Let $\mathbf{x}$ be a $p$-dimensional vector, and $\mathbf{z}$ is a $p$-dimensional standard normal vector whose elements are i.i.d. Now we assume for some scaling parameter $n \to \infty$ (and fixed $p$), $$ \mathbf{x} \xrightarrow{\mathcal{L}} \mathbf{z}, $$ where $\xrightarrow{\mathcal{L}}$ is the convergence in law (or convergence in distribution). Then my question is that the following.
Question.
Can we say that for any path $(p,n) \to (\infty, \infty)$, $$ \frac{1}{p} \mathbf{x}^\top \mathbf{x} \xrightarrow{\mathbb{P}} 1?$$
For one path: $n\to \infty$ first, then $p \to \infty$ is quite obvious, but I'm not sure how to handle for any arbitrary path.
I look forward your any help.
Thanks,
It would not work for all paths. Here is a counterexample, where we first let $p\to\infty$ and then let $n\to\infty$:
Given iid $B_i\sim \text{Bern}(1/n), i=1,...,p$, each independent of $\bf z$, define the $i$th element of $\bf x$ as
$$x_i=z_i+nB_i.$$
Note that as $n\to \infty$,
$$nB_i \xrightarrow{P}0\implies nB_i \xrightarrow{D}0\underbrace{\implies}_{\text{Slutsky's theorem}} x_i\xrightarrow{D}z_i \underbrace{\implies}_{x_i \text{ iid}} {\bf x} \xrightarrow{D} {\bf z}.$$
WLLN implies as $p\to \infty$, $${1 \over p} {\bf x}^\top {\bf x} \xrightarrow{P} E[x_i^2].$$
However, note that $$E[x_i^2]=E[z_i^2]+n^2E[B_i]+2n\underbrace{E[B_iz_i]}_{=0 \text{ by indep}}=1+n\overset{n\to \infty}{\not\to} 1.$$