What I know is that:
$$Var(X), Var(Y) < \infty$$
I am to prove the following thesis:
$$E(XY) < \infty$$
I tried to solve the problem using this dependence:
$$
\begin{split}
\infty &> Var(X) + Var(Y) \\
&= Var(X + Y) + 2Cov(X, Y) \\
&= Var(X + Y) + 2\left(E(XY) - E(X)E(Y)\right)
\end{split}
$$
I don't know if it's a good attempt however and if it is I don't know how to finish the proof.
-
-
Edit
So after some calculations I do have: $$E\big((X+Y)^2 \big) - E(X+Y)^2 + 2\big(E(XY) - E(X)E(Y) \big) \\ = E(X^2)+2E(XY)+E(Y^2) -E(X)^2 -2E(X)E(Y) -E(Y)^2 +2E(XY) - 2E(X)E(Y) \\ = E(X^2)+4E(XY)+E(Y^2) -E(X)^2 -4E(X)E(Y) -E(Y)^2 \\ = E(X^2)+4E(XY)+E(Y^2) - \big( E(X)^2 +4E(X)E(Y) +E(Y)^2\big)$$ And I can use the inequality given in the post but it don't simplify anything. Where can I go from here?
Hint:
For any two real numbers $a$ and $b$,
$$|ab|\leq\frac{a^2+b^2}{2}$$
Note that $V(X)$ finite implies $E(X^2)$ is finite.