suppose that $x_1 , x_2, \ldots$ are independent poisson (mean${}=1$)
1) show that $\frac {y_n -n }{\sqrt n} \to z$ in distribution as $n \to \infty$ where $z$ belong to $N(0,1)$
where $y_n = x_1 +x_2 +x_3 + \cdots +x_n$
2) deduce that $e^{-n} \sum_{n=1}^\infty (\frac{n^k}{k!}) \to 1/2$
in part one I use characterstic function of $s_n =\frac {y_n -n }{\sqrt n}$
the last step of my work $\exp(-t\sqrt n) ϕ_{y_n}\left(\frac t {\sqrt n} \right)$ but this not equal to characteristic of normal
please help me !
Best Answer
Note that 1) is a direct consequence of the central limit theorem, but maybe you are not allowed to use that fact?
The characteristic function of $\frac{y_n - n}{\sqrt{n}}$ can be computed to be $\exp(n(e^{it/\sqrt{n}}-1) - it\sqrt{n})$. To show the exponent tends to $-t^2/2$ you can do l'Hôpital's rule (or recognize the limit as a derivative of a particular function). $$\lim_{n \to \infty} [n(e^{it/\sqrt{n}}-1) - it\sqrt{n}] \overset{x := 1/\sqrt{n}}{=} \lim_{x \to 0} \frac{e^{itx} - 1 - itx}{x^2} = \frac{it}{2} \lim_{x \to 0} \frac{e^{itx} - 1}{x} = - \frac{t^2}{2}.$$
For 2), (with kimchi lover's correction), note that it suffices to show $P(y_n \ge n) \to 1/2$ because $y_n \sim \text{Poisson}(n)$. $$P(y_n \ge n) = P\left(\frac{y_n -n}{\sqrt{n}} \ge 0\right) \to P(z \ge 0) = \frac{1}{2} $$