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I've been reading some discussions regarding the difference between undefined and infinite moments like this answer.

My book states that the kth moment only exists when $E(|X|^k)$ is finite. If $E(|X|^k)\to\infty$, does this mean that, for $k$ even, then the kth moment diverges to $\infty$, and for $k$ odd, kth moment is either an indeterminant form like $\infty-\infty$ or diverges to $\pm\infty$?

Yandle
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  • Since you have absolute values under $E$ it can not possibly be $-\infty$ or $\infty-\infty$. It is either zero, or positive finite, or positive infinite. You are probably confusing moments with absolute moments. – Conifold Jun 24 '19 at 05:50
  • @Conifold Sorry that was a typo, the last expectation should be the kth moment $E(X^k)$, not the absolute moment – Yandle Jun 24 '19 at 20:21
  • All of them should be that for the post to make sense, including the one in the title. – Conifold Jun 24 '19 at 20:23
  • @Conifold My question actually relates to the distinction between the moment and absolute moment when the latter $\to\infty$. My book states that $\mu_k = E(X^k)=\int_{-\infty}^{\infty}x^kf_X(x)dx$ only exists when $E(|X|^k)=\int_{-\infty}^{\infty}|x|^kf_X(x)dx$ is finite. I am curious on the possible values for $\mu_k=E(X^k)$ if $E(|X|^k)\to\infty$. – Yandle Jun 24 '19 at 20:34

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$E|Y| <\infty$ iff $EY^{+} <\infty$ and $EY^{-} <\infty$. If both the conditions fail then $EY$ is not defined and $E|Y|=\infty$. If one of them is finite and the other one is $\infty$ then $EY=\infty$ or $EY=-\infty$. Apply this to $Y=X^{k}$. Note that for $k$ even there is no chance of $EX^{k}$ being $-\infty$.

Kavi Rama Murthy
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