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Proof that if higher moment exists then lower moment also exists

For a random variable $X$, lets say I know $E[X^k]$ is finite and I know that $E[X]$ is finite. Can I say that all moments between first and kth moment are finite?

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bdeonovic
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Yes. In fact, you don't even need to know that $E[X]$ is finite: if you know that the $k$-th moment $E[X^k]$ is finite, then all lower moments must be finite.

You can see this using Jensen's inequality, which says that for any convex function $\varphi$ and random variable $X$, $$\varphi(E[X]) \leq E[\varphi(X)].$$ Now, suppose we know that $E[X^k]$ is finite, and we want to check whether $E[X^m]$ is finite where $m < k$. Let $\varphi(X) = X^{k/m}$. Since $k>m$, this is a convex function. Then by Jensen's inequality, $$(E[X^m])^{k/m} \leq E[(X^m)^{k/m}] = E[X^k].$$ We know that $E[X^k]$ is finite, so $(E[X^m])^{k/m}$ must also be finite, and so $E[X^m]$ is finite.

In summary: given that the $k$-th moment is finite, all lesser moments must also be finite.

Jonathan Christensen
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    This lacks rigor, starting from the very formulation of Jensen's inequality... which should say something to the extent of "... provided the necessary expectations are finite". – StasK Jan 25 '13 at 02:32
  • @StasK Guilty as charged. I was written an exposition, not a textbook. – Jonathan Christensen Jan 25 '13 at 02:34
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    Thanks guys, I tried searching for this but was having trouble coming up with the right queries. – bdeonovic Jan 25 '13 at 02:36