If two different statistics converge to a same function, does it mean both of them converge to a same distribution? Let $A(x)$ be a (deterministic) function, and let $A_n(x)$ and $A_n^*(x)$ be two different estimators of $A(x)$. Define $B_n=\sup|A_n-A|$ and $B_n^*=\sup|A_n^*-A|$. If for a sequence $(c_n)$ and a distribution function $F$


*

*$B_n=o_p\left(\frac{1}{c_n}\right)$

*$\lim\Pr(c_n B_n\leq x)=F(x)$

*$\sup|A_n-A_n^*|=o_p\left(\frac{1}{c_n}\right)$
does it mean $\lim\Pr(c_n B_n^*\leq x)=F(x)$?
My answer is "yes", but my proof is kind of strange. Since we have
\begin{eqnarray*}
B_n^*\leq\sup|A_n^*-A_n|+\sup|A_n-A|=o_p\left(\frac{1}{c_n}\right),
\end{eqnarray*}
then it is abvious that $|B_n-B_n^*|=o_p\left(\frac{1}{c_n}\right)$. From this, if $n$ is large enough, then
\begin{eqnarray*}
\Pr(c_n B_n^*\leq x)=\Pr(c_n B_n\leq x)\rightarrow F(x).
\end{eqnarray*}
I think, my intuition is right, but the proof is too heuristic and not firm. Do you guys have any idea what is my mistake and how to fix it? Thank you so much
 A: If $|B_n-B_n^{*}| = O_p \left( \frac{1}{c_n} \right)$ then $|c_nB_n-c_nB_n^{*}| = O_p \left( 1 \right)$. Since $|c_nB_n-c_nB_n^{*}|$ does not converge in probability to zero for large $n$, you cannot conclude that $Pr(c_nB_n^{*}\leq x)$ converges towards $Pr(c_nB_n\leq x)$.
Wikipedia defines the notation $o_p$ as follows:
$X_n=o_p(a_n) \ \text{ if } \ \lim_{n\rightarrow \infty}\mathrm{Pr}\left(\left|\frac{X_n}{a_n}\right|\geq \epsilon\right)=0$.
With this notation it can be deduced that $F(x)=0$ when $x<0$, that $F(x)=1$ when $x>0$ and that $\mathrm{Pr}(c_nB_n*\leq x)=F(x)$ when $x \ne 0$. However there exists a counterexample in the case when $x=0$.
Let $A(x)=0$, $A_n(x)=1$, $A_n^* (x)=0$ and $c_n=\frac{1}{n}$. It can be verified that all three conditions hold. Note that $c_nB_n=\frac{1}{n}$ and hence that $F(0)=\lim_{n\rightarrow\infty}\mathrm{Pr}(\frac{1}{n}\leq 0)=0$. On the other hand, $\mathrm{Pr}(c_nB_n^* \leq 0)=1$ since $B_n^*=0$. It follows that $\lim_{n\rightarrow\infty}\mathrm{Pr}(c_nB_n^*\leq 0)=1 \neq F(0)$.
