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Show that if the random variables X and Y are independent and for some $p > 0$: $E(|X+Y|^p)<\infty$, then $E(|X|^p)<\infty$ and $E(|Y|^p)<\infty$. I'd share thoughts or working out, but unfortunately I have none worth sharing.

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Use fubini to show $\mathbb E \vert X + y \vert^p < \infty$ for almost every y –  mike Aug 29 '12 at 16:05
    
Sorry, but I don't understand this. –  Hargrove Aug 29 '12 at 17:00
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I remember this problem from Chung's book. You can write this with joint distributions. $\mathbb E \vert X + Y \vert^p < = \int \int \vert x + y \vert^P d \mu(x) d\mu(y) = \int (\int \vert x + y \vert^P d \mu(x))d\mu(y) < \infty$. Since the integral is finite, the integrand must be finite for a.e. y. For any such y $\int \vert x + y \vert^P d \mu(x) = \mathbb E \vert X + y \vert^p < \infty$. –  mike Aug 29 '12 at 17:53
    
@mike $X$ and $Y$ doesn't necessarily share the same distribution, but the proof is also ok. –  Ahriman Aug 29 '12 at 18:22
    
Does $\mathbb{E}(|X+y|^p) < \infty$ immediately imply that $\mathbb{E}(|X|^p)<\infty$? –  Hargrove Aug 30 '12 at 16:19

1 Answer 1

up vote 3 down vote accepted

Prove that for every $t> 0$ and $\epsilon > 0$, one has $P(|X+Y| > t) \ge P(|X| > t + \epsilon) P(|Y| \le \epsilon)$.

Then use the formula which gives absolute moments of a random variable $Z$ in terms of an integral of the distribution function $P(|Z| > t)$ : that is $E|Z|^p = \int_0^{\infty} p t^{p-1} P(|Z| >t)dt$.

EDIT : Further hints

For the first part, you just have to show that $\{|X| > t+ \epsilon, |Y| \le \epsilon \} \subset \{|X+Y| > t\}$ and then use independance.

For the second part, use the recalled formula to prove that $E|X|^p$ is finite, using the majoration $P(|X|>t) \le \frac{P(|X+Y| > t- \epsilon)}{P(|Y| \le \epsilon)}$, valid for all $t > \epsilon$.

(you can choose once for all $\epsilon > 0$ such that $P(|Y| \le \epsilon) > 0$ (why ?))

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+1. One could take $\epsilon=1$, to emphasize that this is merely a fixed value--but this is a matter of taste. –  Did Aug 29 '12 at 15:19
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@did No, one has to choose an $\epsilon$ such that $P(|Y| \le \epsilon) > 0$. –  Ahriman Aug 29 '12 at 15:20
    
Right. I was sloppy. Thanks. –  Did Aug 29 '12 at 15:22
    
I appreciate your help, but I still don't get it. Sorry. –  Hargrove Aug 29 '12 at 17:50
    
@Hargrove I gave further details. Hard to guess exactly where you need help. Did you understand the main lines of the proof ? Are you only stucked on technical details ? –  Ahriman Aug 29 '12 at 18:15

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