Chebychev's inequality for two random variables

In summary, the problem involves showing that the probability of the absolute value of the sum of two random variables, X1 and X2, being greater than or equal to a certain value is less than or equal to a specific expression involving their common variance and correlation coefficient. The approach suggested is to find the variance and mean for the sum of X1 and X2, which would convert the problem to one that can be solved using Chebychev's inequality.
  • #1
rayge
25
0
(I wasn't sure how to title this, it's just that the statement resembles Chebychev's but with two RV's.)

Homework Statement

Let [itex]\sigma_1^1 = \sigma_2^2 = \sigma^2[/itex] be the common variance of [itex]X_1[/itex] and [itex]X_2[/itex] and let [roh] (can't find the encoding for roh) be the correlation coefficient of [itex]X_1[/itex] and [itex]X_2[/itex]. Show for [itex]k>0[/itex] that

[itex]P[|(X_1-\mu_1) + (X_2-\mu_2)|\geq k\sigma]\leq2(1+[roh])/k^2[/itex]

Homework Equations


Chebychev's inequality:
[itex]P(|X-\mu|\geq k\sigma) \leq 1/k^2[/itex]

The Attempt at a Solution



I'm really only looking for a place to start. I can try working backwords, and expanding [roh] into its definition, which is [itex]E[(X_1-\mu_1)(X_2-\mu_2)]/\sigma_1\sigma_2[/itex], but I really don't know how to evaluate that. I was wondering about using Markov's inequality and substituting [itex]u(X_1,X_2)[/itex] for [itex]u(X_1)[/itex], but of course there's no equation linking [itex]X_1[/itex] and [itex]X_2[/itex]. Feeling stumped. Any suggestions welcome!
 
Last edited:
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  • #2
I would try to find the variance and mean (okay, mean is obvious) for X1+X2. I think this converts Chebychev's inequality to the one you have to show.

The greek letter is rho, and \rho gives ##\rho##.
 

1. What is Chebychev's inequality for two random variables?

Chebychev's inequality for two random variables is a mathematical formula that provides an upper bound for the probability that two random variables deviate from their mean by a certain amount. It is also known as the Chebychev-Cantelli inequality and is commonly used in statistics and probability theory.

2. How is Chebychev's inequality for two random variables calculated?

Chebychev's inequality for two random variables is calculated by taking the sum of the variances of the two variables and dividing it by the square of the difference between their means. The resulting value is then multiplied by the probability of the two variables being within a specified range of their means.

3. What is the purpose of using Chebychev's inequality for two random variables?

The purpose of using Chebychev's inequality for two random variables is to determine the probability that both variables deviate from their means by a certain amount. It can be used to determine the likelihood of a certain outcome occurring in a given situation.

4. Can Chebychev's inequality for two random variables be applied to any type of random variables?

Yes, Chebychev's inequality for two random variables can be applied to any type of random variables, as long as they have a well-defined mean and variance. This includes both discrete and continuous random variables.

5. What are the limitations of Chebychev's inequality for two random variables?

Chebychev's inequality for two random variables is a conservative bound and may overestimate the probability of the two variables deviating from their means. It also assumes that the two variables are independent, which may not always be the case in real-world situations. Additionally, it only provides an upper bound and does not give an exact probability value.

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