# Limit of c.d.f. of Poisson($n$) when $n$ goes to infinity

I'm trying to prove that $\lim_{n\rightarrow\infty}F_{X}(n)=1/2$ when $X\sim \text{Poisson}(n)$ without success. Could someone help me ?

• Hint: Use the CLT. Jun 16, 2013 at 22:32
• I thought I had to use incomplete gamma functions properties since $F_{X}(n)=\frac{\Gamma(n+1,n)}{n!}$.
– plb
Jun 16, 2013 at 22:39
• I suppose you could do that, but one reason to consider my hint is that you will then see why an analogous result is true in much greater generality. Jun 16, 2013 at 22:44
• To add to cardinal's hint, $X = Y_1+Y_2+\cdots + Y_n$ where the $Y_i$ are independent Poisson$(1)$ random variables. Jun 16, 2013 at 23:17

You know that your variable $X_{n}$ takes a value between $0$ and $n$ with probability $Pr(X_{n}\leq n)$. If your random variables $Y_{n}$ are independent, then it holds that $\sum_{i=1}^{n}Y_{i}\sim X \sim Poi(n)$, hence $$Pr \left(\frac{Y_{1} + Y_{2}+...+Y_{n} - n}{\sqrt{n}}\leq 0\right) = Pr\left(X_{n} \leq n \right)$$ If you then apply the central limit theorem to the left-hand side expression, you will see that $$\lim_{n\to \infty} Pr \left(\frac{Y_{1} + Y_{2}+...+Y_{n} - n}{\sqrt{n}}\leq 0\right)$$ converges in distribution towards a Normal distribution with $N(0,1)$ (check this). To get your answer, you then just have to see what is the probability that $P(N(0,1)) \leq 0)$