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