# If higher order moment is finite, then lower order moment finite? [duplicate]

I'm trying to prove the theorem below.

If $E(|X|^n)<\infty$ for some positive integer $n$, then $E(|X^k|)<\infty$ for every positive integer $k$ such that $k<n$

Here's what I've tried.

[Step 1] set inequality about $|X|^k$ and $|X|^n$

if $|X|\ge1$, then $|X|^k\le|X|^n$, while if $|X|<1$, then $|X|^k\le1$.

consequently, $|X|^k \le \max(1, |X|^n) \cdots (1)$.

[Step 2] prove the right-hand side of (1) has finite expectation

if $E(|X|^n)<\infty$, then $E(|X|^n) = \int^{1}_{-1}|X|^nf(x)dx \;+\; \int_{|X|\ge1}|X|^nf(x)dx$

and it follows that $\int_{|x|\ge1}|x|^nf(x)dx<\infty$ as a difference of two finite quantities.

Thus, $E(|X|^k))\le E\left\{\max\left(1, |X|^n\right)\right\} = \int^{1}_{-1}f(x)dx \;+\; \int_{|X|\ge1}|X|^nf(x)dx < \infty$

[Question 1] In [Step 1], Why contains "=" in $|X|^k\le1$?

As I think, "If $|X| < 1$, then $|X|^k<1$" is right.

[Question 2] In [Step 2], what are 'two finite quantities' referring to?

$E(|X|^n)$ and $\int^1_{-1}|X|^nf(x)dx$?

## marked as duplicate by whuber♦ self-study StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 13 '18 at 15:16

• A broad generalization of this inequality is proven at stats.stackexchange.com/questions/234048/…. I also recall this question was asked and answered before, but I'm unable to turn up the thread in a search. – whuber Mar 11 '18 at 16:04

[Question 1] : (Since k>0,"=" is just an implication)

[Question 2] : we only need to know $E(|X|^n)$ is some finite value for the proof. while $\int^1_{-1}|X|^nf(x)dx <=1$ can be derived.

below is the Proof: given, $E(|X|^n)$ is finite. some $k$, $0<k<=n$.

if $|X|<1$ then $|X|^k≤1$, So $\int^1_{-1}|X|^kf(x)dx <= \int^1_{-1}1f(x)dx <=1$ (by definition)

if $|X|>=1$ then $|X|^k≤|X|^n$, So $\int_{|X|\ge1}|X|^kf(x)dx <=\int_{|X|\ge1}|X|^nf(x)dx <= E(|X|^n)$

from above 2 cases, we get $E(|X|^k)<= 1 + E(|X|^n)$. hence $E(|X|^k)$ is finite.

• Your limitation of the integral endpoints to $\pm 1$ trivializes the problem: it's then impossible for any positive moment to be infinite, period. The interest lies in its generality: that is, its applicability to all random variables. – whuber Mar 11 '18 at 16:06