Convergence in distribution of a sequence of contrasts applied to a random vector which is converging to a multivariate normal

Question

Suppose that we know that $$\Sigma_n^{-1/2}(\hat{\theta}_n - \theta^*_n) \overset{\text{d}}{\longrightarrow} N_{k}(0, I_{k})$$ where $\hat{\theta}_n$ is a random vector of length $k$, $\theta_n^*$ is a deterministic vector of length $k$, and $\Sigma_n$ is a deterministic diagonal matrix whose entries are all positive for all $n$. Suppose also that we have a sequence of contrasts taking the form of a vector $u_n \in \mathbb{R}^k$ such that $\Vert u_n \Vert_2^2 = 1$ for all $n$. Note that $\theta_n^*$ does not necessarily converge to anything, so we can't say that $\hat{\theta}_n$ is consistent for $\theta_n^*$, but we do know that $\hat{\theta}_n - \theta^*_n \overset{\text{p}}{\rightarrow} 0$.

I am interested in the asymptotic distribution of $(u_n^\top \Sigma_n u_n)^{-1/2} (u_n^\top \hat{\theta}_n - u_n^\top \theta_n^*)$.

As a heuristic argument, if we let ourselves be sloppy and incorrect with our asymptotics by letting $n$ appear on the right hand side of the limiting distribution, then we can begin by saying that our original convergence implies that $$\hat{\theta}_n - \theta^*_n \overset{\text{d}}{\longrightarrow} N_k(0, \Sigma_n),$$ which then implies that $$u_n^\top \hat{\theta}_n - u_n^\top \theta_n^* \overset{\text{d}}{\longrightarrow} N(0, u_n^\top \Sigma_n u_n),$$ giving a result I would love to have, which is that $(u_n^\top \Sigma_n u_n)^{-1/2} (u_n^\top \hat{\theta}_n - u_n^\top \theta_n^*) \overset{\text{d}}{\longrightarrow} N(0,1)$.

However, obviously the above is just a heuristic and isn't a real proof. I know that the result would clearly be false for a general $u_n$, because the dependence on $n$ would allow the entries of $u_n$ to explode at a rate that makes the final result impossible. However, the fact that $\Vert u_n \Vert_2^2 = 1$ means that it may be okay?

If it changes much, I would be okay with letting $u$ not depend on $n$ and just having that $\Vert u \Vert_2^2 = 1$.

Thanks!

Answer 1

If $\|u_n\|_2$ is bounded, then you have the convergence you need.

For fixed $u$ this is the Mann-Wold theorem: convergence in distribution of a vector is equivalent to convergence of all one-dimensional projections.

For $u_n\to u$: vector dot product $(u_n,\hat\theta)\mapsto u^t_n\hat\theta$ is continuous, so use the continuous mapping theorem.

Finally, for bounded $u_n$, take subsequences. Any subsequence of a bounded sequence has a convergent subsubsequence. Along this subsubsequence $\Sigma_n^{-1/2}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$. But that is just to say that $\Sigma_n^{-1/2}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$ for the whole sequence.