# LLN and CLT on a sum with a random number of terms

Consider a random sequence $(N_n)_{n \geq 1} \subset \mathbb{N}^*$ such that $\mathbb{P}( \lim N_n = \infty ) = 1$

Let $(X_n)_{n \geq 1}$ be an iid sequence with $m = \mathbb{E}[X_1]$ and $\sigma^2 = \mathbb{V}[X_1]$

We define $Z_n = \sqrt{\frac{n}{\sigma^2}}( \overline{X_n} - m )$

$\bullet$ Show that $( \overline{X}_{N_n})_{n \geq 1} \rightarrow m \:$ almost surely

$\bullet$ Show that $(Z_{N_n})$ converges in distribution towards $\mathcal{N}(0,1)$

How could I justify neatly this exercise?

I've tried writing a few things ,

Let $A = \{ \lim N_n = \infty \}$ , $\forall \omega \in A$, we get : $\frac{1}{N_n( \omega) }\sum_{k=1}^{N_n( \omega) } X_k (\omega) \rightarrow m$ by LLN ?

With the CLT we get $Z_n \rightarrow \mathcal{N}(0,1)$ and by convergence of measures it follows that

For $\epsilon > 0$

So $\exists K_1 \geq 1, \forall n \geq K_1, | \mathbb{E}[f(Z_n)] - \mathbb{E}[f(Y)] | \leq \epsilon/2 \: \: \:$ with f continuous and bounded

And as $N_n \rightarrow \infty$ in probability this means that

$\exists K_2 > 0 \geq 1, \forall n \geq K_2, \mathbb{P}(N_n < K_1) \leq \epsilon/ f(2M)$

So, $\forall n \leq K_2,$ $| \mathbb{E}[f(Z_{N_n}) \mathbb{1}_{ (N_n < K_1) \cup ( (N_n \geq K_1) } ] - \mathbb{E}[f(Y)] | \leq \epsilon /2 + | \mathbb{E}[f(Z_{N_n}) \mathbb{1}( N_n < K_1) ]$ ..... ?

What would be a correct way of writing all this? Thanks in advance

• What do you mean exactly by a "random sequence $N_n$"? – Math1000 Aug 1 '18 at 8:51
• @Math1000 $N_n$ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process. – Psylex Aug 6 '18 at 9:47