decision) ( R, Q) mark as a point on the plane – income is postponed upward vertically, and risk – to the right horizontally (Fig. 10.1). We received four points and continue the analysis of example 2.

The higher the point ( R, Q), the more profitable the operation; the further the point to the right, the more risky it is. This means you need to choose a point higher and to the left. In our case, the Pareto set consists of only one third operation.

To find the best operation, a suitable weighing formula is sometimes used, which for the operation Q with characteristics ( R, Q) gives one number by which the best operation is determined. For example, let the weighing formula be f (Q)=2Q–R. Then for the operations (decisions) of Example 2 we have: f (Q 1)=2*29/6– 20/6=6,33; f (Q 2)=4,33; f (Q 3)=12,83; f (Q 4)=0.33. It can be seen that the third operation is the best, and the fourth – the worst.

Chapter 2. CHARACTERISTICS OF PROBABILISTIC FINANCIAL

OPERATIONS

The financial transaction is called probabilistic , if there is a probability of each outcome. The profit of such an operation – the difference between the final and initial monetary estimates – is a random variable. For such an operation, it is possible to introduce a quantitative risk assessment that is consistent with our intuition.

2.1. Quantitative risk assessment

The previous chapter defined a risky operation as one that has at least two outcomes that are not equivalent in the decision maker’s preference system. In the context of this chapter, instead of the decision maker, you can also use the term “investor” or something similar, reflecting the interest of the person conducting the operation (possibly passively) in its success.

When examining the risk of surgery, we encounter a fundamental statement.

Statement.

Quantitative assessment of the risk of surgery is only possible with a probabilistic characterization of multiple surgical outcomes.

Example 1.

Let's consider two probabilistic operations:

Undoubtedly, the risk of the first operation is less than the risk of the second operation. As for which operation the decision maker will choose, it depends on his appetite for risk (such issues are discussed in detail in the addendum to Part 2).

2.2. Risk of a separate operation

Since we want to quantify the riskiness of an operation, and this cannot be done without a probabilistic characteristic of the operation, we will assign probabilities to its outcomes and evaluate each outcome by the income that the decision maker receives from this outcome. As a result, we get a random variable Q, which it is natural to call the incidental income of the operation, or simply random income . For now, let’s limit ourselves to a discrete random variable (d.r.v.):

Where q j - income, and R j – the probability of this income.

The operation and the random variable representing it – We will identify random income if necessary, choosing from these two terms the more convenient in a particular situation.

Now you can apply the apparatus of probability theory and find the following characteristics of the operation.

Average expected income – mathematical expectation r.v. Q, i.e. M [Q ]=q 1 p 1 +…+q n p n, also denoted m Q, Q, the name is also used efficiency of the operation .

Variance of operation - dispersion r.v. Q, i.e. D [Q ]=M [(Q - m Q) 2 ], also denoted D Q.

Standard deviation s.v. Q, i.e. [ Q ]=√(D [E ]), denoted by

Also σ Q.

Note that the average expected return, or operational efficiency, like the standard deviation, is measured in the same units as income.

Let us recall the fundamental meaning of the mathematical expectation of r.v.

The arithmetic mean of the values ​​taken as r.v. in a long series of experiments, approximately equal to its mathematical expectation. It is becoming increasingly accepted to assess the riskiness of the entire operation using the standard deviation of the random variable of income Q, i.e. through σ Q. This is the main quantification in this book.

So, risk of surgery called number σ Q – standard deviation of random operation income Q. Also designated r Q.

Example 2.

Let's find the risks of the first and second operations from example 1:

First, we calculate the mathematical expectation of r.v. Q 1:

T 1 =– 5*0.01+25*0.99=24.7. Now let's calculate the variance using the formula D 1 =M [Q 1 2 ]-m 1 2 . We have M [Q 1 2 ]= 25*0.01+625*0.99=619. Means, D 1 =619– (24.7)2=8.91 and finally r 1 =2,98.

Similar calculations for the second operation give m 2 =20; r 2 =5. As “intuition suggested,” the first operation is less risky.

The proposed quantitative risk assessment is fully consistent with the intuitive understanding of risk as the degree of dispersion of the outcomes of the operation – After all, dispersion and standard deviation (the square root of the dispersion) are the essence of measures of such dispersion.

Other risk measures.

In our opinion, standard deviation is the best measure of the risk of an individual operation. In ch. 1 discusses the classical scheme of decision-making under conditions of uncertainty and risk assessment in this scheme. It is useful to get acquainted with: other risk measures. In most cases, these meters – simply the probabilities of undesirable events.

2.3. Some common risk measures

Let the distribution function be known F random income operation Q. Knowing it, you can give meaning to the following questions and answer them.

1. What is the probability that the operation’s income will be less than the specified one? s. You can ask by – to another: what is the risk of receiving less than the specified income? Answer: F (s).

2. What is the probability that the operation will be unsuccessful, i.e. her income will be less than the average expected income m ?

Answer: F (m) .

3. What is the probability of losses and what is their average expected size? Or what is the risk of losses and their assessment?

4. What is the ratio of average expected loss to average expected income? The lower this ratio, the lower the risk of ruin if the decision maker has invested all his funds in the operation.

When analyzing operations, the decision maker wants to have more income and less risk. Such optimization problems are called two-criteria. When analyzing them, there are two criteria - income and risk – often “collapsed” into one criterion. This is how, for example, the concept arises relative risk of surgery . The fact is that the same value of the standard deviation σ Q, which measures the risk of an operation, is perceived differently depending on the value of the average expected return T Q , therefore the value σ Q / T Q is sometimes called the relative risk of surgery. This risk measure can be interpreted as a convolution of a two-criteria problem

σ Q →min,

T Q →max,

those. maximize average expected return while minimizing risk.

2.4. Risk of ruin

This is the name for the probability of such large losses that the decision maker cannot compensate and which, therefore, lead to his ruin.

Example 3.

Let the random income of the operation Q has the following distribution series, and losses of 35 or more lead to the ruin of the decision maker. Therefore, the risk of ruin as a result of this operation is 0.8;

The severity of the risk of ruin is assessed precisely by the value of the corresponding probability. If this probability is very small, it is often neglected.

2.5. Risk indicators in the form of ratios.

If the decision maker's funds are equal WITH, then if the losses exceed U above WITH there is a real risk of ruin. To prevent this attitude TO 1 = U / WITH , called risk coefficient , limited by a special number ξ 1 . Operations for which this coefficient exceeds ξ1 are considered particularly risky. The probability is also often taken into account R losses U and then consider the risk coefficient TO 2 = R Y/ WITH , which is limited by another number ξ 2 (it is clear that ξ 2 ≤ ξ 1). In financial management, inverse relationships are more often used. WITH / U And WITH /(RU), which are called risk coverage coefficients and which are limited from below by the numbers 1/ ξ 1 and 1/ ξ 2.

This is precisely the meaning of the so-called Cook’s coefficient, equal to the ratio:

The Cook's Ratio is used by banks and other financial companies. Probabilities act as scales when “weighing” – risks of loss of the relevant asset.

2.6. Credit risk

This is the probability of non-repayment of the loan taken on time.

Example 4.

The loan request statistics are as follows: 10% – government agencies, 30% – other banks and others – individuals. The probabilities of non-repayment of the loan taken are respectively: 0.01; 0.05 and 0.2. Find the probability of non-return of the next loan request. The head of the credit department was informed that a message about the non-repayment of the loan had been received, but the client's name was poorly printed in the fax message. What is the probability that this loan will not repay – is it a bank?

Solution. We will find the probability of non-return using the total probability formula. Let N 1 - the request came from a government agency, N 2 – from the bank, N 3 – from an individual and A - non-repayment of the loan in question. Then

R (A)= R (N 1)R H1 A + R (N 2)R H2 A + R (N h) P H3 A = 0,1*0,01+0,3*0,05+0,6*0,2=0,136.

We find the second probability using Bayes' formula. We have

R A N 2 =R (N 2)R H2 A / R (A)= 0,015/0,136=15/136≈1/9.

How in reality all the data given in this example are determined, for example, conditional probabilities R H1 A? Based on the frequency of loan defaults for the corresponding group of clients. Let individuals take out only 1000 loans and not return 200. So the corresponding probability R H3 A estimated as 0.2. Relevant Data – 1000 and 200 are taken from the bank's information database.

Chapter 3. GENERAL RISK REDUCTION METHODS

As a rule, they try to reduce the risk. There are many methods for this. A large group of such methods is associated with the selection of other operations. Such that the overall operation has less risk.

3.1. Diversification

Recall that the variance of the sum of uncorrelated random variables is equal to the sum of the variances. From this follows the following statement underlying the diversification method.

Statement 1.

Let ABOUT 1 ,...,ABOUT n – uncorrelated operations with efficiencies e 1 ,..., e n and risks r 1 ,...,r 2 . Then the operation “arithmetic mean” ABOUT =(ABOUT 1 +...+O n) / P has efficiency e =(e 1 +...+e n)/ n and risk r =√(r 1 2 +…r 2n)/ n .

Proof of this statement – a simple exercise on the properties of mathematical expectation and dispersion.

Corollary 1.

Let the operations be uncorrelated and a≤ e i and b ≤ r i ≤ c with for everyone i =1,..,n. Then the efficiency of the “arithmetic mean” operation is no less A(i.e. the least of the efficiency of operations), and the risk satisfies the inequality b √ n ≤ r ≤c √ n and thus, with increasing n decreases. So, with an increase in the number of uncorrelated operations, their arithmetic average has an efficiency within the range of the efficiencies of these operations, and the risk definitely decreases.

This output is called diversification effect(diversity) and is essentially the only reasonable rule for working in financial and other markets. The same effect is embodied in folk wisdom – "Don't put all your eggs in one basket." The principle of diversification states that it is necessary to carry out various, unrelated operations, then the efficiency will be averaged, and the risk will definitely decrease.

You need to be careful when applying this rule. Thus, it is impossible to refuse the uncorrelated nature of operations.

Proposal 2.

Let us assume that among the operations there is a leading one with which all the others are in a positive correlation. Then the risk of the “arithmetic mean” operation does not decrease with an increase in the number of summed operations.

Indeed, for simplicity we accept a stronger assumption, namely, that all operations ABOUT i ; i =1,...,n, just copy the operation O 1 in which – then scales, i.e. O i = k i O 1 and all proportionality factors k i are positive. Then the operation “arithmetic mean” ABOUT =(O 1 +...+O n)/ n there is just an operation O 1 to scale

and the risk of this operation

Therefore, if operations are approximately the same in scale, i.e. k i ≈1, then

We see that the risk of the arithmetic mean operation does not decrease with increasing number of operations.

3.2. Hedging

In the effect of diversification, the decision maker constituted a new operation out of several at his disposal. When hedging (from English. hedge - fence) The decision maker selects or even specially designs new operations in order to reduce the risk by performing them together with the main one.

Example 1.

According to the contract, the Russian company must receive a large payment from the Ukrainian company in six months. The payment is equal to 100,000 hryvnia (approximately 600 thousand rubles) and will be made in hryvnia. The Russian company has concerns that over these six months the hryvnia exchange rate will fall against the Russian ruble. The company wants to insure against such a fall and enters into a forward contract with one of the Ukrainian banks to sell it 100,000 hryvnia at the rate of 6 rubles. per hryvnia. Thus, no matter what happens during this time with the ruble exchange rate – hryvnia, the Russian company will not bear the cost – for this loss.

This is the essence of hedging. In diversification, independent (or uncorrelated) transactions were of greatest value. When hedging, operations are selected that are strictly related to the main one, but, so to speak, of a different sign, or more precisely, negatively correlated with the main operation.

Indeed, let O 1 – main operation, its risks r 1 , O 2 – some additional surgery, its risk r 2 , ABOUT - operation – sum, then the variance of this operation D =r 1 2 +2k 12 r 1 r 2 +r 2 2 where k- correlation coefficient of the effectiveness of the main and additional operations. This variance can be less than the variance of the main operation only if this correlation coefficient is negative (more precisely: it should be 2 k 12 r 1 r 2 +r 2 2 <0, т.е. k 1 2 <–r 2 /(2r 1)).

Example 2.

Let the decision maker decide to carry out the operation O 1 .

He is advised to undergo surgery at the same time S, strictly related to ABOUT. In essence, both operations must be depicted with the same set of outcomes.

Let us denote the total operation by ABOUT, this operation is the sum of operations O 1 and S. Let's calculate the characteristics of the operations:

M [O 1 ]=5, D [O 1 ]=225, r 1 =15;

M [S ]=0, D [S ]=25;

M [O ]=5, D [O ]=100, r =10.

The average expected effectiveness of surgery remained unchanged, but the risk decreased due to the strong negative correlation of additional surgery S in relation to the main operation.

Of course, in practice it is not so easy to select an additional operation that is negatively correlated with the main one, and even with zero efficiency. Usually, a small negative efficiency of an additional operation is allowed and because of this, the efficiency of the total operation becomes less than that of the main one. The extent to which a reduction in efficiency is allowed per unit of risk reduction depends on the decision maker’s attitude to risk.

3.3. Insurance

Insurance can be considered as a type of hedging. Let's clarify some terms.

Policyholder(or insured) – the one who insures.

Insurer - the one who insures.

Sum insured - the amount of money for which the property, life, and health of the policyholder are insured. This amount is paid by the insurer to the policyholder upon the occurrence of an insured event. Payment of the insurance amount is called insurance compensation .

Insurance payment paid by the policyholder to the insurer.

Let us denote the insurance amount ω , insurance payment s, probability of an insured event R . Let us assume that the insured property is valued at z. According to insurance rules ω≤ z.

Thus, we can propose the following scheme:

Thus, insurance seems to be the most profitable measure in terms of risk reduction, if not for the insurance payment. Sometimes the insurance payment forms a significant part of the insured amount and represents a substantial amount.

3.4. Quality risk management

Risk – such a complex concept that it is often impossible to quantify it. Therefore, qualitative risk management methods, without quantitative assessment, are widely developed. These include many banking risks. The most important of them – These are credit risk and the risks of illiquidity and insolvency.

1. Credit risk and ways to reduce it . When issuing a loan (or loan), there is always a fear that the client will not repay the loan. Preventing default, reducing the risk of loan default – This is the most important task of the bank's credit department. What ways are there to reduce the risk of loan default?

The department must constantly systematize and summarize information on loans issued and their repayment. Information on loans issued should be systematized according to the size of the loans issued, and a classification of clients who took out a loan should be constructed.

The department (the bank as a whole) must maintain the so-called credit history of its clients, including potential ones (i.e. when, where, what loans the client took and how they were repaid). So far in our country, the majority of clients do not have their own credit history.

There are various ways to secure a loan, for example, the client gives something as collateral and if he does not repay the loan, then the bank becomes the owner of the collateral;

The bank must have clear instructions for issuing a loan (to whom can a loan be issued and for what period);

Clear authority for issuing credit must be established. Let's say, an ordinary department employee can issue a loan of no more than $1000, loans up to $10,000 can be issued by the head of the department, over $10,000, but not more than $100,000, can be issued by the vice president for finance, and loans over $100,000 can only be issued by the board of directors (read novel A . Hayley "Moneychangers");

To issue particularly large and dangerous loans, several banks unite and jointly issue this loan;

There are insurance companies that insure loan default (but there is a point of view that loan default is not subject to insurance – This is the risk of the bank itself);

There are external restrictions on the issuance of loans (for example, established by the Central Bank); say, it is not allowed to issue a very large loan to one client;

2. Risks of illiquidity , insolvency and ways to reduce it . They say that a bank's funds are sufficiently liquid if the bank is able to quickly and without any significant losses ensure payment to its clients of funds that they entrusted to the bank on a short-term basis. Illiquidity risk – this is the risk of not being able to cope with it. However, this risk is named only for brevity; its full name is – risk of imbalance balance sheet in terms of liquidity .

All bank assets according to their liquidity are divided into three groups:

1) first-class liquid funds (cash, bank funds in a correspondent account with the Central Bank, government securities, bills of large reliable companies;

2) liquid funds (expected short-term payments to the bank, some types of securities, some tangible assets that can be sold quickly and without large losses, etc.);

3) illiquid funds (overdue loans and bad debts, many tangible assets of the bank, primarily buildings and structures).

When analyzing illiquidity risk, first-class liquid funds are taken into account first.

They say that a bank is solvent if it is able to pay off all its customers, but this may require some large and lengthy transactions, including the sale of equipment, buildings owned by the bank, etc. Insolvency risk arises when it is unclear whether the bank will be able to pay.

Bank solvency depends on so many factors. The Central Bank sets a number of conditions that banks must comply with to maintain their solvency. The most important of them are: limiting the bank’s liabilities; refinancing of banks by the Central Bank; reserving part of the bank's funds in a correspondent account with the Central Bank.

The risk of illiquidity leads to possible unnecessary losses for the bank: in order to pay the client, the bank may have to borrow money from other banks at a higher interest rate than under normal conditions. The risk of insolvency may well lead to bank bankruptcy.

Practical part

Let's assume that a decision maker has the opportunity to compose an operation from four uncorrelated operations, the efficiencies and risks of which are given in the table.

Let's consider several options for composing operations from these operations with equal weights.

1. The operation consists of only the 1st and 2nd operations. Then e 12 =(3+5)/2=4;

r 12 = √ (2 2 +4 2)/2≈2,24

2. The operation consists of only the 1st, 2nd and 3rd operations.

Then e 123 =(3+5+8)/3=5,3; r 123 =√(2 2 +4 2 +6 2)/3≈2,49.

3. The operation is made up of all four operations. Then

e 1– 4 =(3+5+8+10)/4=6,5; r 1– 4 =√(2 2 +4 2 +6 2 +12 2)/4≈ 3,54.

It can be seen that when composing an operation from an increasing number of operations, the risk grows very slightly, remaining close to the lower limit of the risks of the component operations, and the efficiency each time is equal to the arithmetic average of the component efficiencies.

The principle of diversification is applied not only to averaging operations carried out simultaneously, but in different places (averaging in space), but also carried out sequentially in time, for example, when repeating one operation over time (averaging over time). For example, a completely reasonable strategy is to buy shares of some stable company on January 20th of each year. Thanks to this procedure, the inevitable fluctuations in the stock price of this company are averaged out and this is where the diversification effect is manifested.

Theoretically, the effect of diversification is only positive – efficiency averages out and risk decreases. However, efforts to conduct a large number of operations and monitor their results can, of course, negate all the benefits of diversification.

CONCLUSION

This course work examines theoretical and practical issues and risk problems.

The first chapter discusses the classic scheme for assessing financial transactions under conditions of uncertainty.

The second chapter provides an overview of the characteristics of probabilistic financial transactions. Financial risks are understood as credit, commercial, exchange transaction risks and the risk of unlawful application of financial sanctions by state tax inspectorates.

Chapter three shows general risk mitigation techniques. Examples of high-quality risk management are given.

Bibliography

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