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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.

Seeking Knowledge
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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
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    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
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    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](http://stats.stackexchange.com/questions/18668/need-an-example-of-rv-with-a-mean-and-no-second-moment)! (In the latter case, the usage is more conventional.) :) – cardinal Nov 20 '11 at 16:16
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    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

2 Answers2

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I believe it follows from Hölder's inequality:

LeelaSella
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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](http://en.wikipedia.org/wiki/Jensen%27s_inequality) -- might that work for non-absolute moments too? – onestop Nov 20 '11 at 14:41
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    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
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    @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
  • I would rather say it follows from Jensen's inequality: $\mathbb{E}[|X|^i]=\mathbb{E}[\{|X|^r\}^{i/r}]\le\{\mathbb{E}[|X|^r]\}^{i/r}$ when $i\le r$, since the power is then a concave function. And the existence of the expectation $\mathbb{E}[|X|^i]$ is equivalent to the existence of the expectation $\mathbb{E}[X^i]$. – Xi'an Nov 21 '11 at 07:21
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    @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
  • @cardinal: uh-uh, I had this memory from my 1983 measure theory course that $\mathbb{E}[X]$ was finite if and only if $\mathbb{E}[|X|]$ was finite, assuming $X$ is measurable. Dusty memories, I presume... – Xi'an Nov 22 '11 at 06:40
  • @Xi'an: Well, *that* is certainly true. :) But, it's not at all equivalent to your previous comment. (Existence and finiteness of the Lebesgue integral are different notions.) – cardinal Nov 22 '11 at 12:43
  • Look at theorem 1.4.1 of the book Characteristic Functions by Eugene Lukacs. The theorem says that the algebraic moment of order $k$ exists if and only if the absolute moment of order $k$ exists. This follows from basic properties of Riemann-Stieltjes Integrals. – Kasun Fernando Jan 18 '19 at 22:31
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My solution lacks rigor, but here's a rough sketch:
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 Aden
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  • well as you wrote it does not solve the problem – Seeking Knowledge Nov 20 '11 at 13:12
  • @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