Expected value, variance and covariance with discrete random variables

Question

I wamted to check my solutions for this problem: The joint distribution of the random variables $X$ and $Y$ is given by:

Calculate $E[X], E[Y], Var(X),Var(Y),Cov(X,Y)$

So for $E[X]=1/8+1/4+2(1/8+1/2)=13/8 \\ E[Y]=1/4+2(1/4+1/2)=7/4\\ E[X^2]=3/8+4*5/8=23/8 \\ E[Y^2]=1/4+12/4=13/4$

So we have for $Var(X)=E[X^2]-E[X]^2=23/8-(13/8)^2=15/64 \\ Var(Y)=E[Y^2]-E[Y]^2=13/4-(7/4)^2=3/16$

for $E[XY]=1/8+2*1/4+2*1/8+2*1/2=15/8$

And therefore $Cov(X,Y)=E[XY]-E[X]E[Y]=15/8-(13/8)(7/4)=-31/32$

Is everything right or have I done any mistake? Thanks for any help

Answer 1

I checked $E[X]$ and it is correct. I only skimmed the calculations for the rest and they all seem like they should be correct -- the method looks right although I haven't taken the time to independently compute the arithmetic.