If $E[|X|^k]$ exists, does $E[|X|^m]$ exist for $0 \leq m \leq k$?

Question

The question is:

If $E[|X|^k]$ exists, does $E[|X|^m]$ exist for $0 \leq m \leq k$? If so, can you prove it?

Here $X$ is just a generic random variable.

I don't know how to solve this problem. I thought about approaching it by first defining $g(x, i) = |x|^i$ for some $i \in \mathbb{R}^+$.

$$ E[g(x,i)] = \int_{x_{\min}}^{x_{\max}} g(x,i) f_X(x) \\ E[g(x,k)] = \int_{x_{\min}}^{x_{\max}} g(x,k) f_X(x) = c $$

where $c$ is some constant. I don't know if this is the right direction or not? Could someone advise?

@KaviRamaMurthy Ah I'm not familiar with that inequality, but I'll look into it. Is there also a more first principles approach to solving this? — user5965026, Apr 02 '21 at 00:22

score 1 · Accepted Answer · answered Apr 02 '21 at 00:26

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Yes. $$ \begin{align*} E[|X|^m] &= \int_{-\infty}^{\infty} |x|^m f_X(x) dx \\ &= \int_{ |x| \leq 1 } |x|^m f_X(x) dx + \int_{ |x| > 1 } |x|^m f_X(x) dx \\ &\leq 1 + \int_{ |x| > 1 } |x|^k f_X(x) dx \leq 1 + E[|X|^k] < \infty \end{align*} $$

answered Apr 02 '21 at 00:26

Anon

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$X$ needs not admit a probability density function. – Danny Pak-Keung Chan Apr 02 '21 at 00:32
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@DannyPak-KeungChan If you swap out the integrals for sums, don't we get an analogous result for pmfs? – user5965026 Apr 02 '21 at 00:42
Is correct, we do (just use pmf instead of pdf in that case). – Anon Apr 02 '21 at 00:54
Does your result change if it was a piecewise pdf instead of a connected one? – user5965026 Apr 02 '21 at 01:02
No, I've made no assumptions on the pdf apart from the two standard ones: always non-negative, and integral over R is 1. – Anon Apr 02 '21 at 01:03
Wait, why is $\int_{|x| > 1} |x|^k f_X(x) dx \leq E[|X|^k] = \int_{-\infty}^{\infty} |x|^k f_X(x) dx$? – user5965026 Apr 02 '21 at 02:04
LHS $\leq \int_{\mathbb{R}} |x|^k f_X(x) dx = $ RHS. – Anon Apr 02 '21 at 02:05
Hmm, but what if we have $X \sim U(-0.5, 2)$? Then $E[|X|^k] < \int_{|x| > 1} |x|^k f_X(x)dx$? – user5965026 Apr 02 '21 at 02:09
Let us continue this discussion in chat. – Anon Apr 02 '21 at 02:11

If $E[|X|^k]$ exists, does $E[|X|^m]$ exist for $0 \leq m \leq k$?

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