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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 ?

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    $\begingroup$ Hint: Use the CLT. $\endgroup$
    – cardinal
    Jun 16, 2013 at 22:32
  • $\begingroup$ I thought I had to use incomplete gamma functions properties since $F_{X}(n)=\frac{\Gamma(n+1,n)}{n!}$. $\endgroup$
    – plb
    Jun 16, 2013 at 22:39
  • $\begingroup$ 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. $\endgroup$
    – cardinal
    Jun 16, 2013 at 22:44
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    $\begingroup$ To add to cardinal's hint, $X = Y_1+Y_2+\cdots + Y_n$ where the $Y_i$ are independent Poisson$(1)$ random variables. $\endgroup$ Jun 16, 2013 at 23:17

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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 $P(N(0,1)) \leq 0)$

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