Let there be a random variable X with mathematical expectation m and dispersion D, while both of these parameters are unknown. Over magnitude X produced N independent experiments, which resulted in a set of N numerical results x 1 , x 2 , …, x N. As an estimate of the mathematical expectation, it is natural to propose the arithmetic mean of the observed values

(1)

Here as x i specific values ​​(numbers) obtained as a result of N experiments. If we take others (independent of the previous ones) N experiments, then, obviously, we will get a different value. If you take more N experiments, we will get one more new value . Denote by X i random variable resulting from i th experiment, then the realizations X i will be the numbers obtained as a result of these experiments. It is obvious that the random variable X i will have the same probability distribution density as the original random variable X. We also assume that the random variables X i and Xj are independent at i, not equal j(various independent relative to each other experiments). Therefore, we rewrite formula (1) in a different (statistical) form:

(2)

Let us show that the estimate is unbiased:

Thus, the mathematical expectation of the sample mean is equal to the true mathematical expectation of the random variable m. This is a fairly predictable and understandable fact. Therefore, the sample mean (2) can be taken as an estimate of the mathematical expectation of a random variable. Now the question arises: what happens to the variance of the expectation estimate as the number of experiments increases? Analytical calculations show that

where is the variance of the estimate of the mathematical expectation (2), and D- true variance of the random variable X.

From the above, it follows that with increasing N(number of experiments) the variance of the estimate decreases, i.e. the more we summarize the independent implementations, the closer to the expected value we get the estimate.

Mathematical variance estimates

At first glance, the most natural estimate seems to be

(3)

where is calculated by formula (2). Let's check if the estimate is unbiased. Formula (3) can be written as follows:

We substitute expression (2) into this formula:

Let's find the mathematical expectation of the variance estimate:

(4)

Since the variance of a random variable does not depend on what the mathematical expectation of the random variable is, we will take the mathematical expectation equal to 0, i.e. m = 0.

	(5)
at .	(6)

The need to estimate the mathematical expectation based on test results appears in problems where the result of the experiment is described by a random variable and the quality indicator of the object under study is taken to be the mathematical expectation of this random variable. For example, the mathematical expectation of the uptime of a system can be taken as a reliability indicator, and when evaluating the efficiency of production, the mathematical expectation of the number of good products, etc.

The problem of estimating the mathematical expectation is formulated as follows. Suppose that to determine the unknown value of the random variable X, it is supposed to make n independent and free from systematic errors measurements X v X 2 ,..., X p. It is required to choose the best estimate of the mathematical expectation.

The best and most common estimate of the mathematical expectation in practice is the arithmetic mean of the test results

also called statistical or sample mean.

Let us show that the estimate t x satisfies all the requirements for the evaluation of any parameter.

1. It follows from expression (5.10) that

i.e. score t "x- unbiased estimate.

2. According to the Chebyshev theorem, the arithmetic mean of the test results converges in probability to the mathematical expectation, i.e.

Consequently, estimate (5.10) is a consistent estimate of the expectation.

3. Estimation variance t x, equal

As the sample size increases, n decreases indefinitely. It is proved that if a random variable X is subject to the normal distribution law, then for any P the variance (5.11) will be the minimum possible, and the estimate t x- effective estimation of mathematical expectation. Knowing the variance of the estimate makes it possible to make a judgment about the accuracy of determining the unknown value of the mathematical expectation using this estimate.

As an estimate of the mathematical expectation, the arithmetic mean is used if the measurement results are equally accurate (variances D, i = 1, 2, ..., P are the same in every dimension). However, in practice, one has to deal with tasks in which the measurement results are not equal (for example, during testing, measurements are made by different instruments). In this case, the estimate for the mathematical expectation has the form

where is the weight of the i-th measurement.

In formula (5.12), the result of each measurement is included with its own weight With.. Therefore, the evaluation of the measurement results t x called weighted average.

It can be shown that estimate (5.12) is an unbiased, consistent, and efficient estimate of the expectation. The minimum variance of the estimate is given by

When conducting experiments with computer models, similar problems arise when estimates are found from the results of several series of tests and the number of tests in each series is different. For example, two series of tests were carried out with a volume p 1 and n 2 , according to the results of which the estimates t xi and t x _. In order to improve the accuracy and reliability of determining the mathematical expectation, the results of these series of tests are combined. To do this, use the expression (5.12)

When calculating the coefficients C, instead of the variances D, their estimates obtained from the test results in each series are substituted.

A similar approach is also used in determining the probability of a random event occurring based on the results of a series of tests.

To estimate the mathematical expectation of a random variable X, in addition to the sample mean, other statistics can be used. Most often, members of the variational series are used for these purposes, i.e. order statistics, on the basis of which estimates are built,

satisfying the main of the requirements, namely consistency and unbiasedness.

Assume that the variation series contains n = 2k members. Then, any of the averages can be taken as an estimate of the mathematical expectation:

Wherein to-e the average

is nothing else than the statistical median of the distribution of the random variable X, since the obvious equality takes place

The advantage of the statistical median is that it is free from the influence of anomalous observations, which is inevitable when using the first average, that is, the average of the smallest and largest number of variation series.

With an odd sample size P = 2k- 1 statistical median is its middle element, i.e. to-th member of the variation series Me = x k.

There are distributions for which the arithmetic mean is not an effective estimate of the mathematical expectation, for example, the Laplace distribution. It can be shown that for the Laplace distribution, the effective estimate of the mean is the sample median.

It is proved that if a random variable X has a normal distribution, then with a sufficiently large sample size, the distribution law of the statistical median is close to normal with numerical characteristics

From a comparison of formulas (5.11) and (5.14) it follows that the dispersion of the statistical median is 1.57 times greater than the dispersion of the arithmetic mean. Therefore, the arithmetic mean as an estimate of the mathematical expectation is as much more effective than the statistical median. However, due to the simplicity of calculations, insensitivity to anomalous measurement results (“contamination” of the sample), in practice, the statistical median is nevertheless used as an estimate of the mathematical expectation.

It should be noted that for continuous symmetric distributions, the mean and the median are the same. Therefore, the statistical median can serve as a good estimate of the mathematical expectation only for a symmetrical distribution of the random variable.

For skewed distributions, the statistical median Me has a significant bias relative to the mathematical expectation, therefore, it is unsuitable for its estimation.

Let there be a random variable X with mathematical expectation m and dispersion D, while both of these parameters are unknown. Over magnitude X produced N independent experiments, which resulted in a set of N numerical results x 1 , x 2 , …, x N. As an estimate of the mathematical expectation, it is natural to propose the arithmetic mean of the observed values

(1)

Here as x i specific values ​​(numbers) obtained as a result of N experiments. If we take others (independent of the previous ones) N experiments, then, obviously, we will get a different value. If you take more N experiments, we will get one more new value . Denote by X i random variable resulting from i th experiment, then the realizations X i will be the numbers obtained as a result of these experiments. It is obvious that the random variable X i will have the same probability distribution density as the original random variable X. We also assume that the random variables X i and Xj are independent at i, not equal j(various independent relative to each other experiments). Therefore, we rewrite formula (1) in a different (statistical) form:

(2)

Let us show that the estimate is unbiased:

Thus, the mathematical expectation of the sample mean is equal to the true mathematical expectation of the random variable m. This is a fairly predictable and understandable fact. Therefore, the sample mean (2) can be taken as an estimate of the mathematical expectation of a random variable. Now the question arises: what happens to the variance of the expectation estimate as the number of experiments increases? Analytical calculations show that

where is the variance of the estimate of the mathematical expectation (2), and D- true variance of the random variable X.

From the above, it follows that with increasing N(number of experiments) the variance of the estimate decreases, i.e. the more we summarize the independent implementations, the closer to the expected value we get the estimate.

Mathematical variance estimates

At first glance, the most natural estimate seems to be

(3)

where is calculated by formula (2). Let's check if the estimate is unbiased. Formula (3) can be written as follows:

We substitute expression (2) into this formula:

Let's find the mathematical expectation of the variance estimate:

(4)

Since the variance of a random variable does not depend on what the mathematical expectation of the random variable is, we will take the mathematical expectation equal to 0, i.e. m = 0.

	(5)
at .	(6)

Let a random variable with unknown mathematical expectation and variance be subjected to independent experiments that yielded results - . Let us calculate consistent and unbiased estimates for the parameters and .

As an estimate for the mathematical expectation, we take the arithmetic mean of the experimental values

. (2.9.1)

According to the law of large numbers, this estimate is wealthy , with magnitude in probability. The same estimate is unbiased , insofar as

. (2.9.2)

The variance of this estimate is

. (2.9.3)

It can be shown that for a normal distribution, this estimate is effective . For other laws, this may not be the case.

Let us now estimate the variance. Let us first choose a formula for estimating statistical dispersion

. (2.9.4)

Let us check the consistency of the variance estimate. Let's open the brackets in the formula (2.9.4)

.

For , the first term converges in probability to the quantity , in the second - to . Thus, our estimate converges in probability to the variance

,

hence she is wealthy .

Let's check unbiasedness estimates for the quantity . To do this, we substitute expression (2.9.1) into formula (2.9.4) and take into account that random variables independent

,

. (2.9.5)

Let us pass in formula (2.9.5) to fluctuations of random variables

Expanding the brackets, we get

,

. (2.9.6)

Let us calculate the mathematical expectation of the value (2.9.6), taking into account that

. (2.9.7)

Relation (2.9.7) shows that the value calculated by formula (2.9.4) is not an unbiased estimator for dispersion. Its mathematical expectation is not equal, but somewhat less. Such an estimate leads to a downward systematic error. To eliminate such a bias, it is necessary to introduce a correction by multiplying not the value . Then such a corrected statistical variance can serve as an unbiased estimate for the variance

. (2.9.8)

This estimate is just as consistent as the estimate , because for .

In practice, instead of estimate (2.9.8), it is sometimes more convenient to use an equivalent estimate related to the second initial statistical moment

. (2.9.9)

Estimates (2.9.8), (2.9.9) are not efficient. It can be shown that in the case of a normal distribution they will be asymptotically efficient (when will tend to the minimum possible value).

Thus, it is possible to formulate the following rules for processing limited statistical material. If in independent experiments the random variable takes the values with unknown mathematical expectation and variance , then to determine these parameters, one should use approximate estimates

(2.9.10)

End of work -

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Actions on events
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Combinations
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The formula for adding probabilities for incompatible events
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Probability Addition Formula for Arbitrary Events
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Random continuous variables
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Theorems on numerical characteristics of random variables
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Uniform distribution law
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Normal distribution law
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Exponential distribution law
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Systems of random variables
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Conditional laws of distribution
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Normal distribution of a system of two random variables
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Let there be a random variable X, and its parameters are the mathematical expectation a 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 of statistical estimation - the method of substitution and analogy - consists in the fact that as an estimate of one or another numerical characteristic (average, variance, etc.) of the general population, the corresponding characteristic of the sample distribution is taken - the sample characteristic.

By the substitution method as an estimate of the mathematical expectation a 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 a, 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.

Hence,

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

,

As , 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 a 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 variation series as c). Then

, .

interval estimation

Above, we considered the question of estimating an unknown parameter a 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, in order to get an idea of ​​the proximity between a parameter and its estimate, so-called interval estimates are introduced in mathematical statistics.

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 a and s, i.e. X = N( a, s). The mathematical expectation in this case is equal to a. 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 the critical values ​​of the one-tailed F-criterion. Therefore, if a two-sided criterion is applied (H 1: s x 2 ¹ s y 2), then the right-hand critical point F cr (a / 2, n 1, n 2) is searched for by the significance level a / 2 (half the specified one) and the number of degrees freedom n 1 and n 2 (n 1 - 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.

Decision. 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:

u i	m i	m i u i	m i u i 2	m i (u i +1) 2		v i	n i	n i v i	n i v i 2	n i (v i +1) 2
-3		-3				-1		-2
å	-
						å	-

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 новой технологиях принимаем.

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.