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Let $S_n=X_1+...+X_n$ where $X_i\sim Poisson(1)$ and iid. I need to compute $\mathbb P\{S_n\leq n\}$ when $n\to \infty $.

My teacher said that it converge in lay to $\mathbb P\{\mathcal N(0,1)\leq 0\}=1/2$, but since $S_n\geq 0$ (since it's $poisson(n)$, don't we have that $$\mathbb P\{S_n\leq n\}=\mathbb P\{0\leq S_n\leq n\}=\mathbb P\{-\frac{1}{\sqrt n}\leq (S_n-n)/\sqrt n\leq 0 \}\to \mathbb P\{\mathcal N(0,1)=0\}=0\ \ ?$$

user380364
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1 Answers1

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$$0 \le S_n \le n$$ implies $$- n \le S_n - n \le 0$$ which in turn implies $$-\sqrt{n} \le \frac{S_n - n}{\sqrt{n}} \le 0.$$

heropup
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