Find $E(XY)$ assuming no independence with $E(X) = 4$, $E(Y) = 10$, $V(X) = 5$, $V(Y) = 3$, $V(X+Y) = 6$. I am having trouble finding a way to solve this problem. I understand this would be simple if $X$ and $Y$ were independent however, I believe they are not since $V(X) + V(Y) \neq  8$.
Find $E(XY)$ with $E(X) = 4, E(Y) = 10, V(X) = 5, V(Y) = 3, V(X+Y) = 6$.
 A: Use the formulas:
$V(X+Y)=V(X)+V(Y)+2Cov(X,Y)$
$Cov(X,Y)=E[XY]-E[X]E[Y]$
Can you solve now?
A: Hint: Note that $XY=(1/2)[(X+Y)^2-X^2-Y^2]$ and expected value is linear, and the variances are expected values of squares.
EDIT: Just to spell it out, from $XY=(1/2)[(X+Y)^2-X^2-Y^2]$ you get
$$E(XY)=(1/2)[E[(X+Y)^2]-E(X^2)-E(Y^2)] \\
=(1/2)[V(X+Y)-V(X)-V(Y)]=(1/2)[6-5-3]=-1.$$
Note that with this approach we did not need to use $E(X),E(Y)$ (or the covariance which depends on them).
I must apologize: variances are expected values of squared differences to the mean. So one does need to know $E(X),E(Y)$. Altogether the other answer is faster, but we can get to expected values of squares using $E(Z^2)=V(Z)+E(Z)^2$ and fix the above. That is, we have $E(X^2)=5+4^2=21,\ E(Y^2)=3+10^2=103,$ and $E[(X+Y)^2]=6+14^2=202$ so that
$$E(XY)=(1/2)[E[(X+Y)^2]-E(X^2)-E(Y^2)] \\
=(1/2)[202-21-103]=39,$$
in agreement with the other (less involved) answer.
A: Here is how I would deal with this problem!
First, we must realize that:
$\mathrm {Var}(X+Y) = \mathrm {Var}(X) + \mathrm {Var}(Y) + 2\mathrm{Cov}(X,Y)$
$\mathrm{Cov}(X,Y) = \mathrm E(XY) - \mathrm E(X)\mathrm E(Y)$
These two relationships will help us to solve this problem. 
Using the first equation, we can see that we are given 3 of the 4 unknowns. Solving for the $\mathrm{Cov}(X,Y)$ we get its value as $-1$. 
Now using the second equation for the definition of the covariance, we see that the only variables quantity that we do not have is $\mathrm E(XY)$. Solving for this quantity yields $39$. 
Hope that helps and please be mindful, this is my first answer on Math.Stackexchange :)
