# covariance of random variables

Suppose X, Y, W are independent random variables such that X ∼ GAM(2,3), Y ∼ N(1,4) and W ∼ BIN(10,1/4). Let U = 2X − 3Y and V = Y − W . Find cov(U, V ).

I know that cov(U, V) = E(U, V) - E(U)E(V). I've found E(U) and E(V) but I don't know how to find E(U, V).

• You know, or can look up, the mean and variance of the various distributions. But in the second simpler way in the answer, all we need is the cov of $Y$ and $Y$, which is the variance of $Y$. We were told that $Y$ has variance $4$, so there is no work to do. – André Nicolas Apr 19 '16 at 6:11

## 1 Answer

You mean that you don't know how to find $E(UV)$. This is $E((2X-3Y)(Y-W))$. Expand the product, and use the linearity of expectation. We find that $$E(UV)=2E(XY)-2E(XW)-3E(Y^2)+3E(YW).$$ Now you can use independence and your knowledge of the various distributions to finish the calculation of $E(UV)$.

However, this is not the best way to find the covariance of $U$ and $V$. Instead, use the bilinearity of covariance. We have $$\text{Cov}(UV)=2\text{Cov}(X,Y)-2\text{Cov}(X,W)-3\text{Cov}(Y,Y)+3\text{Cov}(Y,W).$$ Almost all these covariances are $0$!

• Almost all of them are 0? Aren't they all 0 since X, Y, W are all independent? – clover20 Apr 19 '16 at 6:11
• The Cov of $Y$ and $Y$ is not $0$, it is the variance of $Y$. The other three are $0$. – André Nicolas Apr 19 '16 at 6:13