1. ## Difficult problem

2. Let X1 and X2 be independent random variables, with means μ1 and μ2 and variances σ1^2 and σ2^2, respectively. Find the Covariance of Y1 = X1 and Y2 = X1 – X2.

So for this one is the covariance of y 1 just the standard formula of covariance without any manipulation? And how about y2?

2. Originally Posted by wolverine21
2. Let X1 and X2 be independent random variables, with means μ1 and μ2 and variances σ1^2 and σ2^2, respectively. Find the Covariance of Y1 = X1[snip]
Cov(X1, X1) = Var(X1)

3. Originally Posted by wolverine21
2. Let X1 and X2 be independent random variables, with means μ1 and μ2 and variances σ1^2 and σ2^2, respectively. Find the Covariance of Y1 = X1 and Y2 = X1 – X2.

So for this one is the covariance of y 1 just the standard formula of covariance without any manipulation? And how about y2?
Because $X_1$ and $X_2$ their joint pdf is the product of their individial pdf's.

$E( (Y_1-\bar{Y_1})(Y_2-\bar{Y_2})=\iint (x_1-\bar{x_1})(x_1-x_2-\bar{x_1}+\bar{x_2}) p(x_1)p(x_2) dx_1 dx_2$

Which you should be able to complete

CB

4. Originally Posted by mr fantastic
Cov(X1, X1) = Var(X1)
Your snip is in the wrong place, the covariance of $Y_1$ and $Y_2$ is required.

CB

5. Originally Posted by CaptainBlack
Your snip is in the wrong place, the covariance of $Y_1$ and $Y_2$ is required.

CB

6. Originally Posted by CaptainBlack
Because $X_1$ and $X_2$ their joint pdf is the product of their individial pdf's.

$E( (Y_1-\bar{Y_1})(Y_2-\bar{Y_2})=\iint (x_1-\bar{x_1})(x_1-x_2-\bar{x_1}+\bar{x_2}) p(x_1)p(x_2) dx_1 dx_2$

Which you should be able to complete

CB

Wait so is this joint pdf I find the final answer?

7. Originally Posted by wolverine21
Wait so is this joint pdf I find the final answer?
No, you need to complete the double integral, you don't have to know the actual pdf's for X_1 and X_1 it is sufficient to know their means and variances and that they are independant (so Cov(X_1,X_2)=0 amoung other things).

CB