Summation of independent random variables. Distribution of the sum of two random independent variables Probability of the sum of random variables

Let's use the above general method to solve one problem, namely to find the law of distribution of the sum of two random variables. There is a system of two random variables (X,Y) with distribution density f(x,y). Let's consider the sum of random variables X and Y: and find the law of distribution of the value Z. To do this, let's build a line on the xOy plane, the equation of which is (Fig. 7). This is a straight line cutting off segments equal to z on the axes. A straight line divides the xOy plane into two parts; to the right and above it; to the left and below.

Region D in this case is the lower left part of the xOy plane, shaded in Fig. 7. According to formula (16) we have:

Differentiating this expression with respect to the variable z included in the upper limit of the internal integral, we obtain:

This is the general formula for the distribution density of the sum of two random variables.

For reasons of symmetry of the problem with respect to X and Y, we can write another version of the same formula:

which is equivalent to the first and can be used instead.

An example of the composition of normal laws. Let's consider two independent random variables X and Y, subject to normal laws:

It is required to produce a composition of these laws, that is, to find the law of distribution of the quantity: .

Let us apply the general formula for the composition of distribution laws:

If we open the brackets in the exponent of the integrand and add similar terms, we get:

Substituting these expressions into the formula we have already encountered

after transformations we get:

and this is nothing more than a normal law with a center of dispersion

and standard deviation

The same conclusion can be reached much more easily using the following qualitative reasoning.

Without opening the parentheses and without making any transformations in the integrand (17), we immediately come to the conclusion that the exponent is quadratic trinomial relative to x species

where the value z is not included in coefficient A at all, in coefficient B it is included to the first power, and in coefficient C it is squared. Keeping this in mind and applying formula (18), we come to the conclusion that g(z) is an exponential function, the exponent of which is a square trinomial with respect to z, and the distribution density; This type corresponds to the normal law. Thus, we; we come to a purely qualitative conclusion: the law of distribution of the value z must be normal. To find the parameters of this law - and - we will use the theorem of addition of mathematical expectations and the theorem of addition of variances. According to the theorem of addition of mathematical expectations. By the theorem of addition of variances or from which formula (20) follows.

Moving from standard deviations to probable deviations proportional to them, we obtain: .

Thus, we came to the following rule: when combining normal laws, a normal law is obtained again, and the mathematical expectations and variances (or squares of probable deviations) are summed up.

The rule for the composition of normal laws can be generalized to the case of an arbitrary number of independent random variables.

If there are n independent random variables: subject to normal laws with centers of dispersion and standard deviations, then the value is also subject to the normal law with parameters

Instead of formula (22), you can use an equivalent formula:

If a system of random variables (X, Y) is distributed according to a normal law, but the values ​​X, Y are dependent, then it is not difficult to prove, just as before, based on the general formula (6.3.1), that the law of distribution of a value is also a normal law. The scattering centers are still added algebraically, but for standard deviations the rule becomes more complex: , where, r is the correlation coefficient of the X and Y values.

When adding several dependent random variables, which in their entirety are subject to the normal law, the distribution law of the sum also turns out to be normal with the parameters

or in probable deviations

where is the correlation coefficient of the quantities X i, X j, and the summation applies to all different pairwise combinations of quantities.

We have become convinced of a very important property normal law: with the composition of normal laws, a normal law is obtained again. This is the so-called “stability property”. A distribution law is called stable if the composition of two laws of this type again results in a law of the same type. We showed above that the normal law is stable. Very few distribution laws have the property of stability. The law of uniform density is unstable: by combining two laws of uniform density in sections from 0 to 1, we obtained Simpson's law.

The stability of the normal law is one of essential conditions its widespread use in practice. However, in addition to the normal one, some other distribution laws also have the property of stability. A feature of the normal law is that with composition it is sufficient large number of practically arbitrary distribution laws, the total law turns out to be as close to normal as desired, regardless of what the distribution laws of the terms were. This can be illustrated, for example, by composing the three laws of uniform density in areas from 0 to 1. The resulting distribution law g(z) is shown in Fig. 8. As can be seen from the drawing, the graph of the function g(z) is very similar to the graph of the normal law.

A decision maker can use insurance to reduce the adverse financial impact of certain types of random events.

But this consideration is very general, since the decision maker could mean either an individual seeking protection from damage to property, savings or income, or an organization seeking protection from the same type of damage.

In fact, such an organization may turn out to be Insurance Company, which is looking for ways to protect itself from financial loss due to too many insurance claims occurring to an individual client or its insurance portfolio. This type of protection is called reinsurance.

Let us consider one of two models (namely individual risk model) widely used in determining insurance rates and reserves, as well as in reinsurance.

Let us denote by S the amount of accidental losses of the insurance company for some part of its risks. In this case S is a random variable for which we must determine a probability distribution. Historically, for distributions of r.v. S there were two sets of postulates. The individual risk model determines S in the following way:

where r.v. means losses caused by the insurance object with the number i, A n denotes the total number of insurance objects.

It is usually assumed that they are independent random variables, since in this case mathematical calculations are simpler and information about the nature of the relationship between them is not required. The second model is the collective risk model.

The individual risk model under consideration does not reflect changes in the value of money over time. This is done to simplify the model, and that is why the title of the article refers to a short time interval.

We will consider only closed models, i.e. those in which the number of insurance objects n in formula (1.1) is known and fixed at the very beginning of the time interval under consideration. If we introduce assumptions about the presence of migration from or to the insurance system, we obtain an open model.

Random variables describing individual payments

First, let us recall the basic provisions regarding life insurance.

When insuring in case of death for a period of one year, the insurer undertakes to pay the amount b, if the policyholder dies within a year from the date of concluding the insurance contract, and does not pay anything if the policyholder lives this year.

The probability of an insured event occurring during a specified year is indicated by .

A random variable describing insurance payments has a distribution that can be specified either by a probability function

(2.1)

or the corresponding distribution function

(2.2)

From formula (2.1) and from the definition of moments we obtain

(2.4)

These formulas can also be obtained by writing X as

where is a constant value paid in case of death, and is a random variable taking the value 1 upon death and 0 otherwise.

Thus, and , and the mean and variance of r.v. are equal to and respectively, and the mean value and variance of the r.v. are equal to and , which coincides with the formulas written above.

A random variable with a range of values ​​(0,1) is widely used in actuarial models.

In textbooks on probability theory it is called indicator, Bernoulli random size or binomial random variable in a single trial design.

We will call her indicator for reasons of brevity, and also because it indicates the occurrence, or non-occurrence, of the event in question.

Let's move on to searching for more general models, in which the amount of the insurance payment is also a random variable and several insured events may occur in the time interval under consideration.

Health insurance, car and other property insurance, and insurance civil liability immediately provide many examples. Generalizing formula (2.5), we put

where is a random variable describing insurance payments in the considered time interval, r.v. denotes the total amount of payments in this interval and r.v. is an indicator for the event that at least one insured event has occurred.

Being an indicator of such an event, r.v. records the presence () or lack () insured events in this time interval, but not the number of insured events in it.

Probability will still be denoted by .

Let's discuss several examples and determine the distribution of random variables in a certain model.

Let's first consider death insurance for a period of one year with an additional payment if death occurs as a result of an accident.

To be certain, let’s assume that if death occurred as a result of an accident, the payment amount will be 50,000. If death occurs for other reasons, the payment amount will be 25,000.

Suppose that for a person of a given age, health status, and profession, the probability of death as a result of an accident during the year is 0.0005, and the probability of death from other causes is 0.0020. In formula form it looks like this:

Summing over all possible values, we get

,

Conditional distribution c. V. provided it has the form

Let's now consider collision insurance (compensation paid to the car owner for damage to his car) with an unconditional deductible of 250 and a maximum payout of 2,000.

For clarity, let's assume that the probability of one insured event occurring during the time period under consideration for an individual is 0.15, and the probability of more than one collision is zero:

, .

The unrealistic assumption that no more than one insured event can occur during one period is made in order to simplify the distribution of r.v. .

We abandon this assumption in the next section after we look at the distribution of multiple claims.

Since this is the amount of the insurer's payments, and not the damage caused to the car, we can consider two characteristics, and .

First, the event includes those collisions in which the damage is less than the unconditional deductible, which is 250.

Secondly, the distribution of r.v. will have a "clump" of probabilistic mass at the point maximum size insurance payments, which is equal to 2000.

Let us assume that the probability mass concentrated at this point is 0.1. Let us further assume that the value of insurance payments in the range from 0 to 2000 can be modeled by a continuous distribution with a density function proportional to (In practice, the continuous curve that is chosen to represent the distribution of insurance benefits is the result of studies of the levels of benefits in the previous period.)

Summarizing these assumptions about the conditional distribution of r.v. provided we arrive at the distribution mixed type, having a positive density in the range from 0 to 2000 and some “clump” of probabilistic mass at the point 2000. This is illustrated by the graph in Fig. 2.2.1.

The distribution function of this conditional distribution looks like this:

Fig.2.1. Distribution function r.v. In condition I = 1

Let's calculate expected value and variance in our car insurance example in two ways.

First, we write out the distribution of r.v. and use it to calculate and . Denoting by the distribution function r.v. , we have

For x<0

This is a mixed distribution. As shown in Fig. 2.2, it has both a discrete (“clump” of probabilistic mass at point 2000) and a continuous part. Such a distribution function corresponds to a combination of the probability function

Rice. 2.2. Distribution function r.v. X = IB

and density functions

In particular, and . That's why .

There are a number of formulas connecting the moments of random variables with conditional mathematical expectations. For the mathematical expectation and for the variance, these formulas have the form

(2.10)

(2.11)

It is understood that the expressions on the left sides of these equalities are calculated directly from the r.v. distribution. . When calculating expressions on the right sides, namely and , the conditional distribution of r.v. is used. at a fixed value of r.v. .

These expressions are thus functions of r.v. , and we can calculate their moments using the r.v. distribution. .

Conditional distributions are used in many actuarial models, and this allows the formulas above to be applied directly. In our model. Considering r.v. in quality and r.v. as , we get

(2.12)

, (2.14)

, (2.15)

and consider the conditional mathematical expectations

(2.16)

(2.17)

Formulas (2.16) and (2.17) are defined as a function of r.v. , which can be written as the following formula:

Since at , then (2.21)

For we have and (2.22)

Formulas (2.21) and (2.22) can be combined: (2.23)

Thus, (2.24)

Substituting (2.21), (2.20) and (2.24) into (2.12) and (2.13), we get

Let us apply the obtained formulas for calculations in the example of car insurance (Fig. 2.2). Since the density function r.v. The condition given is expressed by the formula

and P(B=2000|I=1)= 0.1, we have

Finally, believing q= 0.15, from formulas (2.25) and (2.26) we obtain the following equalities:

To describe a different insurance situation, we can propose other models for r.v. .

Example: Model for the number of deaths due to aviation accidents

As an example, consider a model for the number of deaths resulting from aviation accidents over a one-year period of operation of an airline.

We can start with a random variable describing the number of deaths for one flight, and then sum such random variables over all flights for the year.

For one flight, the event will indicate the occurrence of a plane crash. The number of deaths that this disaster entailed will be represented by the product of two random variables and , where is the aircraft load factor, i.e., the number of persons on board at the time of the accident, and is the proportion of deaths among those on board.

The number of deaths is presented in this way, since separate statistics for quantities and are more accessible than statistics for r.v. . So, Although the proportion of fatalities among persons on board and the number of persons on board are likely to be related, as a first approximation it can be assumed that r.v. and independent.

Sums of independent random variables

In the individual risk model, insurance payments made by an insurance company are represented as the sum of payments to many individuals.

Let us recall two methods for determining the distribution of a sum of independent random variables. Let us first consider the sum of two random variables, the sample space of which is shown in Fig. 3.1.

Rice. 2.3.1. Event

The straight line and the area under the straight line represent an event. Therefore, the r.v. distribution function S has the form (3.1)

For two discrete non-negative random variables, we can use the total probability formula and write (3.1) in the form

If X And Y independent, the last sum can be rewritten as

(3.3)

The probability function corresponding to this distribution function can be found using the formula

(3.4)

For continuous non-negative random variables, the formulas corresponding to formulas (3.2), (3.3) and (3.4) have the form

When either one or both random variables X And Y have a mixed distribution (which is typical for individual risk models), the formulas are similar, but more cumbersome. For random variables that can also take negative values, the sums and integrals in the above formulas are taken over all values ​​of y from to .

In probability theory, the operation in formulas (3.3) and (3.6) is called the convolution of two distribution functions and and is denoted by . The convolution operation can also be defined for a pair of probability functions or density functions using formulas (3.4) and (3.7).

To determine the distribution of the sum of more than two random variables, we can use iterations of the convolution process. For , where are independent random variables, denotes the distribution function of the r.v., and is the distribution function of the r.v. , we will get

Example 3.1 illustrates this procedure for three discrete random variables.

Example 3.1. The random variables , , and are independent and have distributions that are determined by columns (1), (2) and (3) of the table below.

Let us write down the probability function and the r.v. distribution function.

Solution. The table uses the notation introduced before the example:

Columns (1)-(3) contain available information.

Column (4) is derived from columns (1) and (2) using (3.4).

Column (5) is derived from columns (3) and (4) using (3.4).

The definition of column (5) completes the determination of the probability function for r.v. . Its distribution function in column (8) is the set of partial sums of column (5), starting from the top.

For clarity, we have included column (6), the distribution function for column (1), column (7), which can be obtained directly from columns (1) and (6) using (2.3.3), and column (8), defined similarly for columns (3) and (7). Column (5) can be determined from column (8) by sequential subtraction.

Let's move on to consider two examples with continuous random variables.

Example 3.2. Let r.v. has a uniform distribution on the interval (0,2), and let r.v. does not depend on r.v. and has a uniform distribution over the interval (0,3). Let us define the r.v. distribution function.

Solution. Since the distributions of r.v. and continuous, we use formula (3.6):

Then

Sample space r.v. and is illustrated in Fig. 3.2. The rectangular area contains all possible values ​​of the pair and . The event of interest to us, , is depicted in the figure for five values s.

For each value the straight line intersects the axis Y at the point s and a straight line at point . The function values ​​for these five cases are described by the following formula:

Rice. 3.2. Convolution of two uniform distributions

Example 3.3. Let us consider three independent r.v. . For r.v. has an exponential distribution and . Let us find the r.v. density function. , using the convolution operation.

Solution. We have

Using formula (3.7) three times, we get

Another method for determining the distribution of a sum of independent random variables is based on the uniqueness of the generating function of moments, which for r.v. is determined by the relation .

If this mathematical expectation is finite for everyone t from some open interval containing the origin, then is the only generating function of the moments of the r.v. distribution. in the sense that there is no other function other than , which would be the generating function of the moments of the r.v. distribution. .

This uniqueness can be used as follows: for the sum

If independent, then the mathematical expectation of the product in formula (3.8) is equal to ..., So

Finding an explicit expression for the only distribution that corresponds to the moment generating function (3.9) would complete the finding of the r.v. distribution. . If it is not possible to indicate it explicitly, then you can search for it using numerical methods.

Example 3.4. Let's consider the random variables from Example 3.3. Let us define the r.v. density function. , using the generating function of moments r.v. .

Solution. According to equality (3.9), which can be written in the form using the method of decomposition into simple fractions. The solution is . But is the generating function of the moments of the exponential distribution with the parameter, so the density function of the r.v. looks like

Example 3.5. When studying random processes, an inverse Gaussian distribution was introduced. It is used as the r.v. distribution. IN, the amount of insurance payments. The density function and the generating function of the moments of the inverse Gaussian distribution are given by the formulas

Let us find the distribution of r.v. , where r.v. are independent and have the same inverse Gaussian distributions.

Solution. Using formula (3.9), we obtain the following expression for the generating function of the moments of the r.v. :

The generating function of moments corresponds to a unique distribution, and we can verify that it has an inverse Gaussian distribution with parameters and .

Approximations for sum distribution

The central limit theorem provides a method for finding numerical values ​​for the distribution of a sum of independent random variables. Typically this theorem is formulated for a sum of independent and identically distributed random variables, where .

For any n, the r.v. distribution where = , has a mathematical expectation of 0 and variance of 1. As is known, the sequence of such distributions (for n= 1, 2, ...) tends to the standard normal distribution. When n Greatly this theorem is applied to approximate the distribution of r.v. normal distribution with mean μ and dispersion. Similarly, the distribution of the amount n random variables are approximated by a normal distribution with mean and variance.

The effectiveness of such an approximation depends not only on the number of terms, but also on the proximity of the distribution of terms to normal. Many elementary statistics courses state that n must be at least 30 for the approximation to be reasonable.

However, one of the programs for generating normally distributed random variables used in simulation modeling implements a normal random variable as the average of 12 independent random variables uniformly distributed over the interval (0,1).

In many individual risk models, the random variables included in the amounts are not equally distributed. This will be illustrated with examples in the next section.

The central limit theorem also applies to sequences of unequally distributed random variables.

To illustrate some applications of the individual risk model, we use the normal approximation of the distribution of the sum of independent random variables to obtain numerical solutions. If , That

and further, if r.v. are independent, then

For the application in question, we only need:

• find the means and variances of random variables modeling individual losses,
• sum them up in order to obtain the average and variance of losses of the insurance company as a whole,
• use the normal approximation.

Below we illustrate this sequence of actions.

Applications to insurance

This section illustrates the use of the normal approximation with four examples.

Example 5.1. A life insurance company offers a one-year death insurance policy with payouts of 1 and 2 units to individuals with a 0.02 or 0.01 probability of death. The table below shows the number of persons nk in each of the four classes formed in accordance with the payment b k and the likelihood of an insured event occurring qk:

The insurance company wants to collect from this group of 1,800 individuals an amount equal to the 95th percentile of the distribution of total insurance benefits for this group. In addition, she wants each person's share of this amount to be proportional to the person's expected insurance benefit.

The share of the person with number whose average payment is equal to should be . From the 95th percentile requirement it follows that . The amount of excess, , is a risk premium and is called a relative risk premium. Let's do the math.

Solution. The value is determined by the relation = 0.95, where S = X 1 + X 2 + ... + X 1800. This probability statement is equivalent to the following:

In accordance with what was said about the central limit theorem in Sect. 4, we approximate the r.v. distribution. standard normal distribution and use its 95th percentile, from which we get:

For the four classes into which policyholders are divided, we obtain the following results:

Thus,

Therefore, the relative risk premium is

Example 5.2. Clients of a car insurance company are divided into two classes:

The truncated exponential distribution is defined by the distribution function

This is a mixed type distribution with a density function , and a “clump” of probabilistic mass at the point L. The graph of this distribution function is shown in Fig. 5.1.

Rice. 5.1. Truncated exponential distribution

As before, the probability that the total amount of insurance payments exceeds the amount collected from policyholders should be equal to 0.05. We will assume that the relative risk premium should be the same in each of the two classes considered. Let's calculate.

Solution. This example is very similar to the previous one. The only difference is that the amounts of insurance payments are now random variables.

We first obtain expressions for the moments of the truncated exponential distribution. This will be a preparatory step for applying formulas (2.25) and (2.26):

Using the parameter values ​​given in the condition and applying formulas (2.25) and (2.26), we obtain the following results:

So, S, the total amount of insurance payments, has moments

The condition for the definition remains the same as in example 5.1, namely,

Using again the approximation of the normal distribution, we obtain

Example 5.3. The insurance company's portfolio includes 16,000 death insurance contracts for a period of one year according to the following table:

The probability of an insured event q for each of 16,000 clients (these events are assumed to be mutually independent) is 0.02. The company wants to establish its own retention rate. For each policyholder, the level of own retention is the value below which this company (ceding company) makes payments independently, and payments exceeding this value are covered under a reinsurance agreement by another company (reinsurer).

For example, if the deductible level is 200,000, then the company reserves coverage up to 20,000 for each policyholder and purchases reinsurance to cover the difference between the insurance claim and the amount of 20,000 for each of the 4,500 policyholders whose insurance benefits exceed the amount of 20,000 .

As a decision criterion, the company chooses to minimize the probability that insurance claims retained plus the amount paid for reinsurance will exceed the amount of 8,250,000. Reinsurance costs 0.025 per unit of coverage (i.e. 125% of expected the amount of insurance payments per unit is 0.02).

We believe that the portfolio under consideration is closed: new insurance contracts concluded during the current year will not be taken into account in the described decision-making process.

Partial solution. Let us first carry out all the calculations, choosing 10,000 as the unit of payment. For illustration, let us assume that c. V. S is the amount of payments left on own deduction, has the following form:

To these insurance payments left on own deduction, S, the amount of reinsurance premiums is added. In total, the total amount of coverage under this scheme is

The amount left on own deduction is equal to

Thus, the total reinsured value is 35,000-24,000=11,000 and the cost of reinsurance is

This means that with a deduction level equal to 2, the insurance payments left on deduction plus the cost of reinsurance amount to . The decision criterion is based on the probability that this total will exceed 825,

Using the normal distribution, we find that this value is approximately 0.0062.

The average values ​​of insurance payments for excess loss insurance, as one of the types of reinsurance, can be approximated using the normal distribution as the distribution of total insurance payments.

Let the total insurance payments X have a normal distribution with mean and variance

Example 5.4. Let's consider an insurance portfolio, as in example 5.3. Let us find the mathematical expectation of the amount of insurance payments under an excess of loss insurance contract if

(a) there is no individual reinsurance and the unconditional franchise is set at 7,500,000

(b) own deductible is established in the amount of 20,000 for individual insurance contracts and the amount of the unconditional deductible for the portfolio is 5,300,000.

Solution.

(a) In the absence of individual reinsurance and the transition to 10,000 as a monetary unit

application of formula (5.2) gives

which amounts to 43,770 in original units.

(b) In Example 5.3, we obtained the mean and variance of total premiums at an individual deductible level of 20,000 to be 480 and 784, respectively, using 10,000 as a unit. Thus =28.

application of formula (5.2) gives

which amounts to 4140 in original units.

Let us use the general method outlined above to solve one problem, namely, to find the law of distribution of the sum of two random variables. There is a system of two random variables (X,Y) with distribution density f(x,y).

Let's consider the sum of random variables X and Y: and find the law of distribution of the value Z. To do this, let's build a line on the xOy plane, the equation of which is (Fig. 6.3.1). This is a straight line cutting off segments equal to z on the axes. Straight divides the xOy plane into two parts; to the right and above her ; to the left and below

Region D in this case is the lower left part of the xOy plane, shaded in Fig. 6.3.1. According to formula (6.3.2) we have:

This is the general formula for the density distribution of the sum of two random variables.

For reasons of symmetry of the problem with respect to X and Y, we can write another version of the same formula:

It is required to produce a composition of these laws, that is, to find the law of distribution of the quantity: .

Let us apply the general formula for the composition of distribution laws:

Substituting these expressions into the formula we have already encountered

and this is nothing more than a normal law with a center of dispersion

The same conclusion can be reached much more easily using the following qualitative reasoning.

Without opening the parentheses and without making any transformations in the integrand (6.3.3), we immediately come to the conclusion that the exponent is a square trinomial with respect to x of the form

where the value z is not included in coefficient A at all, in coefficient B it is included to the first power, and in coefficient C it is squared. Keeping this in mind and applying formula (6.3.4), we come to the conclusion that g(z) is an exponential function, the exponent of which is a square trinomial with respect to z, and the distribution density; This type corresponds to the normal law. Thus, we; we come to a purely qualitative conclusion: the law of distribution of the value z must be normal. To find the parameters of this law - and - we will use the theorem of addition of mathematical expectations and the theorem of addition of variances. According to the theorem of addition of mathematical expectations . By the theorem of addition of variances or whence formula (6.3.7) follows.

Moving from standard deviations to probable deviations proportional to them, we obtain:
.

Thus, we came to the following rule: when combining normal laws, a normal law is obtained again, and the mathematical expectations and variances (or squares of probable deviations) are summed up.

The rule for the composition of normal laws can be generalized to the case of an arbitrary number of independent random variables.

If there are n independent random variables: subject to normal laws with centers of dispersion and standard deviations, then the value is also subject to the normal law with parameters

If a system of random variables (X, Y) is distributed according to a normal law, but the values ​​X, Y are dependent, then it is not difficult to prove, just as before, based on the general formula (6.3.1), that the law of distribution of a value is also a normal law. The scattering centers are still added algebraically, but for standard deviations the rule becomes more complex: , where r is the correlation coefficient of the X and Y values.

When adding several dependent random variables, which in their entirety are subject to the normal law, the distribution law of the sum also turns out to be normal with the parameters

where is the correlation coefficient of the quantities X i, X j, and the summation extends to all different pairwise combinations of quantities.

We have become convinced of a very important property of the normal law: with the composition of normal laws, a normal law is again obtained. This is the so-called “stability property”. A distribution law is called stable if the composition of two laws of this type again results in a law of the same type. We showed above that the normal law is stable. Very few distribution laws have the property of stability. The law of uniform density is unstable: by combining two laws of uniform density in sections from 0 to 1, we obtained Simpson's law.

The stability of the normal law is one of the essential conditions for its widespread use in practice. However, in addition to the normal one, some other distribution laws also have the property of stability. A feature of the normal law is that when a sufficiently large number of practically arbitrary distribution laws are composed, the total law turns out to be as close to normal as desired, regardless of what the distribution laws of the terms were. This can be illustrated, for example, by composing the three laws of uniform density in areas from 0 to 1. The resulting distribution law g(z) is shown in Fig. 6.3.1. As can be seen from the drawing, the graph of the function g(z) is very similar to the graph of the normal law.

In practice, there is often a need to find the law of distribution of a sum of random variables.

Let there be a system (Хь Х 2) two continuous s. V. and their sum

Let's find the distribution density c. V. U. In accordance with the general solution of the previous paragraph, we find the region of the plane where x+ x 2 (Fig. 9.4.1):

Differentiating this expression with respect to y, we obtain p.r. random variable Y = X + X 2:

Since the function φ (x b x 2) = Xj + x 2 is symmetric with respect to its arguments, then

If with. V. X And X 2 are independent, then formulas (9.4.2) and (9.4.3) will take the form:

In the case when independent s. V. X x And X 2, talk about the composition of distribution laws. Produce composition two laws of distribution - this means finding the law of distribution of the sum of two independent s. c., distributed according to these laws. To denote the composition of distribution laws, the symbolic notation is used

which essentially denotes formulas (9.4.4) or (9.4.5).

Example 1. The operation of two technical devices (TD) is considered. At first, the TU works; after its failure (failure), it is included in the operation of TU 2. Failure-free operation times TU L TU 2 - X x And X 2 - are independent and distributed according to exponential laws with parameters A,1 and X 2. Therefore, time Y trouble-free operation of a technical device consisting of technical equipment! and TU 2, will be determined by the formula

It is required to find p.r. random variable Y, i.e. a composition of two demonstrative laws with parameters and X 2.

Solution. Using formula (9.4.4) we obtain (y > 0)

If there is a composition of two exponential laws with the same parameters (?ts = X 2 = Y), then in expression (9.4.8) we obtain an uncertainty of type 0/0, revealing which we obtain:

Comparing this expression with expression (6.4.8), we are convinced that the composition of two identical exponential laws (?ts = X 2 = X) represents Erlang's second-order law (9.4.9). When combining two exponential laws with various parameters X x and A-2 receive generalized Erlang's second order law (9.4.8). ?

Problem 1. The law of distribution of the difference of two s. V. System s. V. (X and X 2) has a joint p.r./(x b x 2). Find p.r. their differences Y= X - X 2.

Solution. For system with. V. (X b - X 2) etc. will be/(x b - x 2), i.e. we replaced the difference with the sum. Therefore, p.r. random variable Will have the form (see (9.4.2), (9.4.3)):

If With. V. X x iX 2 are independent, then

Example 2. Find p.r. the difference between two independent exponentially distributed s. V. with parameters X x And X 2.

Solution. Using formula (9.4.11) we obtain

Rice. 9.4.2 Rice. 9.4.3

Figure 9.4.2 shows a p.r. g(y). If we consider the difference of two independent exponentially distributed s. V. with the same parameters (A-i= X 2 = A,), That g(y) = /2 - already familiar

Laplace's law (Fig. 9.4.3). ?

Example 3. Find the law of distribution of the sum of two independent s. V. X And X 2, distributed according to Poisson's law with parameters a x And a 2.

Solution. Let's find the probability of the event (X x + X 2 = t) (t = 0, 1,

Therefore, s. V. Y= X x + X 2 distributed according to Poisson's law with parameter a x2) - a x + a 2. ?

Example 4. Find the law of distribution of the sum of two independent s. V. X x And X 2, distributed according to binomial laws with parameters p x ri p 2, p respectively.

Solution. Let's imagine s. V. X x as:

Where X 1) - event indicator A Wu's experience:

Distribution series p. V. X,- has the form

We will make a similar representation for s. V. X 2: where X] 2) - event indicator A in y"-th experience:

Hence,

where is X? 1)+(2) if event indicator A:

Thus, we have shown that s. V. Test the amount (u + n 2) event indicators A, from which it follows that s. V. ^distributed according to the binomial law with parameters ( p x + p 2), r.

Note that if the probabilities R are different in different series of experiments, then as a result of the addition of two independent s. in., distributed according to binomial laws, it turns out c. c., distributed not according to the binomial law. ?

Examples 3 and 4 are easily generalized to an arbitrary number of terms. When combining Poisson's laws with parameters a b a 2, ..., a t again we get Poisson's law with the parameter a (t) = a x + a 2 + ... + and t.

When composing binomial laws with parameters (p p p); (i 2, R) , (p t, p) again we get a binomial law with parameters (“(“), R), Where n (t) = n + n 2 + ... + p t.

We have proven important properties of Poisson's law and the binomial law: the “stability property”. The distribution law is called sustainable, if the composition of two laws of the same type results in a law of the same type (only the parameters of this law differ). In Subsection 9.7 we will show that the normal law has the same property of stability.