The law of distribution of the sum of two random variables. Composition of distribution laws

In practice, it often becomes necessary to find the distribution law for the sum of random variables.

Let there be a system (X b X 2) two continuous s. in. and their sum

Let us find the distribution density c. in. U. In accordance with common solution of the previous paragraph, we find the area of ​​the plane where x + x 2 (Fig. 9.4.1):

Differentiating this expression with respect to y, we obtain an ap. random variable Y \u003d X + X 2:

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

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

In the case when independent c. in. x x and X 2, talk about the composition of distribution laws. Produce composition two distribution laws - this means finding the distribution law for the sum of two independent c. c., distributed according to these laws. The symbolic notation is used to designate the composition of distribution laws

which is essentially denoted by formulas (9.4.4) or (9.4.5).

Example 1. We consider the work of two technical devices(THAT). First, TU works after its failure (failure) is included in the operation of TU 2. Uptime TU TU TU 2 - x x and X 2 - are independent and distributed according to exponential laws with parameters A,1 and X 2 . Therefore, the time Y trouble-free operation of the TU, consisting of TU! and TU 2 will be determined by the formula

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

Solution. By formula (9.4.4) we get (y > 0)

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

Comparing this expression with expression (6.4.8), we are convinced that the composition of two identical exponential laws (?c = X 2 = x) is the second-order Erlang law (9.4.9). When composing two exponential laws with different parameters x x and A-2 get second-order generalized Erlang law (9.4.8). ?

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

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

If a With. in. X x iX 2 independent, then

Example 2. Find a f.r. the difference of two independent exponentially distributed s. in. with parameters x x and X 2 .

Solution. According to the formula (9.4.11) we get

Rice. 9.4.2 Rice. 9.4.3

Figure 9.4.2 shows a p. g(y). If we consider the difference of two independent exponentially distributed s. in. with the same settings (A-i= X 2 = BUT,), then g(y) \u003d / 2 - already familiar

Laplace's law (Fig. 9.4.3). ?

Example 3. Find the distribution law for the sum of two independent c. in. X and X 2, distributed according to the Poisson law with parameters a x and a 2 .

Solution. Find the probability of an event (X x + X 2 = t) (t = 0, 1,

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

Example 4. Find the distribution law for the sum of two independent c. in. x x and X 2, distributed according to binomial laws with parameters p x ri p 2 , p respectively.

Solution. Imagine with. in. x x as:

where X 1) - event indicator BUT wu "th experience:

Distribution range with. in. X,- has the form

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

Consequently,

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

Thus, we have shown that in. Father-in-law amount (u + n 2) event indicators BUT, whence it follows that s. in. ^distributed according to the binomial law with parameters ( n x + n 2), p.

Note that if the probabilities R in different series of experiments are different, then as a result of adding two independent s. c., 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 composing Poisson's laws with parameters a b a 2 , ..., a t Poisson's law is again obtained with the parameter a (t) \u003d a x + a 2 + ... + and t.

When composing binomial laws with parameters (n r); (i 2 , R) , (n t, p) again we get the binomial law with parameters (“(“), R), where n (t) \u003d u + n 2 + ... + etc.

We have proved 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 stability property.

An extremely important object of probability theory is the sum of independent random variables. It is the study of the distribution of sums of independent random variables that laid the foundation for the development of analytical methods of probability theory.

Distribution of the sum of independent random variables

In this section, we will obtain a general formula that allows us to calculate the distribution function of the sum of independent random variables, and consider several examples.

Distribution of the sum of two independent random variables. Convolution formula

independent random variables with distribution functions

respectively

Then the distribution function F sums of random variables

can be calculated using the following formula ( convolution formula)

To prove this, we use Fubini's theorem.

The second part of the formula is proved similarly.

Distribution density of the sum of two independent random variables

If the distributions of both random variables have densities, then the density of the sum of these random variables can be calculated by the formula

If the distribution of a random variable (or ) has a density, then the density of the sum of these random variables can be calculated by the formula

To prove these assertions, it suffices to use the definition of density.

Multiple convolutions

The calculation of the sum of a finite number of independent random variables is performed using the sequential application of the convolution formula. Sum distribution function k independent identically distributed random variables with a distribution function F

called k–fold convolution of the distribution function F and denoted

Examples of calculating the distribution of sums of independent random variables

In this paragraph, examples of situations are given, when summing random variables, the form of the distribution is preserved. The proofs are exercises on summation and calculation of integrals.

Sums of independent random variables. Normal distribution

Sums of independent random variables. Binomial distribution

Sums of independent random variables. Poisson distribution

Sums of independent random variables. Gamma distribution

Poisson process

a sequence of independent identically distributed random variables having an exponential distribution with parameter

Random sequence of points

on the non-negative semi-axis is called Poisson (point) process.

Let us calculate the distribution of the number of points

Poisson process in the interval (0,t)

equivalents, so

But the distribution of the random variable

is an Erlang distribution of order k, so

Thus, the distribution of the number of points of the Poisson process in the interval (o,t) is a Poisson distribution with the parameter

The Poisson process is used to simulate the moments of occurrence of random events - the process of radioactive decay, the moments of calls to the telephone exchange, the moments of the appearance of customers in the service system, the moments of equipment failure.

DISTRIBUTION FUNCTIONS

Let x(k) is some random variable. Then for any fixed value x a random event x(k) x defined as the set of all possible outcomes k such that x(k) x. In terms of the original probability measure given on the sample space, distribution functionP(x) defined as the probability assigned to a set of points k x(k) x. Note that the set of points k satisfying the inequality x(k) x, is a subset of the set of points that satisfy the inequality x(k) . Formally

It's obvious that

If the range of values ​​of the random variable is continuous, which is assumed below, then probability density(one-dimensional) p(x) is determined by the differential relation

(4)

Consequently,

(6)

In order to be able to consider discrete cases, it is necessary to admit the presence of delta functions in the composition of the probability density.

EXPECTED VALUE

Let the random variable x(k) takes values ​​from the range from -  to + . Mean(otherwise, expected value or expected value) x(k) is calculated using the corresponding passage to the limit in the sum of products of values x(k) on the probability of these events occurring:

(8)

where E- mathematical expectation of the expression in square brackets by index k. The mathematical expectation of a real single-valued continuous function is defined similarly g(x) from a random variable x(k)

(9)

where p(x)- probability density of a random variable x(k). In particular, taking g(x)=x, we get mean square x(k) :

(10)

Dispersionx(k) defined as the mean square of the difference x(k) and its average value,

i.e. in this case g(x)= and

By definition, standard deviation random variable x(k), denoted , there is a positive value square root from dispersion. The standard deviation is measured in the same units as the mean.

MOST IMPORTANT DISTRIBUTION FUNCTIONS

UNIFORM (RECTANGULAR) DISTRIBUTION.

Let us assume that the experiment consists in a random selection of a point from the interval [ a,b] , including its endpoints. In this example, as the value of a random variable x(k) you can take the numeric value of the selected point. The corresponding distribution function has the form

Therefore, the probability density is given by the formula

In this example, the calculation of the mean and variance using formulas (9) and (11) gives

NORMAL (GAUSSIAN) DISTRIBUTION

, - arithmetic mean, - RMS.

The value of z corresponding to the probability P(z)=1-, i.e.

CHI - SQUARE DISTRIBUTION

Let - n independent random variables, each of which has a normal distribution with zero mean and unit variance.

Chi-squared random variable with n degrees of freedom.

probability density .

DF: 100 - percentage points - distributions are denoted by , i.e.

mean and variance are equal

t - STUDENT DISTRIBUTIONS

y, z are independent random variables; y - has - distribution, z - normally distributed with zero mean and unit variance.

value - has t- Student's distribution with n degrees of freedom

DF: 100 - percentage point t - distribution is indicated

Mean and variance are equal

F - DISTRIBUTION

Independent random variables; has - distribution with degrees of freedom; distribution with degrees of freedom. Random value:

,

F is a distributed random variable with and degrees of freedom.

,

DF: 100 - percentage point:

The mean and variance are equal:

DISTRIBUTION OF THE AMOUNT

TWO RANDOM VARIABLES

Let x(k) and y(k) are random variables having a joint probability density p(x,y). Find the probability density of the sum of random variables

At a fixed x we have y=z–x. That's why

At a fixed z values x run the interval from – to +. That's why

(37)

whence it can be seen that in order to calculate the desired density of the sum, one must know the original joint probability density. If a x(k) and y(k) are independent random variables having densities and, respectively, then and

(38)

EXAMPLE: THE SUM OF TWO INDEPENDENT, UNIFORMLY DISTRIBUTED RANDOM VARIABLES.

Let two random independent variables have densities of the form

In other cases Let us find the probability density p(z) of their sum z= x+ y.

Probability Density for i.e. for Consequently, x less than z. In addition, is not equal to zero for By formula (38), we find that

Illustration:

The probability density of the sum of two independent, uniformly distributed random variables.

RANDOM CONVERSION

VALUES

Let x(t)- random variable with probability density p(x), let it go g(x) is a single-valued real continuous function of x. Consider first the case when the inverse function x(g) is also a single-valued continuous function of g. Probability Density p(g), corresponding to a random variable g(x(k)) = g(k), can be determined from the probability density p(x) random variable x(k) and derivative dg/dx under the assumption that the derivative exists and is different from zero, namely:

(12)

Therefore, in the limit dg/dx#0

(13)

Using this formula, follows on its right side instead of a variable x substitute the appropriate value g.

Consider now the case when the inverse function x(g) is valid n-valued function of g, where n is an integer and all n values ​​are equally probable. Then

(14)

EXAMPLE:

DISTRIBUTION OF THE HARMONIC FUNCTION.

Harmonic function with fixed amplitude X and frequency f will be a random variable if its initial phase angle  =  (k)- random value. In particular, let t fixed and equal t o, and let the harmonic random variable have the form

Let's pretend that  (k) has a uniform probability density p( ) kind

Find the probability density p(x) random variable x(k).

In this example, the direct function x( ) unambiguously, and the inverse function  (x) ambiguous.

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). Consider the sum of random variables X and Y: and find the law of distribution of the value Z. To do this, we construct a line on the xOy plane, the equation of which is (Fig. 7). This is a straight line that cuts off segments equal to z on the axes. The straight line divides the xy plane into two parts; to the right and above it; left and below.

Area D in this case-- 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 inner integral, we get:

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. Consider two independent random variables X and Y, subject to normal laws:

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

We apply the general formula for the composition of distribution laws:

If we open the brackets in the exponent of the integrand and bring like terms, we get:

Substituting these expressions into the formula we have already encountered

after transformations we get:

and this is nothing but a normal law with a dispersion center

and standard deviation

The same conclusion can be reached much more easily with the help of the following qualitative reasoning.

Without opening the brackets and without performing transformations in the integrand (17), we immediately come to the conclusion that the exponent is a square trinomial with respect to x of the form

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

Passing from root-mean-square deviations to probable deviations proportional to them, we will receive: .

Thus, we have come to the following rule: when normal laws are composed, a normal law is again obtained, and the mathematical expectations and variances (or squared probable deviations) are summed up.

The composition rule for 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 scattering centers and standard deviations, then the value is also subject to the normal law with parameters

Instead of formula (22), an equivalent formula can be used:

If the system of random variables (X, Y) is distributed according to the normal law, but the quantities X, Y are dependent, then it is easy to prove, just as before, based on the general formula (6.3.1), that the distribution law of the quantity is also a normal law. Scattering centers are still added algebraically, but for standard deviations the rule becomes more complicated: , where, r is the correlation coefficient of X and Y values.

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

or probable deviations

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 seen a very important property of the normal law: when normal laws are combined, one again obtains a normal law. This is the so-called "stability property". A distribution law is said to be stable if, by composing two laws of this type, a law of the same type is again obtained. We have shown above that the normal law is stable. Very few distribution laws have the property of stability. The law of uniform density is unstable: when composing two laws of uniform density in sections from 0 to 1, we obtained Simpson's law.

The stability of a normal law is one of essential conditions its widespread use in practice. However, the property of stability, in addition to the normal one, is also possessed by some other distribution laws. A feature of the normal law is that, in composition, it is sufficient a large number practically arbitrary distribution laws, the total law turns out to be arbitrarily close to normal, regardless of what the distribution laws of the terms were. This can be illustrated, for example, by composing the composition of three laws of uniform density in sections 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.