finite variance but infinite higher moments

Question

Is it possible to find a positive random variable with finite variance such as \begin{equation*} \mathbb{E}(X^{2+\varepsilon})=+\infty \end{equation*} for all $\varepsilon > 0$ ?

Equivalently, is it possible to find a random variable $X$ for which \begin{equation*} \{p\in\mathbb{R}^{+}, \mathbb{E}(X^{p})=+\infty\} \end{equation*} is a non-empty open set?

Equivalently again, is it true or false that \begin{equation*}L^{1}=\bigcup_{p>1}L^{p} \enspace \text{?} \end{equation*}

Answer 1

Consider the series $$ S = \sum_{n=1}^\infty \frac{1}{n\log^2 n}. $$ Since $$ \int_e^\infty \frac{dx}{x\log^2 x} = \left.-\frac{1}{\log x}\right|_e^\infty = 1, $$ the series converges.

Define $q_n = 1/(n^3\log^2 n)$ and $Q = \sum_{n=1}^\infty q_n$. Define a random variable $X$ which is equal to $n \geq 1$ with probability $p_n = Q^{-1}/(n^3\log^2 n)$. Then $$ \mathbb{E}[X^2] = Q^{-1} \sum_{n=1}^\infty \frac{n^2}{n^3 \log^2n} = Q^{-1} S < \infty. $$ On the other hand, for all $\epsilon > 0$ we have $$ \mathbb{E}[X^{2+\epsilon}] = Q^{-1} \sum_{n=1}^\infty \frac{1}{n^{1-\epsilon} \log^2 n} \geq \Omega(Q^{-1}) \sum_{n=1}^\infty \frac{1}{n^{1-\epsilon/2}} = \infty. $$

Answer 2

Any two-parameter Pareto distribution where $2 < \alpha \leq 3$ will have that property, as only the moments of the Pareto less than its shape parameter exist.