Show that X+Y has a finite second moment

[kingwinner]

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!

 

[statdad]

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 X, have finite second moments (variances exist) then they have finite moments of every lower order. For your specific case:

<br /> E[|XY|]^2 \le E(X^2) E(Y^2) &lt; \infty<br />

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

 

[kingwinner]

Hi statdad,

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 X, 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:
<br /> E[|XY|]^2 \le E(X^2) E(Y^2) &lt; \infty<br />
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!

 

[kingwinner]

I am still stuck on this problem and would appreicate if anyone could help me out...

 

[statdad]

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

<br /> E[|XY|^2] \le E(X^2) E(Y^2)<br />

 

[kingwinner]

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

<br /> E[|XY|^2] \le E(X^2) E(Y^2)<br />

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)

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!

 

[statdad]

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

<br /> 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] &lt; \infty<br />

so the existence of a finite second moment gives the existence of the finite first moment.

 

[kingwinner]

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

<br /> 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] &lt; \infty<br />

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

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!

 

[gel]

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.