# ISI PCB NC$9$ Limiting Distribution of Bernoulli to Poisson

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

• 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 $S_n = \sum_i^{n^2} X_i$