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Show that X+Y has a finite second moment

  1. Jul 15, 2009 #1
    Prove that if X and Y have finite second moments (i.e. E(X^2) and E(Y^2) are finite), then X+Y has a finite second moment.

    (X+Y)^2 ≤ X^2 + Y^2 + 2|XY|
    => E[(X+Y)^2] ≤ E(X^2) + E(Y^2) + 2E(|XY|)

    I don't understand the (probably incomplete) proof. On the right side, E(X^2) and E(Y^2) are finite, but how can we know whether E(|XY|) is finite or not?

    Thanks for explaining!
  2. jcsd
  3. Jul 15, 2009 #2


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    This uses a classic inequality (really from measure theory, but applied to probability).

    Essentially, if both [tex] X, [/tex] have finite second moments (variances exist) then they have finite moments of every lower order. For your specific case:

    E[|XY|]^2 \le E(X^2) E(Y^2) < \infty

    where the RHS is finite because of the assumptions about the second-order moments being finite.
  4. Jul 15, 2009 #3
    Hi statdad,

    Is it in any way related to the Cauchy-Schwartz inequality?

    I don't see how this fact can be applied to our problem. E(|XY|), E(X^2), and E(Y^2) are all second moments, right?

    For the left side of the inequality, do you mean E(|XY|^2) or [E(|XY|)]^2 ?

    Thanks for your help!:smile:
  5. Jul 20, 2009 #4
    I am still stuck on this problem and would appreicate if anyone could help me out...
  6. Jul 20, 2009 #5


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    Sorry for the delay - no excuse on my part.

    Yes, as you pointed out, the moment-inequality is from the function version of the C-S inequality. My post should read

    E[|XY|^2] \le E(X^2) E(Y^2)
  7. Jul 20, 2009 #6
    That's OK, don't worry.

    But the version of C-S inequality that I've seen in Wikipedia is the following:
    |E(XY)|^2 ≤ E(X^2) E(Y^2)

    We have:
    (X+Y)^2 ≤ X^2 + Y^2 + 2|XY|
    E[(X+Y)^2] ≤ E(X^2) + E(Y^2) + 2E(|XY|)
    We are given that E(X^2) and E(Y^2) are both finite, but how can we show that E(|XY|) is finite?
    For E(|XY|), here we have the absolute value inside the expectation, but for the left side of the C-S inequality, the absolute value is outside.

    Thanks for your help!
    Last edited by a moderator: Apr 18, 2017
  8. Jul 20, 2009 #7


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    Consider [tex] X [/tex] - the proof for [/tex] Y [/tex] is similar.

    E[|X|]^2 = E[|X| \cdot 1 ]^2 = \left(\int |x| \cdot 1 \, dF(x)\right)^2 \le \left(\int |x|^2 \, dF(x)\right)^2 \cdot \left(\int 1 \, dF(x)\right)^2 = E[X^2] < \infty

    so the existence of a finite second moment gives the existence of the finite first moment.
  9. Jul 21, 2009 #8
    ??But E(|X|2) = E(X2) always, no? (since |X|2 = X2)

    Also, I don't see how the C-S inequality 3c2d62f6a6a33c74752cd006b8034541.png would necessarily imply that E(|XY|2) ≤ E(X2)E(Y2) as you said in post #5. And how can we use this to prove that E(|XY|) is finite? Could you please explain this part?

    Thanks a lot!:smile:
    Last edited by a moderator: Apr 28, 2017
  10. Jul 21, 2009 #9


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    A quick method is |X+Y| <= 2 max(|X|,|Y|), so (X+Y)^2 <= 4 max(X^2,Y^2) <= 4(X^2+Y^2)
    => E[(X+Y)^2] <= 4E[X^2] + 4E[Y^2] < infinity
    the 4 can be replaced by 2, but this is enough.
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