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.



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?

  • 1
    $\begingroup$ $Var(X+Y)=E((X+Y)^2)-E(X+Y)^2$ 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


For any two real numbers $a$ and $b$,


Note that $V(X)$ finite implies $E(X^2)$ is finite.

  • $\begingroup$ Why $V(X) < \infty \rightarrow E(X^2) < \infty$? $\endgroup$ – Hendrra Apr 10 '18 at 19:27
  • 1
    $\begingroup$ Think of how $V$ is defined $\endgroup$ – TSF Apr 10 '18 at 19:30
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.