CLT with random (Binomial) number of summands

Question

Say we have iid real-valued $X_1,X_2,\ldots$ with expectation $\theta\in\mathbb{R}$ and variance $\sigma^2>0$. Then clearly $$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i-\theta\right)\xrightarrow[n\rightarrow\infty]{\mathcal{L}}Z\sim\mathcal{N}(0,\sigma^2),$$ but what happens if instead of $n$ we have some $Z_n\sim\mathrm{Bin}(n,p)$, $p\in(0,1)$, independent of the $X_i$? In order to have everything well-defined, say we consider $$\sqrt{Z_n+1}\left(\frac{1}{Z_n+1}\sum_{i=1}^{Z_n+1} X_i-\theta\right).$$ Through the tower property for conditional expectation and using $$E[1/(Z_n+1)]=\frac{1-(1-p)^{n+1}}{p(n+1)}=:v_n,$$ we see that $$E\left[\frac{1}{Z_n+1}\sum_{i=1}^{Z_n+1} X_i\right]=\theta$$ and $$\mathrm{Var}\left[\frac{1}{Z_n+1}\sum_{i=1}^{Z_n+1} X_i\right]=v_n\cdot\sigma^2,$$ so that I would expect $$\sqrt{n+1}\left(\frac{1}{Z_n+1}\sum_{i=1}^{Z_n+1} X_i-\theta\right)\xrightarrow[n\rightarrow\infty]{\mathcal{L}}Z\sim\mathcal{N}(0,\sigma^2/p)$$ (since $(n+1)v_n$ converges to $1/p$). But is this correct and how can I justify it?

Answer 1

Use characteristic functions. Conditioned on the sequence $\{Z_n\}$ the characteristic function converges to $e^{-t^{2}\sigma ^{2}/2}$ since, by SLLN, $Z_n \to \infty $ a.s..; you can take expectation and apply Dominated Convergence Theorem.