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Let $X_i\sim (i.i.d.)$, Bernoulli($\frac{\lambda}{n}$), $n\ge \lambda\ge 0$.
$Y_i\sim (i.i.d.)$, Poisson($\frac{\lambda}{n}$). $\{X_i\}$ and $\{Y_i\}$ are independent.
Define $T_n=\sum_{i=1}^{n^2}X_i$ and $S_n=\sum_{i=1}^{n^2}Y_i$.
Find the limiting distribution of $\frac{T_n}{S_n}$ as $n\to\infty$.

My solution:
Let $p=\frac{\lambda}{n}$
$T_n=\sum_{i=1}^{n^2}X_i=\binom{n^2}{k}p^{k}(1-p)^{n^2-k}$ for some $n\ge k\ge 0$.
Similarly, $S_n=\sum_{i=1}^{n^2}X_i=\binom{n^2}{k}p^{k}(1-p)^{n^2-k}$ for some $n\ge k\ge 0$

Hence, $\lim_{n\to\infty}\frac{T_n}{S_n}=1.$

Is there any mistake in the solution? Please help me to correct it. Thanks in advance.

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  • $\begingroup$ Please add the self-study tag & read its wiki. $\endgroup$ Jul 16 at 20:32
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    $\begingroup$ I don't think what you've written for $S_n$ is correct because it has a binomial pmf instead of a Poisson pmf. In the statement of the problem, $S_n$ is a sum of $Y_i$, but you've written $S_n = \sum_i^{n^2} X_i$ $\endgroup$
    – Sycorax
    Jul 16 at 20:43
  • $\begingroup$ Also asked at math.stackexchange.com/q/4199781/321264. $\endgroup$ Jul 16 at 21:06
  • $\begingroup$ Work with the characteristic functions instead. $\endgroup$
    – whuber
    Jul 16 at 21:31

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