# What is the limit distribution of $\frac{S_{N_n}}{\sigma \sqrt{a_n}}$ as $n\rightarrow \infty$.

Let $X_1, X_2, X_3,...$ be iid with $\mathbb E[X_i]=0$ and $\operatorname{Var}[X_i]=\sigma^2>0$, and let $S_n = \Sigma_{i=1}^{n} X_i$. Let $N_n$ be a sequence of integer valued random variables independent of $X_i$, $i \geq 1$, and let $a_n$ be a sequence of positive integers with $\frac{N_n}{a_n}\rightarrow 1$ in probability and $a_n\rightarrow \infty$ as $n \rightarrow \infty$. What is the limit distribution of $\frac{S_{N_n}}{\sigma \sqrt{a_n}}$ as $n\rightarrow \infty$.

It looks like a CLT question but now I have trouble dealing with the ${S_{N_n}}$, besides I didn't use the condition "$\frac{N_n}{a_n}\rightarrow 1$ in probability". Any hints will be appreciated.

• Hints: 1. If $Z_n\to Z$ in distribution, $(N_n)$ is independent of $(Z_n)$, and $N_n\to\infty$ in probability, then $Z_{N_n}\to Z$ in distribution. 2. If $U_n\to U$ in distribution and $K_n\to1$ in probability then $K_nU_n\to U$ in distribution. Hence... – Did May 25 '16 at 15:44
• @Did Hence $\frac{S_{N_n}}{\sigma \sqrt{a_n}}=\frac{S_{N_n}}{\sigma \sqrt{N_n}} \times \frac{\sqrt{N_n}}{ \sqrt{a_n}}$, which converges to $\mathbb N(0,1)$? – hil316 May 25 '16 at 17:06
• Indeed.   – Did May 25 '16 at 17:13