1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.