Convergence Question Suppose that $X_m \in \mathbb{R}^d$ and $W_m \in \mathbb{R}^{d \times d}$ be a sequence of random variables such that the following asymptotic statements are true
\begin{equation*}
   \begin{aligned}
        \sqrt{m}\left(X_{m}-X_{0}\right) &\stackrel{d}{\rightarrow} Z \sim \mathcal{N}(0, \Omega) \\
         W_{m} &\stackrel{p}{\rightarrow} W_{0}
    \end{aligned}
\end{equation*}
where $W_0$ is a constant. I want to find the asymptotic distribution of $\sqrt{m}\left(W_{m} X_{m}-W_{0} X_{0}\right)$, so I proceed as follows.
\begin{equation*}   
   \begin{aligned}
        \sqrt{m}\left(W_{m} X_{m}-W_{0} X_{0}\right)&=\sqrt{m}\left(W_{m} X_{m}-W_{m} X_{0}+W_{m} X_{0}-W_{0} X_{0}\right) \\
        &= \sqrt{m}\left(W_{m} X_{m}-W_{m} X_{0}\right)+\sqrt{m}\left(W_{m}-W_{0}\right) X_{0}
    \end{aligned}
\end{equation*}
Using Slutsky's theorem, the first term has the following asymptotic distribution.
\begin{equation} 
  \sqrt{m}\left(W_{m} X_{m}-W_{m} X_{0}\right) \stackrel{d}{\longrightarrow} W_0Z \sim \mathcal{N}\left(0, W_0 \Omega W_0^{\top}\right).
\end{equation}
For the second term we have,
\begin{equation} 
 \sqrt{m}\left(W_{m}-W_{0}\right) X_{0}=o_{p}(1).
\end{equation}
So the claim is that, $\sqrt{m}\left(W_{m} X_{m}-W_{0} X_{0}\right) \stackrel{d}{\longrightarrow} W_0 Z$. I have the following 3 questions

*

*Is it ok to add/subtract terms as I did by adding and subtracting $W_mX_0$ and splitting the expression into 2 separate terms

*Is it ok to say the second term is random variable that converges to zero ?

*Indeed I don't know the rate at which $W_m \stackrel{p} {\rightarrow} W_0$. What happens under the more stringent condition, $W_m \stackrel{a.s.} {\longrightarrow} W_0$ ?

 A: *

*Yes.


*No. You don't say why $\sqrt{m}(W_m-W_0)X_0$ would be $o_p(1)$, and it's not true in general.  The fact that $W_m\stackrel{p}{\to}W_0$ doesn't imply any rate of convergence. Suppose, for example, that $W_0=17$ and that $W_m$ has mean $17+m^{-1/4}$ and variance $m^{-1/4}$. Then $W_m$ converges to $W_0$ in probability but $\sqrt{m}(W_m-W_0)$ does not converge.
Now, the question whether that's a local counterexample or a global counterexample: is the result false or is it just that the proof is wrong?
At least if $W_0$ is constant the result is still true, as you can see using almost-sure representations. By the almost-sure representation theorem, there are versions of $X_m$, $X$, $Z$, and $W_m$, possibly defined on a different probability space, such that
$$\sqrt{m}(X_m-X_0)\stackrel{as}{\to}Z$$
and
$$W_m\stackrel{as}{\to}W_0$$
For these versions
$$\sqrt{m}(W_mX_m-W_0X_0)\stackrel{as}{\to}W_0Z$$
and so for these versions
$$\sqrt{m}(W_mX_m-W_0X_0)\stackrel{d}{\to}W_0Z$$
and since that's a result depending only on distributions for each $m$
$$\sqrt{m}(W_mX_m-W_0X_0)\stackrel{d}{\to}W_0Z$$
for any sequences having those distributions.
[the proof doesn't go through if $W_0$ is non-degenerate random variable; I don't know if the result is true, but I suspect not]
