Show that X+Y has a finite second moment

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Discussion Overview

The discussion revolves around the proof that if random variables X and Y have finite second moments, then the sum X+Y also has a finite second moment. Participants explore the implications of inequalities related to expectations and the conditions under which these expectations are finite.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants present the inequality (X+Y)^2 ≤ X^2 + Y^2 + 2|XY| and derive E[(X+Y)^2] ≤ E(X^2) + E(Y^2) + 2E(|XY|), questioning the finiteness of E(|XY|).
  • Others argue that if X and Y have finite second moments, then they also have finite moments of lower orders, suggesting E[|XY|]^2 ≤ E(X^2)E(Y^2) is finite.
  • A participant seeks clarification on whether E(|XY|) is finite and how it relates to the Cauchy-Schwarz inequality, noting the difference in placement of absolute values in the expectations.
  • Some participants discuss the implications of the Cauchy-Schwarz inequality, with one suggesting that E[|XY|^2] ≤ E(X^2)E(Y^2) could be used to show finiteness.
  • Another participant points out that the existence of a finite second moment implies the existence of a finite first moment, but questions the application of the Cauchy-Schwarz inequality in this context.
  • A later reply proposes a different approach, suggesting that |X+Y| ≤ 2 max(|X|,|Y|) could lead to a bound on E[(X+Y)^2].

Areas of Agreement / Disagreement

Participants express uncertainty about the finiteness of E(|XY|) and the application of the Cauchy-Schwarz inequality. There is no consensus on how to definitively prove that X+Y has a finite second moment based on the provided arguments.

Contextual Notes

Participants highlight the need for clarity regarding the assumptions made about the moments of X and Y, and the implications of inequalities used in the discussion. The relationship between different forms of the Cauchy-Schwarz inequality is also a point of contention.

kingwinner
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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!
 
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kingwinner said:
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!

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:

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

where the RHS is finite because of the assumptions about the second-order moments being finite.
 
Hi statdad,

statdad said:
This uses a classic inequality (really from measure theory, but applied to probability).
Is it in any way related to the Cauchy-Schwartz inequality?

Essentially, if both [tex]X,[/tex] have finite second moments (variances exist) then they have finite moments of every lower order.
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 your specific case:
[tex] E[|XY|]^2 \le E(X^2) E(Y^2) < \infty[/tex]
where the RHS is finite because of the assumptions about the second-order moments being finite.
For the left side of the inequality, do you mean E(|XY|^2) or [E(|XY|)]^2 ?


Thanks for your help!:smile:
 
I am still stuck on this problem and would appreicate if anyone could help me out...
 
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

[tex] E[|XY|^2] \le E(X^2) E(Y^2)[/tex]
 
statdad said:
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

[tex] E[|XY|^2] \le E(X^2) E(Y^2)[/tex]
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)
3c2d62f6a6a33c74752cd006b8034541.png

http://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality#Probability_theory

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

[tex] 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[/tex]

so the existence of a finite second moment gives the existence of the finite first moment.
 
statdad said:
Consider [tex]X[/tex] - the proof for [/tex] Y [/tex] is similar.

[tex] 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[/tex]

so the existence of a finite second moment gives the existence of the finite first moment.
??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:
kingwinner said:
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!

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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