Find the limiting distribution of $T_n/S_n$ as $n\rightarrow \infty$

Asked May 05 '23 at 08:42

Active May 05 '23 at 08:42

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Let, $X_i~(i.i.d.)$ Bernoulli ($\lambda/n$), $n\geq\lambda\geq0. $

$Y_i~(i.i.d.)$ Poisson ($\lambda/n$), ${X_i}$ and ${Y_i} $ are independent.

Let,$\sum_{i=1}^{n^2}X_i=T_n$ and $\sum_{i=1}^{n^2}Y_i=S_n$ (say).

Find the limiting distribution of $T_n/S_n$ as $n\rightarrow \infty. $

My attempt:

$E(X_i) =\lambda/n $ and $Var(X_i) =\lambda/n(1-\lambda/n)$

Also $E(Y_i) =\lambda/n=Var(Y_i) $

Thus $E(T_n) =n\lambda=E(s_n) $ And $Var(T_n) =\lambda(n-\lambda) $ $Var(S_n) =n\lambda$ Then I can't proceed further. Please suggest something....

asked May 05 '23 at 08:42

Arnab seal

1

This is equivalent to using $X_i' \sim_{\text{iid}} \text{Binomial}(n, \lambda)$, $Y_i' \sim_{\text{iid}} \text{Poisson}(\lambda)$, $T_n' = \sum_{i = 1}^n X_i'$, $S_n' = \sum_{i = 1}^n Y_i'$ and finding the dis. lim. of $T_n'/S_n'$ – balddraz May 05 '23 at 09:14
Then can you tell me the final answer? – Arnab seal May 09 '23 at 15:22
No please do the work yourself – balddraz May 09 '23 at 19:03

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