Let $Y_n,X_n$ be a sequence of r.v. Does $\sup_z \Big|P(Y_nLet $Y_n,X_n$ be a sequence of random variables. Does $\sup_z \Big|P(Y_n<z\ |\ X_n) - \Phi(z)\Big|\xrightarrow[]{p}0$ implies $\{Y_n|X_n=x_n\}\xrightarrow[]{d}N(0,1)?$ where $x_n$ is any sequence of real numbers within the support of $X_n$. Note that $P(Y_n<z\ |\ X_n)$ is a random variable.
By asymptotic equivalence, the above implies for all value of $z$, $P(Y_n<z\ |\ X_n)\xrightarrow[]{d}\Phi(z)$ which implies $P(Y_n<z\ |\ X_n)\xrightarrow[]{p}\Phi(z)$ for all $z$. This is as far as I could go.
This question is closely related to this other question.
 A: Let $\mathcal{P}(\mathbb{R})$ denote the space of Borel probability measures on $\mathbb{R}$ equipped with the topology of convergence in distribution. Also, ​for each $n$, let $\mu_n(\cdot) = \mathbf{P}(Y_n \in \cdot \mid X_n)$ denote the regular conditional distribution of $Y_n$ given $X_n$. Note that each $\mu_n$ is a $\mathcal{P}(\mathbb{R})$-valued random variable.
Then we claim that $\mu_n$ converges weakly to $\mathcal{N}(0, 1)$ in probability, in the sense that for each neighborhood $U$ of $\mathcal{N}(0, 1)$,
$$ \lim_{n\to\infty} \mathbf{P}(\mu_n \notin U) = 0. $$
By the Portmanteau theorem, this is equivalent to showing that for each bounded, continuous $f : \mathbb{R} \to \mathbb{R}$,
$$ \mu_n [f] = \mathbf{E}[f(Y_n) \mid X_n] \to \mathbf{E}[f(Z)] \quad \text{in probability}, $$
where $Z \sim \mathcal{N}(0, 1)$.

To this end, write
$$ D_n(g) = \left| \mu_n[g] - \mathbf{E}[g(Z)] \right| \qquad\text{and}\qquad M_n
= \sup_{z\in\mathbb{R}} \left| D_n(\mathbf{1}_{(-\infty, z)}) \right|. $$
Then each $M_n$ defines a random variable, and the assumption corresponds to the fact that $M_n \stackrel{p}\to 0$. Also, for each $-\infty \leq a < b \leq +\infty$, we have
\begin{align*}
\mu_n([a, b))
= \mu_n[\mathbf{1}_{[a, b)}]
&\leq D_n(\mathbf{1}_{[a, b)}) + \mathbf{P}(Z \in [a, b)) \\
&\leq D_n(\mathbf{1}_{(-\infty, b)}) + D_n(\mathbf{1}_{(-\infty, a)}) + \mathbf{P}(Z \in [a, b)) \\
&\leq 2M_n + \mathbf{P}(Z \in [a, b))
\end{align*}
and hence
\begin{align*}
D_n(f \mathbf{1}_{[a, b)})
&\leq \biggl(\sup_{[a, b)} |f| \biggr) \bigl[ \mu_n([a, b)) + \mathbf{P}(Z \in [a, b)) \bigr] \\
&\leq \biggl(\sup_{[a, b)} |f| \biggr) \bigl[ 2M_n + 2\mathbf{P}(Z \in [a, b)) \bigr].
\end{align*}
Now, let $\varepsilon > 0$ be arbitrary, and choose $\eta, R, \delta > 0$ so that

*

*$(2+4\sup|f|)\eta < \varepsilon$,

*$\mathbf{P}(Z \geq R) < \eta$, and

*$|f(x) - f(y)| < \eta$ whenever $x, y \in [-R, R]$ and $|x - y| < \delta$,

Also. let $-R = x_0 < x_1 < \ldots < x_N = R$ be such that $|x_{k+1} - x_k| < \delta$. We first bound $D_n(f)$ as
\begin{align*}
D_n(f)
&\leq D_n(f\mathbf{1}_{(-\infty, -R)}) + D_n(f\mathbf{1}_{[R, \infty)}) + \sum_{k=0}^{N-1} D_n(f\mathbf{1}_{[x_k, x_{k+1})}).
\end{align*}
Then by noting that
\begin{align*}
&D_n(f\mathbf{1}_{(-\infty, -R)}) + D_n(f\mathbf{1}_{[R, \infty)}) \\
&\leq (\sup|f|)(4M_n + 2\mathbf{P}(Z < -R) + 2\mathbf{P}(Z \geq R)) \\
&\leq (\sup|f|)(4M_n + 4\eta)
\end{align*}
and
\begin{align*}
&\sum_{k=0}^{N-1} D_n(f\mathbf{1}_{[x_k, x_{k+1})}) \\
&\leq \sum_{k=0}^{N-1} D_n( (f - f(x_k)) \mathbf{1}_{[x_k, x_{k+1})} ) +  \sum_{k=0}^{N-1}|f(x_k)| D_n(\mathbf{1}_{[x_k, x_{k+1})}) \\
&\leq \eta \sum_{k=0}^{N-1} \bigl[ \mu_n( [x_k, x_{k+1}) ) + \mathbf{P}(Z \in [x_k, x_{k+1})) \bigr] + (\sup |f|) \sum_{k=0}^{N-1} D_n(\mathbf{1}_{[x_k, x_{k+1})}) \\
&\leq  2\eta+ 2 (\sup|f|) N M_n,
\end{align*}
we obtain
$$ D_n(f) \leq (2+4\sup|f|)\eta + (\sup|f|)(2N + 4) M_n. $$
Using this and $M_n \stackrel{p}\to 0$ together, we conclude
$$ \lim_{n\to\infty} \mathbf{P}( D_n(f) > \varepsilon) = 0. $$
Since $\varepsilon > 0$ is arbitrary, this implies that $D_n(f) \stackrel{p}\to 0$ and therefore the desired conclusion follows.

Remark. If you look at the proof closely, all we need is to assume that
$$ \mathbf{P}(Y_n < z \mid X_n) \stackrel{p}\to \mathbf{P}(Z < z), \qquad\text{or equivanetly,}\qquad D_n(\mathbf{1}_{(-\infty, z)}) \stackrel{p}\to 0 $$
for each $z \in \mathbb{R}$. I am not sure if the (seemingly) stronger assumption $M_n \stackrel{p}\to 0$ leads to a stronger conclusion.
