I have the following problem: $X \in Bi(3,\frac{1}{2}), Y \in Ge(\frac{1}{3}), cov(X,Y)=2$.
I have to calculate E(X-Y) and Var(X-Y).
What I've got:
$$cov(X,Y) = E(XY)-EXEY=2 \implies E(XY)=2+EXEY \\ E(X-Y)=EX-EY=3\frac{1}{2}-\frac{1}{\frac{1}{3}}=-\frac{3}{2}\\ Var(X-Y)=E((X-Y)^2)-E(X-Y)^2=E(X^2)+E(Y^2)-2E(XY)-\frac{9}{4}=\\ E(X^2)+E(Y^2)-4-2EXEY-\frac{9}{4}=E(X^2)+E(Y^2)-\frac{61}{4} $$
Assuming what I have is correct, I just have to calculate $E(X^2)$ and $E(Y^2)$ which I don't know how. Of course my calculations might be entirely wrong, and the approach for this problem might be completely different.
Along with their covariance, you are given that the variables have well know distribution which tells you their expectations and variances.
$$\mathsf {Cov}(X,Y)=2\\ \mathsf E(X)=\tfrac 32 \qquad \mathsf {Var}(X)=\tfrac 34\\ \mathsf E(Y) = 3 \qquad \mathsf {Var}(Y)= 6$$
This is useful knowledge! Make good use of it to express the expectation and variance of the difference in terms of what you know.
Such as when, indeed, you used the Linearity of Expectation. $$\mathsf E(X-Y)=\mathsf E(X)-\mathsf E(Y)$$
Similarly, since $\mathsf{Var}(X-Y)~=~\mathsf{Cov}(X-Y,X-Y)$ you may make use of the Bilinearity of Covariance.
$$\mathsf{Var}(X-Y)~=~\mathsf{Var}(X)-2\mathsf{Cov}(X,Y)+\mathsf{Var}(Y)$$
$~\\~\\~$
Which of course is $\mathsf {Var}(X-Y)~{=\mathsf E(X^2)+\mathsf E(Y^2)-\mathsf E(X)^2-\mathsf E(Y)^2-2\mathsf {Cov}(X,Y)\\ = \mathsf E(X^2)+\mathsf E(Y^2)-\tfrac {61}4}$
But just take a step back.