# Does k moments imply k + $\epsilon$ moments?

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).

• 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 Aug 15, 2022 at 10:57
• 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. Aug 16, 2022 at 11:15
• 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. Sep 6, 2022 at 12:49
• @markowitz In this context it is clear that fractional moments are of interest. Not all moments must be integral!
– whuber
Sep 6, 2022 at 15:01

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 $$\varepsilon>0$$, $$\mathbb E\left[Y^{1+\varepsilon}\right]=\sum_{j=2}^\infty j^{1+\varepsilon}\mathbb P(Y=j)=c^{-1}\sum_{j=2}^\infty\frac 1{j^{1-\varepsilon}(\log j)^2}=\infty.$$