Let $X\_1,...,X\_n$ be iid observations from $N(0,1)$. Let $\overline{X}=\dfrac{1}{n}\sum\_{i=1}^n X\_i$ and $S^2=\dfrac{1}{n}\sum\_{i=1}^n (X\_i-\overline{X})^2$. Then is it true that $\sqrt{n}\sup\_x |\Phi(\dfrac{x-\overline{X}}{S})-\Phi(x)|\stackrel{p}{\to}
0$? Here $\Phi(.)$ is standard normal cdf.


Seems something like Berry Esseen bound but I think I am getting it is stochastically bounded and not really going to 0.



Does $\sqrt{n}\sup_x |\Phi(\dfrac{x-\overline{X}}{S})-\Phi(x)|\to0$ in probability?
No, it is not true. First fix any $x$ and consider $\overline{X^2}=\frac1n\sum\_{i=1}^nX\_i^2$, $S^2=\overline{X^2}-\bigl(\overline X\bigr)^2$ and $h(s,t)=\Phi\left(\dfrac{x-s}{\sqrt{t-s^2}}\right)$. 
Then use multivariate Delta method to prove that 
$$
\sqrt{n} \left(\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\right)=\sqrt{n}\biggl(h\biggl(\overline X, \overline{X^2}\biggr)-\underbrace{h(\mathbb EX\_1,\mathbb EX\_1^2)}\_{h(0,1)}\biggr)\xrightarrow{\mathcal D}\mathcal N(0,\sigma\_x^2),
$$
where $\sigma^2\_x = \nabla h(0,1)^T\cdot \Sigma \cdot \nabla h(0,1)$ and $\Sigma$ is a covariance matrix of $(X\_1,X\_1^2)$. 


Here $\text{Var}(X\_1)=1$, $\text{Var}(X\_1^2)=2$, $\text{Cov}(X\_1, X\_1^2)=0$, so
$$
\Sigma=\begin{pmatrix}1 & 0 \cr 0 & 2 \end{pmatrix}
$$
$$
\frac{\partial h}{\partial s}\bigg|\_{s=0, t=1} = -\varphi(x) = -\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
$$
and
$$
\frac{\partial h}{\partial t}\bigg|\_{s=0, t=1} = -\frac12\varphi(x) = -\frac{1}{2\sqrt{2\pi}}e^{-x^2/2}
$$
Please check the derivatives, although their values do not matter much. And 
$$
\sigma^2\_x = \varphi^2(x)\cdot 1+\left(\frac{\varphi(x)}{2}\right)^2\cdot 2 = \frac32\varphi^2(x).
$$
Say, for $x=0$, 
$$
\sqrt{n} \left(\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\right) \xrightarrow{\mathcal D} \mathcal N\biggl(0,\frac{3}{4\pi}\biggr)
$$
For any $\varepsilon>0$
$$
\liminf\_{n\to\infty}\mathbb P\left(\sqrt{n}\sup\_x \biggl|\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\biggr|>\varepsilon\right) \geq \liminf\_{n\to\infty}\mathbb P\left(\sqrt{n} \biggl|\Phi\biggl(\dfrac{0-\overline{X}}{S}\biggr)-\Phi(0)\biggr|>\varepsilon\right) 
$$
$$
=\mathbb P(|Y|>\varepsilon)=2\Phi\left(-\frac{2\sqrt{\pi}\varepsilon}{\sqrt3}\right)>0
$$
where $Y\sim N\bigl(0,\frac{3}{4\pi}\bigr)$.


