# Chebeshev's inequality

I have a exercise, where I have to show that $P[(X_1-\mu_1)+(X_2-\mu_2)\geq k\cdot \sigma ]\leq \frac{2(1+\rho)}{k^2}$ (1) for k being positive.

It's given that $X_1$ and $X_2$ have the same varians $\sigma^2 = \sigma_X^2 = \sigma_Y^2$ and let $\rho$ be given the correlations coefficient.

I figured out from a lot of calculations that the variance of $Y$ is $\sigma_Y^2=2\sigma^2 (1+\rho)$.

I know I have to use the Chebeshev's inequality, but it is given as $P(|X-\mu|\geq k\cdot \sigma)\leq \frac{1}{k^2}$, but I don't know how to show (1).

$X_1$ and $X_2$ are random variables with $\mu_1$ and $\mu_2$ and $\sigma$ as their parameters

Then $X_1$+$X_2$ will have a variance $$Var(Y=X_1+X_2) = Var(X_1)+Var(X_2) + 2 Cov(X_1,X_2)$$

$$\rho = \frac{Cov(X_1,X_2)}{\sigma^2} => Cov(X_1,X_2) = \rho\sigma^2$$

Then $\sigma_Y^2=Var (Y) = \sigma^2 +\sigma^2+ 2\rho\sigma^2 = 2(1+\rho)\sigma^2$

Chebychev's inequality says the upper bound is $\frac{1}{k^2}$ for k standard deviations in $P((X-\mu)\ge k\sigma)$. The standard deviation of Y is $\sqrt{2(1+\rho)}\sigma$. For the following expression $\left(k\sqrt{2(1+\rho)}\right)\sigma$ standard deviations, the upper bound still is $\frac{1}{k^2}$. But what you want is $k\sigma$. Now take the entire remainder of $\left(\frac{k}{\sqrt{2(1+\rho)}}\right)\sqrt{2(1+\rho)}\sigma$ besides $\sqrt{2(1+\rho)}\sigma$ as k' and apply the normal chebyshev to find the upper bound as $\frac{1}{k'^2}$, which when you manipulate gives$\frac{1}{\left(\frac{k}{\sqrt{2(1+\rho)}}\right)^2}$. IF you do some rearrangement , you get the desired result.

This is another way of looking at it.

Thanksgives

• @aa_x, I have given an alternative explanation. – Satish Ramanathan Oct 7 '14 at 20:35
• Oh, so in my case $"k"$ will be $\frac{k}{\sqrt{2(1+\rho})}$? Right? – aa_x Oct 7 '14 at 20:40
• You are absolutely correct. – Satish Ramanathan Oct 7 '14 at 20:41
• Thanks a lot, now I really understand it! – aa_x Oct 7 '14 at 20:41

Misprints all over the place... Once they are corrected you will see that the inequality you recalled for $Y=X_1+X_2$ yields $$P(Y-E(Y)\geqslant k\sigma)\leqslant\frac{\sigma^2_Y}{(k\sigma)^2}=\frac{2(1+\rho)}{k^2}.$$

• I have big difficulties with seeing, how $\frac{1}{k^2}$ becomes $\frac{2(1+\rho)}{k^2}$. Why can we say that $\frac{\sigma_Y^2}{(k\sigma)^2}$? Where does this come from? I don't understand the argument – aa_x Oct 7 '14 at 19:46
• No $1/k^2$ becomes $2(1+\rho)/k^2$. The upper bound is $\sigma^2_Y/(k\sigma)^2$ because for every square integrable $Z$ and positive $t$, $P(Z-E(Z)\gt t)\lt\sigma_Z^2/t^2$. This is a standard application of Bienaymé-Chebyshev inequality that you recalled. – Did Oct 7 '14 at 19:54