Conditional expectation: Is $X/E[X \mid \mathcal{G}] \in L^p$?

Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space, and let $X$ be a strictly positive random variable with finite moments of all orders (i.e. $E[X^q] < \infty$ for all $1 \le q < \infty$). Let $\mathcal{G} \subset \mathcal{F}$ be a sub-$\sigma$-field. For $p > 1$, I would like to know whether it holds that $$E \left[ \left( \frac{X}{E[X \mid \mathcal{G}]} \right)^p \right] < \infty.$$ It clearly holds when $p=1$, when $X$ is $\mathcal{G}$-measurable, and when $X$ is independent of $\mathcal{G}$. But I cannot find a proof or counterexample in general; the inequalities I tried seem to go the wrong direction.

Note that we do not assume $E[X^{-1}] < \infty$.

Answer 1

No, it's not necessarily true. Fix $p>1$. Suppose $a_n > 0$ for all $n\in\mathbb{N}$ and $\sum a_n = 1$. Suppose $0\le q_n \le 1$ for all $n\in\mathbb{N}$ and that $q_n$ is small enough that $$ (*)\ \ \ \ \ \ \ \frac{a_n}{q_n^{p-1}} \ge 2. $$ Also, choose $v_n$ such that $0<v_n<1$ and $$ (**)\ \ \ \ \ \ \ \ a_n q_n \left( \frac{1}{q_n + (1-q_n) v_n} \right)^p \ge 1. $$ Such a choice is always possible by continuity and the fact that, by $(*)$, the condition would hold with plenty of margin if $v_n=0$.

Take $\Omega = \mathbb{N}\times \{H,L\}$ with the discrete $\sigma$-algebra and probability measure given by \begin{eqnarray} \mathbb{P} ( n, L ) &=& a_n (1 - q_n) \\ \mathbb{P} ( n, H ) &=& a_n q_n. \end{eqnarray} Let $X$ be the random variable such that $X(n, H) = 1$ and $X(n,L) = v_n$. Since $0<X\le1$, $X$ is strictly positive and $\mathbb{E}(X^s)\le1$ for all $s\ge 0$. Let $\mathcal{G}$ be the $\sigma$-algebra generated by all events of the form $\{(n, H), (n,L)\}$. Informally, in $\mathcal{G}$, $n$ is known but it is not known whether $H$ or $L$ occured. Then $$ \mathbb{E}(X|\mathcal{G}) = q_n + (1-q_n) v_n, $$ so \begin{eqnarray} \mathbb{E}\left[\left(\frac{X}{\mathbb{E}(X|\mathcal{G})}\right)^p\right] &=& \sum_n a_n q_n \left(\frac{1}{q_n + (1-q_n) v_n}\right)^p + a_n (1-q_n) \left(\frac{v_n}{q_n + (1-q_n) v_n}\right)^p. \end{eqnarray} The series diverges by $(**)$.