# Show that X+Y has a finite second moment

1. Jul 15, 2009

### 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!

2. Jul 15, 2009

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:

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

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

3. Jul 15, 2009

### kingwinner

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 ?

4. Jul 20, 2009

### kingwinner

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

5. Jul 20, 2009

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

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

6. Jul 20, 2009

### kingwinner

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.

Last edited by a moderator: Apr 18, 2017
7. Jul 20, 2009

Consider $$X$$ - 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.

8. Jul 21, 2009

### kingwinner

??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!

Last edited by a moderator: Apr 28, 2017
9. Jul 21, 2009

### gel

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.