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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.

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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?

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    $\begingroup$ $Var(X+Y)=E((X+Y)^2)-E(X+Y)^2$ might help you. $\endgroup$ – TSF Apr 10 '18 at 19:21
  • $\begingroup$ @TonyS.F. Thank you! I managed to solve the problem $\endgroup$ – Hendrra Apr 10 '18 at 19:47
  • $\begingroup$ You could also possibly apply the Cauchy Schwartz inequality to prove the claim. $\endgroup$ – StubbornAtom Apr 10 '18 at 19:51
  • $\begingroup$ My answer was a hint from where to start, it wasn't a hint on how to complete your existing attempt of a proof. If you think about it, the hint given immediately leads to the required result $\endgroup$ – StubbornAtom Apr 10 '18 at 21:02
  • $\begingroup$ Could you use the hint to come to the required conclusion? $\endgroup$ – StubbornAtom Apr 11 '18 at 4:12
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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.

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  • $\begingroup$ Why $V(X) < \infty \rightarrow E(X^2) < \infty$? $\endgroup$ – Hendrra Apr 10 '18 at 19:27
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    $\begingroup$ Think of how $V$ is defined $\endgroup$ – TSF Apr 10 '18 at 19:30
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    $\begingroup$ @Hendrra Because $V(X)=E(X^2)-(E(X))^2$ whenever the expectations exist. The l.h.s is finite thus implies that both $E(X)$ and $E(X^2)$ are finite. $\endgroup$ – StubbornAtom Apr 10 '18 at 19:31
  • $\begingroup$ @StubbornAtom Thanks so much! $\endgroup$ – Hendrra Apr 10 '18 at 19:47
  • $\begingroup$ If I know that $E(X), E(X^2)$ are finite I don't have to use the inequality above, do I? $\endgroup$ – Hendrra Apr 10 '18 at 20:01

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