13.2. Chebyshev Inequality

As a lemma, which is necessary for the proof of theorems belonging to the group of the “law of large numbers,” we will prove one very general inequality, known as the Chebyshev inequality.

Let there be a random variable with mathematical expectation and variance . Chebyshev's inequality states that, whatever the positive number the probability that the magnitude deviate from their expectation by no less than , bounded above by value :

. (13.2.1)

Evidence. 1. Let the value discontinuous, with a number of distribution

We depict the possible values ​​of and its expectation in the form of points on the number axis (fig. 13.2.1).

Fig. 13.2.1.

Let's set some value and calculate the probability that the magnitude deviate from their expectation by no less than :

. (13.2.2)

To do this, set aside from the point right and left along the length segment ; get the segment . Probability (13.2.2) is nothing but the probability that a random point will not fall inside the segment , and outside it:

.

In order to find this probability, you need to sum up the probabilities of all those values which lie outside the segment . We will write this as follows:

(13.2.3)

where is the record under the sum sign means that the summation applies to all those values for which points , lie outside the segment .

On the other hand, we write the expression for the variance of . By definition:

. (13.2.4)

Since all members of the sum (13.2.4) are non-negative, it can only decrease if we do not extend it to all values , but only on some, in particular on those that lie outside the segment :

. (13.2.5)

Replace the expression with the sum sign through . Since for all members of the sum then the amount of such a replacement can also only decrease; means

. (13.2.6)

But according to formula (13.2.3), the sum on the right side of (13.2.6) is nothing more than the probability of a random point falling outside the segment ; Consequently,

,

whence the derived inequality directly follows.

2. In the case where the value is continuous, the proof is carried out in a similar way with the replacement of probabilities element of probability, and finite sums - integrals. Really,

. (13.2.7)

Where - distribution density . Next, we have:

,

where is the sign under the integral means that the integration extends to the outer part of the segment .

Replacing under the sign of the integral through , we get:

,

whence Chebyshev's inequality for continuous quantities follows.

Example. Given a random variable with mathematical expectation and variance . Estimate the top probability of the magnitude deviate from their expectation by no less than .

Decision. Believing in Chebyshev's inequality , we have:

,

i.e. the probability that the deviation of a random variable from its expectation will go beyond the three mean square deviations cannot be greater than .

Note. Chebyshev inequality gives only the upper limit of the probability of a given deviation. Above this limit, probability cannot be under any distribution law. In practice, in most cases the probability that the magnitude go outside the lot , significantly less . For example, for a normal law, this probability is approximately equal to 0.003. In practice, most often we deal with random variables, the values ​​of which only very rarely go beyond . If the distribution law of a random variable is unknown, but only and , in practice, the segment is usually considered a section of practically possible values ​​of a random variable (the so-called “three sigma rule”).