The Chebyshov's theorem (or Chebyshov's inequality) is one of the most important classical results of probability theory. It allows estimating the probability of an event described in terms of a random variable X, by providing us with a bound that does not depend on the distribution of the random variable but on the variance of X.
The theorem is named in honor of the Russian mathematician Pafnuty Chebyshov (also written as Chebychev or Tchebycheff) who, despite not being the first to state the theorem, was the first to give a proof in 1867.
This inequality, or those that due to their characteristics are called Chebyshov's inequality, is used mainly to approximate probabilities by calculating bounds.
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In the study of probability theory it happens that if the distribution function of a random variable X is known, its expected value -or mathematical expectation E (X) - and its variance Var (X) can be calculated, as long as such amounts exist. However, the converse is not necessarily true..
That is, knowing E (X) and Var (X) it is not necessarily possible to obtain the distribution function of X, therefore quantities like P (| X |> k) for some k> 0 are very difficult to obtain. But thanks to Chebyshov's inequality it is possible to estimate the probability of the random variable.
Chebyshov's theorem tells us that if we have a random variable X over a sample space S with a probability function p, and if k> 0, then:
Among the many applications of Chebyshov's theorem, the following can be mentioned:
This is the most common application and is used to give an upper bound for P (| X-E (X) | ≥k) where k> 0, only with the variance and the expectation of the random variable X, without knowing the probability function.
Suppose that the number of products manufactured in a company during a week is a random variable with an average of 50.
If it is known that the variance of a week of production is equal to 25, then what can we say about the probability that this week the production differs by more than 10 from the mean?
Applying Chebyshov's inequality we have:
From this we can obtain that the probability that in the production week the number of articles exceeds the average by more than 10 is at most 1/4.
Chebyshov's inequality plays an important role in proving the most important limit theorems. As an example we have the following:
This law states that given a sequence X1, X2,…, Xn,… of independent random variables with the same average distribution E (Xi) = μ and variance Var (X) = σtwo, and a known mean sample of:
Then for k> 0 we have:
Or, equivalently:
Let's first notice the following:
Since X1, X2,…, Xn are independent, it follows that:
Therefore, it is possible to state the following:
Then, using Chebyshov's theorem we have:
Finally, the theorem results from the fact that the limit on the right is zero when n tends to infinity.
It should be noted that this test was done only for the case in which the variance of Xi exists; that is to say, it does not diverge. Thus we observe that the theorem is always true if E (Xi) exists.
If X1, X2,…, Xn,… is a sequence of independent random variables such that there is some C< infinito, tal que Var(Xn) ≤ C para todo n natural, entonces para cualquier k>0:
Since the sequence of variances is uniformly bounded, we have that Var (Sn) ≤ C / n, for all natural n. But we know that:
Making n tend towards infinity, the following results:
Since a probability cannot exceed the value of 1, the desired result is obtained. As a consequence of this theorem we could mention the particular case of Bernoulli.
If an experiment is repeated n times independently with two possible outcomes (failure and success), where p is the probability of success in each experiment and X is the random variable that represents the number of successes obtained, then for each k> 0 you have to:
In terms of the variance, the Chebyshov inequality allows us to find a sample size n that is sufficient to guarantee that the probability that | Sn-μ |> = k occurs is as small as desired, which allows us to have an approximation to the average.
Precisely, let X1, X2,… Xn be a sample of independent random variables of size n and suppose that E (Xi) = μ and its variance σtwo. Then, by Chebyshov's inequality, we have:
Suppose that X1, X2,… Xn are a sample of independent random variables with Bernoulli distribution, such that they take the value 1 with probability p = 0.5.
What should be the size of the sample to be able to guarantee that the probability that the difference between the arithmetic mean Sn and its expected value (exceeding by more than 0.1), is less than or equal to 0., 01?
We have that E (X) = μ = p = 0.5 and that Var (X) = σtwo= p (1-p) = 0.25. By Chebyshov's inequality, for any k> 0 we have:
Now, taking k = 0.1 and δ = 0.01, we have:
In this way, it is concluded that a sample size of at least 2500 is needed to guarantee that the probability of the event | Sn - 0.5 |> = 0.1 is less than 0.01.
There are several inequalities related to Chebyshov's inequality. One of the best known is the Markov inequality:
In this expression X is a non-negative random variable with k, r> 0.
The Markov inequality can take different forms. For example, let Y be a non-negative random variable (so P (Y> = 0) = 1) and suppose that E (Y) = μ exists. Suppose also that (E (Y))r= μr exists for some integer r> 1. Then:
Another inequality is Gaussian, which tells us that given a unimodal random variable X with mode at zero, then for k> 0,
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