Estimation of mathematical expectation and variance for the sample. Point estimates of mathematical expectation and variance

Let the random sample be generated by the observed random variable ξ, the mathematical expectation and variance which are unknown. As estimates for these characteristics, it was proposed to use the sample mean

and sample variance

. (3.14)

Consider some properties of estimates mathematical expectation and dispersion.

1. Calculate the mathematical expectation of the sample mean:

Therefore, the sample mean is an unbiased estimator for .

2. Recall that the results observations - independent random variables, each of which has the same distribution law as the value , which means that , , . We will assume that the variance is finite. Then, according to the Chebyshev theorem on the law big numbers, for any ε > 0 we have ,

which can be written like this: . (3.16) Comparing (3.16) with the definition of the consistency property (3.11), we see that the estimate is a consistent estimate of the expectation .

3. Find the variance of the sample mean:

. (3.17)

Thus, the variance of the expectation estimate decreases inversely with the sample size.

It can be proved that if the random variable ξ is normally distributed, then the sample mean is an effective estimate of the mathematical expectation, i.e. the variance takes smallest value compared to any other estimate of the mathematical expectation. For other distribution laws of ξ, this may not be the case.

The sample variance is a biased estimate of the variance, since . (3.18)

Indeed, using the properties of the mathematical expectation and formula (3.17), we find

.

To obtain an unbiased estimate of the variance, estimate (3.14) must be corrected, that is, multiplied by . Then we get the unbiased sample variance

. (3.19)

We note that formulas (3.14) and (3.19) differ only in the denominator, and for large values ​​the sample and unbiased variances differ little. However, for a small sample size, relation (3.19) should be used.

To estimate the standard deviation of a random variable, the so-called “corrected” mean is used standard deviation, which is equal to square root from unbiased variance: .

Interval Estimates

In statistics, there are two approaches to estimating unknown parameters of distributions: point and interval. In accordance with point estimation, which was considered in the previous section, only the point near which the estimated parameter is located is indicated. However, it is desirable to know how far this parameter can actually stand from the possible implementation of estimates in different series of observations.

The answer to this question - also approximate - gives another way of estimating the parameters - interval. In accordance with this estimation method, an interval is found that, with a probability close to one, covers an unknown numerical value of the parameter.

The concept of interval estimation

Point Estimation is a random variable and for possible implementations of the sample takes values ​​only approximately equal to the true value of the parameter . The smaller the difference, the more accurate the estimate. In this way, positive number, for which , characterizes the accuracy of the estimate and is called estimation error (or marginal error).

Confidence Probability(or reliability) is called probability β , with which the inequality , i.e.

. (3.20)

Replacing the inequality its equivalent double inequality , or , we get

Interval covering with probability β , , unknown parameter , is called confidence interval (or interval estimation), corresponding to the confidence level β .

A random variable is not only an estimate, but also an error: its value depends on the probability β and, as a rule, from the sample. Therefore, the confidence interval is random and expression (3.21) should be read as follows: “The interval will cover the parameter with the probability β ”, and not like this: “The parameter will fall into the interval with a probability β ”.

Meaning confidence interval consists in the fact that with repeated repetition of the sample volume in the relative proportion of cases equal to β , confidence interval corresponding to the confidence level β , covers the true value of the estimated parameter. So the confidence level β characterizes reliability confidence assessment: the more β , the more likely that the implementation of the confidence interval contains an unknown parameter.

PURPOSE OF THE LECTURE: to introduce the concept of estimating an unknown distribution parameter and give a classification of such estimators; get point and interval estimates of the mathematical expectation and variance.

In practice, in most cases, the law of distribution of a random variable is unknown, and according to the results of observations
it is necessary to evaluate numerical characteristics (for example, mathematical expectation, variance or other moments) or an unknown parameter , which defines the distribution law (distribution density)
random variable under study. So, for an exponential or Poisson distribution, it is enough to evaluate one parameter, and for a normal distribution, two parameters are already to be evaluated - the mathematical expectation and variance.

Types of assessments

Random value
has a probability density
, where is an unknown distribution parameter. As a result of the experiment, the values ​​of this random variable were obtained:
. To make an assessment in essence means that the sample values ​​of a random variable must be associated with a certain value of the parameter , i.e. create some function of the results of observations
, the value of which is taken as an estimate parameter . Index indicates the number of experiments performed.

Any function that depends on the results of observations is called statistics. Since the results of observations are random variables, then the statistics will also be a random variable. Therefore, the estimate
unknown parameter should be considered as a random variable, and its value calculated from experimental given volume, – as one of the possible values ​​of this random variable.

Estimates of distribution parameters (numerical characteristics of a random variable) are divided into point and interval. Point Estimation parameter determined by one number , and its accuracy is characterized by the variance of the estimate. interval estimation called an estimate, which is determined by two numbers, And – by the ends of the interval covering the estimated parameter with a given confidence level.

Classification of point estimates

To make a point estimate of an unknown parameter
is the best in terms of accuracy, it needs to be consistent, unbiased, and efficient.

Wealthy called score
parameter , if it converges in probability to the estimated parameter, i.e.

. (8.8)

Based on the Chebyshev inequality, it can be shown that sufficient condition relation (8.8) is the equality

.

Consistency is an asymptotic characteristic of the estimate for
.

unbiased called score
(estimate without systematic error), the mathematical expectation of which is equal to the estimated parameter, i.e.

. (8.9)

If equality (8.9) is not satisfied, then the estimate is called biased. Difference
called the bias or bias of the estimate. If equality (8.9) is satisfied only for
, then the corresponding estimate is called asymptotically unbiased.

It should be noted that if consistency is an almost obligatory condition for all estimates used in practice (inconsistent estimates are used extremely rarely), then the property of unbiasedness is only desirable. Many commonly used estimators do not have the unbiased property.

In the general case, the accuracy of estimating a certain parameter obtained on the basis of experimental data
, is characterized by the mean square error

,

which can be brought to the form

,

where is the dispersion,
is the square of the estimation bias.

If the estimate is unbiased, then

At final estimates may differ by the mean square of the error . Naturally, the smaller this error, the more closely the evaluation values ​​are grouped around the estimated parameter. Therefore, it is always desirable that the estimation error be as small as possible, i.e., the condition

. (8.10)

Estimate satisfying condition (8.10) is called an estimate with a minimum squared error.

efficient called score
, for which the mean squared error is not greater than the mean squared error of any other estimate, i.e.

where – any other parameter estimate .

It is known that the variance of any unbiased estimate of one parameter satisfies the Cramer–Rao inequality

,

where
– conditional probability distribution density of the obtained values ​​of a random variable with the true value of the parameter .

So the unbiased estimator
, for which the Cramer-Rao inequality becomes an equality, will be effective, i.e., such an estimate has a minimum variance.

Point estimates mathematical expectation and variance

If we consider a random variable
, which has mathematical expectation and dispersion , both of these parameters are assumed to be unknown. Therefore, over a random variable
produced independent experiments that give results:
. It is necessary to find consistent and unbiased estimates of unknown parameters And .

As estimates And usually, the statistical (sample) mean and statistical (sample) variance are chosen respectively:

; (8.11)

. (8.12)

The expectation estimate (8.11) is consistent according to the law of large numbers (Chebyshev's theorem):

.

Mathematical expectation of a random variable

.

Therefore, the estimate is unbiased.

The dispersion of the estimate of the mathematical expectation:

If the random variable
distributed according to the normal law, then the estimate is also effective.

Mathematical expectation of the variance estimate

In the same time

.

Because
, but
, then we get

. (8.13)

In this way,
is a biased estimate, although it is consistent and efficient.

It follows from formula (8.13) that in order to obtain an unbiased estimate
the sample variance (8.12) should be modified as follows:

which is considered "better" than estimate (8.12), although for large these estimates are almost equal to each other.

Methods for obtaining estimates of distribution parameters

Often in practice, based on the analysis of the physical mechanism that generates a random variable
, we can conclude about the law of distribution of this random variable. However, the parameters of this distribution are unknown, and they must be estimated from the results of the experiment, usually presented as a finite sample.
. To solve such a problem, two methods are most often used: the method of moments and the maximum likelihood method.

Method of moments. The method consists in equating the theoretical moments with the corresponding empirical moments of the same order.

Empirical initial moments th order are determined by the formulas:

,

and the corresponding theoretical initial moments th order - formulas:

for discrete random variables,

for continuous random variables,

where is the estimated distribution parameter.

To obtain estimates of the parameters of a distribution containing two unknown parameters And , the system is composed of two equations

where And are the theoretical and empirical central moments of the second order.

The solution of the system of equations is the estimates And unknown distribution parameters And .

Equating the theoretical empirical initial moments of the first order, we obtain that by estimating the mathematical expectation of a random variable
, which has an arbitrary distribution, will be the sample mean, i.e.
. Then, equating the theoretical and empirical central moments of the second order, we obtain that the estimate of the variance of the random variable
, which has an arbitrary distribution, is determined by the formula

.

In a similar way one can find estimates of theoretical moments of any order.

The method of moments is simple and does not require complex calculations, but the estimates obtained by this method are often inefficient.

Maximum likelihood method. The maximum likelihood method of point estimation of unknown distribution parameters is reduced to finding the maximum function of one or more estimated parameters.

Let be
is a continuous random variable, which as a result tests took the values
. To get an estimate of an unknown parameter need to find the value , at which the probability of realization of the obtained sample would be maximum. Because
are mutually independent quantities with the same probability density
, then likelihood function call the argument function :

The maximum likelihood estimate of the parameter this value is called , at which the likelihood function reaches its maximum, i.e., is a solution to the equation

,

which obviously depends on the test results
.

Since the functions
And
reach a maximum at the same values
, then often, to simplify the calculations, they use the logarithmic likelihood function and look for the root of the corresponding equation

,

which is called likelihood equation.

If you need to evaluate several parameters
distribution
, then the likelihood function will depend on these parameters. To find estimates
distribution parameters, it is necessary to solve the system likelihood equations

.

The maximum likelihood method gives consistent and asymptotically efficient estimates. However, the estimates obtained by the maximum likelihood method are sometimes biased, and, in addition, to find the estimates, one often has to solve rather complex systems of equations.

Interval parameter estimates

The accuracy of point estimates is characterized by their dispersion. At the same time, there is no information about how close the obtained estimates are to the true values ​​of the parameters. In a number of tasks, it is required not only to find for the parameter suitable numerical value, but also evaluate its accuracy and reliability. It is necessary to find out what errors the parameter replacement can lead to. its point estimate and with what degree of confidence can we expect that these errors will not go beyond known limits.

Such problems are especially relevant for a small number of experiments. when the point estimate largely random and approximate substitution on the can lead to significant errors.

A more complete and reliable way to estimate the parameters of distributions is to determine not a single point value, but an interval that, with a given probability, covers the true value of the estimated parameter.

Let the results experiments, an unbiased estimate is obtained
parameter . It is necessary to evaluate the possible error. Some sufficiently large probability is chosen
(for example), such that an event with this probability can be considered a practically certain event, and such a value is found , for which

. (8.15)

In this case, the range of practically possible values ​​of the error that occurs when replacing on the , will
, and large absolute errors will appear only with a small probability .

Expression (8.15) means that with probability
unknown parameter value falls into the interval

. (8.16)

Probability
called confidence level, and the interval covering with probability the true value of the parameter is called confidence interval. Note that it is incorrect to say that the parameter value lies within the confidence interval with the probability . The wording used (covers) means that although the estimated parameter is unknown, it has a constant value and therefore does not have a spread, since it is not a random variable.

Let there be a random variable X, and its parameters are the mathematical expectation but and variance are unknown. Over the value of X, independent experiments were carried out, which gave the results x 1, x 2, x n.

Without diminishing the generality of the reasoning, we will consider these values ​​of the random variable to be different. We will consider the values ​​x 1, x 2, x n as independent, identically distributed random variables X 1, X 2, X n .

The simplest method statistical estimation - the method of substitution and analogy - consists in the fact that as an assessment of a particular numerical characteristic (average, variance, etc.) population take the corresponding characteristic of the distribution of the sample - the sample characteristic.

By the substitution method as an estimate of the mathematical expectation but it is necessary to take the mathematical expectation of the distribution of the sample - the sample mean. Thus, we get

To test the unbiasedness and consistency of the sample mean as estimates but, consider this statistic as a function of the chosen vector (X 1, X 2, X n). Taking into account that each of the values ​​X 1, X 2, X n has the same distribution law as the value X, we conclude that the numerical characteristics of these quantities and the value of X are the same: M(X i) = M(X) = a, D(X i) = D(X) = , i = 1, 2, n , where X i are collectively independent random variables.

Consequently,

Hence, by definition, we obtain that is the unbiased estimate but, and since D()®0 as n®¥, then by virtue of the theorem of the previous paragraph is a consistent estimate of the expectation but the general population.

The efficiency or inefficiency of the estimate depends on the form of the distribution law of the random variable X. It can be proved that if the value X is distributed according to the normal law, then the estimate is efficient. For other distribution laws, this may not be the case.

Unbiased estimate of the general variance is the corrected sample variance

,

Because , where is the general variance. Really,

The estimate s -- 2 for the general variance is also consistent, but not efficient. However, in the case of a normal distribution, it is “asymptotically efficient,” that is, as n increases, the ratio of its variance to the minimum possible one approaches indefinitely.

So, given a sample from the distribution F( x) random variable X with unknown mathematical expectation but and dispersion , then to calculate the values ​​of these parameters, we have the right to use the following approximate formulas:

a ,

.

Here x-i- - sampling options, n- i - - frequency options x i , - - sample size.
To calculate the corrected sample variance, the formula is more convenient

.

To simplify the calculation, it is advisable to switch to conditional options (it is advantageous to take the initial variant located in the middle of the interval as c variation series). Then

, .

interval estimation

Above, we considered the question of estimating an unknown parameter but one number. We called such estimates point estimates. They have the disadvantage that, with a small sample size, they can differ significantly from the estimated parameters. Therefore, to get an idea of ​​the closeness between a parameter and its estimate, in mathematical statistics so-called interval estimates are introduced.

Let a point estimate q * be found in the sample for the parameter q. Usually, researchers preassign some sufficiently large probability g (for example, 0.95; 0.99 or 0.999) such that an event with probability g can be considered practically certain, and raise the question of finding such a value e > 0 for which

.

Modifying this equality, we get:

and in this case we will say that the interval ]q * - e; q * + e[ covers the estimated parameter q with probability g.

Interval ]q * -e; q * +e [ is called confidence interval .

The probability g is called reliability (confidence probability) interval estimate.

The ends of the confidence interval, i.e. points q * -e and q * +e are called trust boundaries .

The number e is called assessment accuracy .

As an example of the problem of determining confidence limits, consider the question of estimating the mathematical expectation of a random variable X, which has a normal distribution law with parameters but and s, i.e. X = N( a, s). The mathematical expectation in this case is equal to but. According to the observations X 1 , X 2 , X n calculate the average and evaluation dispersion s 2 .

It turns out that according to the sample data, it is possible to construct a random variable

which has a Student's distribution (or t-distribution) with n = n -1 degrees of freedom.

Let's use Table A.1.3 and find for the given probability g and the number n the number t g such that the probability

P(|t(n)|< t g) = g,

.

After making obvious transformations, we get

The procedure for applying the F-criterion is as follows:

1. An assumption is made about the normal distribution of populations. At a given significance level a, the null hypothesis H 0 is formulated: s x 2 = s y 2 about the equality of the general variances of normal populations under the competing hypothesis H 1: s x 2 > s y 2 .

2. Two independent samples are obtained from the X and Y populations of n x and n y respectively.

3. Calculate the values ​​of the corrected sample variances s x 2 and s y 2 (calculation methods are discussed in §13.4). The larger of the dispersions (s x 2 or s y 2) is designated s 1 2, the smaller - s 2 2.

4. The value of the F-criterion is calculated according to the formula F obs = s 1 2 / s 2 2 .

5. According to the table of critical points of the Fisher - Snedecor distribution, for a given significance level a and the number of degrees of freedom n 1 \u003d n 1 - 1, n 2 \u003d n 2 - 1 (n 1 is the number of degrees of freedom of a larger corrected variance), the critical point is found F cr (a, n 1, n 2).

Note that Table A.1.7 shows critical values one-tailed F-test. Therefore, if a two-sided criterion is applied (H 1: s x 2 ¹ s y 2), then the right-sided critical point F cr (a/2, n 1 , n 2) are searched for by the level of significance a/2 (half the specified value) and the number of degrees of freedom n 1 and n 2 (n 1 is the number of degrees of freedom of greater dispersion). The left-handed critical point may not be found.

6. It is concluded that if the calculated value of the F-criterion is greater than or equal to the critical one (F obs ³ F cr), then the variances differ significantly at a given significance level. Otherwise (F obs< F кр) нет оснований для отклонения нулевой гипотезы о равенстве двух дисперсий.

Task 15.1. The consumption of raw materials per unit of production according to the old technology was:

New technology:

Assuming that the corresponding general populations X and Y have normal distributions, check that the consumption of raw materials for new and old technologies does not differ in variability, if we take the significance level a = 0.1.

Solution. We act in the order indicated above.

1. We will judge the variability of the consumption of raw materials for new and old technologies in terms of dispersion values. Thus, the null hypothesis has the form H 0: s x 2 = s y 2 . As a competing hypothesis, we accept the hypothesis H 1: s x 2 ¹ s y 2, since we are not sure in advance that any of the general variances is greater than the other.

2-3. Find the sample variances. To simplify the calculations, let's move on to conditional options:

u i = x i - 307, v i = y i - 304.

We will arrange all calculations in the form of the following tables:

Control: å m i u i 2 + 2å m i u i + m i = Control: å n i v i 2 + 2å n i v i + n i = 13 + 2 + 9 = 24 = 34 + 20 + 13 = 67

Find the corrected sample variances:

4. Compare the variances. Find the ratio of the larger corrected variance to the smaller one:

.

5. By condition, the competing hypothesis has the form s x 2 ¹ s y 2 , therefore, the critical region is two-sided, and when finding the critical point, one should take significance levels that are half the given one.

According to Table A.1.7, by the significance level a/2 = 0.1/2 = 0.05 and the number of degrees of freedom n 1 = n 1 - 1 = 12, n 2 = n 2 - 1 = 8, we find the critical point F cr ( 0.05; 12; 8) = 3.28.

6. Since F obl.< F кр то гипотезу о равенстве дисперсий расхода сырья при старой и new technologies accept.

Above, when testing hypotheses, it was assumed that the distribution of the random variables under study was normal. However, special studies have shown that the proposed algorithms are very stable (especially with large sample sizes) with respect to the deviation from the normal distribution.

The most important numerical characteristics random variable X are her mathematical expectation m x =M and dispersionσ 2 x = D[x] = M[(X – m x) 2 ] = M –. Number m x is the mean value of the random variable around which the values ​​of the quantities are scattered X, the measure of this spread is the dispersion D[x] And standard deviation:

s x =(1.11)

We will further consider an important problem for the study of an observed random variable. Let there be some sample (we will denote it S) random variable X. It is required to estimate unknown values ​​from the available sample m x And .

The theory of estimates of various parameters occupies a significant place in mathematical statistics. Therefore, let's first consider common task. Let it be required to estimate some parameter a by sample S. Each such evaluation a* is some function a*=a*(S) from the sample values. The sample values ​​are random, so the estimate itself a* is a random variable. You can build multiple various estimates(i.e. functions) a*, but at the same time it is desirable to have a “good” or even “best”, in some sense, assessment. Estimates are usually subject to the following three natural requirements.

1. Unbiased. Mathematical expectation of the estimate a* must equal the exact value of the parameter: M = a. In other words, the score a* should not have a systematic error.

2. Consistency. With an infinite increase in the sample size, the estimate a* should converge to the exact value, that is, as the number of observations increases, the estimation error tends to zero.

3. Efficiency. Grade a* is called efficient if it is unbiased and has the smallest possible error variance. In this case, the scatter of estimates is minimal. a* regarding the exact value and evaluation in in a certain sense is the "most accurate".

Unfortunately, it is not always possible to construct an estimate that satisfies all three requirements simultaneously.

To estimate the mathematical expectation, the estimate is most often used.

= , (1.12)

that is, the arithmetic mean of the sample. If the random variable X has finite m x And s x, then estimate (1.12) is unbiased and consistent. This estimate is effective, for example, if X has a normal distribution (Fig.p.1.4, Appendix 1). For other distributions, it may not be effective. For example, in the case uniform distribution(Fig.p.1.1, Appendix 1) an unbiased, consistent estimate will be

(1.13)

At the same time, estimate (1.13) for a normal distribution will be neither consistent nor efficient, and will even worsen with increasing sample size.

Thus, for each type of distribution of a random variable X you should use your estimate of the mathematical expectation. However, in our situation, the type of distribution can only be known hypothetically. Therefore, we will use estimate (1.12), which is rather simple and has the most important properties of unbiasedness and consistency.

To estimate the mathematical expectation for a grouped sample, the following formula is used:

= , (1.14)

which can be obtained from the previous one, if we consider all m i sample values ​​that fall into i-th interval equal to the representative z i this interval. This estimate is, of course, rougher, but requires much less computation, especially with a large sample size.

To estimate the variance, the most commonly used estimate is:

= , (1.15)

This estimate is not biased and is consistent for any random variable X, which has finite moments up to the fourth order inclusive.

In the case of a grouped sample, an estimate is used:

= (1.16)

Estimates (1.14) and (1.16) are, as a rule, biased and untenable, since their mathematical expectations and the limits to which they converge differ from m x and due to the replacement of all sample values ​​that fall into i-th interval, per interval representative z i.

Note that for large n, coefficient n/(n – 1) in expressions (1.15) and (1.16) is close to unity, so it can be omitted.

Interval estimates.

Let be exact value some parameter is equal to a and found its estimate a*(S) by sample S. Assess a* corresponds to a point on the numerical axis (Fig. 1.5), so this assessment is called point. All estimates considered in the previous section are point estimates. Almost always, by chance

a* ¹ a, and we can only hope that the point a* is somewhere near a. But how close? Any other point estimate will have the same drawback - the absence of a measure of the reliability of the result.

Fig.1.5. Point estimate of the parameter.

More specific in this respect are interval estimates. Interval score is an interval I b \u003d (a, b), in which the exact value of the estimated parameter is located with a given probability b. Interval Ib called confidence interval, and the probability b called confidence level and can be considered as reliability of the estimate.

The confidence interval will be based on the available sample S, it is random in the sense that its boundaries are random a(S) And b(S), which we will calculate from a (random) sample. That's why b there is a probability that the random interval Ib will cover a non-random point a. On fig. 1.6. interval Ib covered the point a, but Ib*- No. Therefore, it is not entirely correct to say that a" falls within the interval.

If the confidence level b large (eg. b = 0.999), then almost always the exact value a is in the constructed interval.

Fig.1.6. Parameter Confidence Intervals a for different samples.

Consider a method for constructing a confidence interval for the mathematical expectation of a random variable X, based on central limit theorem.

Let the random variable X has an unknown mathematical expectation m x And known variance. Then, by virtue of the central limit theorem, the arithmetic mean is:

= , (1.17)

results n independent tests of magnitude X is a random variable whose distribution for large n, close to normal distribution with an average m x and standard deviation . So the random variable

(1.18)

has a probability distribution that can be considered standard normal with distribution density j(t), the graph of which is shown in Fig. 1.7 (as well as in Fig. p. 1.4, Appendix 1).

Fig.1.7. Probability density of a random variable t.

Let the confidence probability be given b And tb- number that satisfies the equation

b \u003d F 0 (t b) - F 0 (-t b) \u003d 2 F 0 (t b),(1.19)

where - Laplace function. Then the probability of falling into the interval (-t b , t b) will be equal to the shaded one in Fig. 1.7. area, and, by virtue of expression (1.19), is equal to b. Consequently

b = P(-t b< < t b) = P( – tb< m x < + t b ) =

=P( – tb< m x < + t b ) .(1.20)

Thus, as a confidence interval, we can take the interval

I b = ( – t b ; + tb ) , (1.21)

since the expression (1.20) means that the unknown exact value m x is in Ib with a given confidence probability b. For building Ib needed according to b to find tb from equation (1.19). Here are some values tb needed in the future :

t 0.9 = 1.645; t 0.95 = 1.96; t 0.99 = 2.58; t 0.999 = 3.3.

When deriving expression (1.21), it was assumed that the exact value of the root-mean-square deviation is known s x. However, it is not always known. Therefore, we use his estimate (1.15) and obtain:

I b = ( – t b ; + t b ). (1.22)

Accordingly, the estimates and obtained from the grouped sample give the following formula for the confidence interval:

I b = ( – t b ; + t b ). (1.23)

Estimates of mathematical expectation and variance.

We got acquainted with the concept of distribution parameters in probability theory. For example, in the normal distribution law given by the probability density function

parameters are but– mathematical expectation and but is the standard deviation. In the Poisson distribution, the parameter is the number a = ex.

Definition. A statistical estimate of an unknown parameter of a theoretical distribution is its approximate value, which depends on the sample data(x 1, x 2, x 3,..., x k ; p 1, p 2, p 3,..., p k), i.e., some function of these quantities.

Here x 1, x 2, x 3,..., x k– feature values, p 1, p 2, p 3,..., p k are the corresponding frequencies. The statistical estimate is a random variable.

Denote by θ is the estimated parameter, and through θ * - its statistical evaluation. Value | θ *–θ | called assessment accuracy. The less | θ *–θ |, the better, the unknown parameter is more precisely defined.

To score θ * had practical value, it should not contain a systematic error and at the same time have the smallest possible variance. In addition, with an increase in the sample size, the probability of arbitrarily small deviations | θ *–θ | should be close to 1.

Let us formulate the following definitions.

1. A parameter estimate is called unbiased if its mathematical expectation is M(θ *) equal to the estimated parameter θ, i.e.

M(θ *) = θ, (1)

and offset if

M(θ *) ≠ θ, (2)

2. An estimate θ* is called consistent if for any δ > 0

(3)

Equality (3) reads as follows: estimate θ * converges in probability to θ .

3. An estimate θ* is called effective if, for a given n, it has the smallest variance.

Theorem 1.The sample mean Х В is an unbiased and consistent estimate of the mathematical expectation.

Proof. Let the sample be representative, i.e., all elements of the general population have the same opportunity to be included in the sample. Feature values x 1 , x 2 , x 3 ,..., x n can be taken as independent random variables X 1, X 2, X 3, ..., X n with the same distributions and numerical characteristics, including those with equal mathematical expectations equal to but,

Since each of the quantities X 1, X 2, X 3, ..., X p has a distribution coinciding with the distribution of the general population, then M(X)= a. That's why

whence it follows that is a consistent estimate M(X).

Using the extremum research rule, we can prove that is also an efficient estimate M(X).