X and Y are independent random variables and their distributions are..

Question

X and Y are independent random variables and their distributions are..

$P(X=1) = 0.1 $ $P(X=2) = 0.2$ $P(X=3) = 0.3 $ $P(X=4) = 0.4 $

$P(Y=4) = 0.4 $ $P(Y=2) = 0.3$ $P(Y=3) = 0.2 $ $P(Y=4) = 0.1$

I want to find the covariance. I know the $Cov(X,Y) = E(XY) − E(X)E(Y)$, but how do I solve for $E(XY)$? Thanks.

Answer 1

$X, Y$ are independent $\implies E[XY] \equiv E[X] \cdot E[Y] \iff E[XY]-E[X]E[Y] \equiv 0 \implies \boxed{\rm{Cov}(X,Y) \equiv0 \ }$

Answer 2

You may verify the result by calculating it numerically:

$E(XY)=\sum_{i,j=1,2,3,4} X_iY_j P(X_i)P(X_j)$

So you sum up all 16 combinations of $X\cdot Y$ weighted by their probability product.

By this I came to $E(XY)=6$.

As $E(X)=3, E(Y)=2$, we have indeed:

$Cov(X,Y)=E(XY)-E(X)E(Y)=6-3\cdot 2=0$