Uniform integrability of conditional quantile functions

Question

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!

Answer 1

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.