# If the n(th) moment exists does it mean all smaller moments exist too?

I would like to prove the following statement:

If the $r$th moment of a random variable $X$ exists and is finite, then all moments $1$ to $r-1$ exist and are finite.

Edit: I mean the raw moments $\mathbb E X^r$. By exists I mean exists and is finite.

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You need to be careful with the word exists. In the standard sense, the statement is false. Also, you should specify whether you mean the raw moments $\mathbb E X^n$ or central moments $\mathbb E (X-\mu)^n$ since the answers are different for each case. –  cardinal Nov 20 '11 at 14:46
Also, I've seen in some statistics texts and other places that state such a result, that they will say exists when what they mean is integrable (i.e., exists and is finite). –  cardinal Nov 20 '11 at 14:47
I would encourage you to please update your question to be more precise, especially since the kind of question you are asking demands a good bit of precision. –  cardinal Nov 20 '11 at 15:31
Thanks for your edit. It is a little strange that you use exists here to mean something different from your use of exists in another very recent and related question! (In the latter case, the usage is more conventional.) :) –  cardinal Nov 20 '11 at 16:16
I have edited your edit to try to clarify. I hope you don't mind. Also, this reads a bit like homework. Please add the homework tag if that is indeed the case. –  cardinal Nov 20 '11 at 16:22
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I believe it follows from Hölder's inequality:

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Isn't that only for absolute moments though? But that link mentions same thing can be proved with Jensen's inequality -- might that work for non-absolute moments too? –  onestop Nov 20 '11 at 14:41
Holder or Jensen will work. I prefer Jensen's since it's a little more natural and easy to see. –  cardinal Nov 20 '11 at 14:43
@onestop: Note that $|\mathbb E X| \leq \mathbb E |X|$ so if $\mathbb E |X|$ is finite, so is $\mathbb E X$. –  cardinal Nov 20 '11 at 14:44
Note the presence of absolute moments in the Holder Inequality application. –  whuber Nov 20 '11 at 17:11
@Xi'an: It is not true that the existence of the expectation $\mathbb E |X|^i$ is equivalent to that of $\mathbb E X^i$. The first always exists as long as $X$ is measurable, though it may be infinite. The finiteness of the former is sufficient for the existence of the latter, but even that is not required. –  cardinal Nov 21 '11 at 16:00
Instead of using the concept of a moment centered around zero, such as $E(X)$ and $E(X^2)$, use the notion of a central moment, defined as $\mu_k=E\left[(X-\mu)^k\right]$.
From your statement, we know the $r$'th moment exists, so we know $\mu_r=E(X-\mu)^r$ exists. Use the binomial formula to expand that binomial, recalling that $(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k$. Then $\mu_r=E(X-\mu)^r = E\left[\sum_{k=0}^r {r \choose k}X^{k}\mu^{r-k}\right]$. Note now that this is a sum of all lower moments from $0, \dots r$, times some coefficient. Thus, in order for the r'th moment to exist, all lower moments must also exist.
Edit: There is a flaw, however, in using the central moments instead: We cannot assume that $\mu$ exists. Is there a way around that?
@Christopher: Regarding your last concern, it depends on what the OP means, which they haven't specified. If they are talking about central moments that this implicitly assumes that $\mu$ exists and is finite. –  cardinal Nov 20 '11 at 14:53