If you have a random variable $X$ such that $\mathbb{E}(|X|^k) < \infty$, does it follow that $\mathbb{E}(|X|^{k+\epsilon}) < \infty$ for some (potentially small) $\epsilon > 0$? If not, what is a good counterexample? I.e. a random variable such that $\mathbb{E}(|X|^k) < \infty$ but $\mathbb{E}(|X|^{k+\epsilon}) = \infty$ for all $\epsilon > 0$.

I know that in general $\mathbb{E}(|X|^{k+\delta}) < \infty$ some $\delta > 0$ would imply that $\mathbb{E}(|X|^k) < \infty$, but I have not been able to find e.g. an example of a random variable which only has second moments (for instance) and not anything higher at all.

I do know that it is possible to have everything up to k moments and not have the kth moment (see e.g. the example in the first answer here: https://math.stackexchange.com/questions/1955968/example-of-random-variable-that-is-integrable-but-have-infinite-second-moment).

  • $\begingroup$ The t distribution for example has $v$th moments when its degrees of freedom are larger than $v$, see e.g. en.wikipedia.org/wiki/Student%27s_t-distribution $\endgroup$ Aug 15, 2022 at 10:57
  • $\begingroup$ Yes that's true but it's not a counterexample. For the $t$-distribution with $v$ degrees of freedom you have all moments up to $v$ but not $v$. But that means that (letting $T$ have a $t$-distribution with $v$ degrees of freedom), for all $k$ such that $\mathbb{E}(|T|^k) < \infty$, you also have $\mathbb{E}(|T|^{k+\epsilon}) < \infty$ for some $\epsilon$ so this isn't a counterexample. $\endgroup$ Aug 16, 2022 at 11:15
  • $\begingroup$ If $k$ and $\epsilon$ are positive integers the counterexample work, your point not. Moments are usually intended as integers: first, second, etc ... I intended my reply (see below) in this sense. $\endgroup$
    – markowitz
    Sep 6, 2022 at 12:49
  • $\begingroup$ @markowitz In this context it is clear that fractional moments are of interest. Not all moments must be integral! $\endgroup$
    – whuber
    Sep 6, 2022 at 15:01

1 Answer 1


It suffices to find a counter-example for $k=1$, that is, a random variable $Y\geqslant 0$ such that $\mathbb E[Y]$ is finite but $\mathbb E\left[Y^{1+\varepsilon}\right]$ is infinite and $X=Y^k$ will give the wanted counter-example.

Let $c=\sum_{j=2}^\infty \left(j \log j\right)^{-2}$ and let $p_j=c^{-1} \left(j \log j\right)^{-2}$, so that $\sum_{j\geqslant 2}p_j=1$. Let $Y$ be a random variable taking the value $j$ (for $j\geqslant 2$) with probability $p_j$. Then $$ \mathbb E[Y]=\sum_{j=2}^\infty j\mathbb P(Y=j)=c^{-1}\sum_{j=2}^\infty\frac 1{j(\log j)^2}<\infty $$ and for $\varepsilon>0$, $$ \mathbb E\left[Y^{1+\varepsilon}\right]=\sum_{j=2}^\infty j^{1+\varepsilon}\mathbb P(Y=j)=c^{-1}\sum_{j=2}^\infty\frac 1{j^{1-\varepsilon}(\log j)^2}=\infty. $$


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