# Uniform integrability of conditional quantile functions

Let $$Z^n$$ be $$\mathbb{R}$$-valued random variables which are uniformly integrable, i.e. $$\lim_{a \to \infty} \sup_{n} E[1_{\{|Z^n| \geq a\}} |Z^n|] = 0.$$ Let $$X^n \to N(0,1)$$ in distribution, and $$X^n$$ uniformly integrable, and consider the conditional quantile evaluated at some fixed $$p \in (0,1)$$, i.e. $$Q^n(p) = \min\{z \in \mathbb{R} \colon \mathbb{P}(Z^n \leq z \mid X^n) \geq p\}.$$ $$\mathbf{Question:}$$ Is $$(Q^n(p))_{n \in \mathbb{N}}$$ also uniformly integrable?

$$\mathbf{Approach}$$: I have shown that $$E[|Z^n| \mid X^n]$$ is u.i. However, I am struggling with bounding the conditional quantile to get something of the sort $$|Q^n(p)| \leq A + B E[|Z^n| \mid X^n],$$ for some $$A, B >0$$, in which case I would be done.

Thanks!

We will assume that $$Z^n$$ is non-negative
Writing the conditional quantile as $$Q^n(p) = \inf\left\{z \in \mathbb{R} \colon \mathbb{P}(Z^n > z \mid X^n) \leqslant 1- p\right\}.$$ and noticing that $$z\mathbb{P}(Z^n > z \mid X^n) =\mathbb E\left[z\mathbf{1}\{Z^n > z\}\mid X^n\right]\leqslant \mathbb E\left[ Z^n \mid X^n\right]$$ we get the inclusion $$\left\{z \in \mathbb{R} \colon \mathbb{P}(Z^n > z \mid X^n) \leqslant 1- p\right\}\supset \left\{z\in\mathbb R: \mathbb E\left[ Z^n \mid X^n\right]\leqslant (1-p)z\right\}$$ hence $$\inf\left\{z \in \mathbb{R} \colon \mathbb{P}(Z^n > z \mid X^n) \leqslant 1- p\right\}\leqslant \inf\left\{ z\in\mathbb R: \mathbb E\left[ Z^n \mid X^n\right]\leqslant (1-p)z\right\}=\frac{ \mathbb E\left[ Z^n \mid X^n\right]}{1-p},$$ which is the wanted inequality.
Note that we do not need any assumption on $$X^n$$, because uniform integrability is preserved by taking conditional expectation, no matter what the conditioning $$\sigma$$-algebra is.