Poisson distribution to normal distribution?statistics

Question

I need to demonstrate why when (lambda)is big enought poisson distribution becomes (aproximation)to normal distribution. Thanks you

Answer 1

The poisson distribution has mean $\lambda$ and variance $\lambda$. So using these as parameters for the normal distribution, we are looking whether

$$ \begin{aligned} e^{-\lambda}\frac{\lambda^x}{x!} \approx \frac{1}{\sqrt{2\pi\lambda}}e^{-\frac{(x-\lambda)^2}{2\lambda}} \\ \frac{e^{-\lambda+x\ln\lambda}}{x!} - \frac{1}{\sqrt{2\pi\lambda}}e^{-\frac{(x-\lambda)^2}{2\lambda}} \approx 0\end{aligned} \qquad \text{for large }\lambda$$

So now we can express this as a limit and evaluate it. If it is $0$, then the approximation is correct for large $\lambda$.

$$ \lim_{\lambda\to\infty}\left( \frac{e^{-\lambda+x\ln\lambda}}{x!} - \frac{1}{\sqrt{2\pi\lambda}}e^{-\frac{(x-\lambda)^2}{2\lambda}} \right) \\ \lim_{\lambda\to\infty} \left( \frac{e^{-\lambda+x\ln\lambda}}{x!}\right) - \lim_{\lambda\to\infty} \left( \frac{1}{\sqrt{2\pi\lambda}}e^{-\frac{(x-\lambda)^2}{2\lambda}}\right) $$

Factor out constants.

$$ \frac{1}{x!}\lim_{\lambda\to\infty} \left( e^{-\lambda+x\ln\lambda}\right) - \frac{1}{\sqrt{2\pi}} \lim_{\lambda\to\infty} \left(\frac{e^{-\frac{(x-\lambda)^2}{2\lambda}}}{\sqrt{\lambda}}\right) $$

Deal with the first limit: Use $\lim_x e^{f(x)} = e^{\lim_x f(x)}$. We have an indeterminate form $\infty -\infty$ so factor out $\lambda$.

$$ \lim_{\lambda\to\infty} \left( -\lambda+x\ln\lambda\right) = \lim_{\lambda\to\infty} \lambda\left( \frac{x\ln\lambda}{\lambda} - 1\right) \\ = \lim_{\lambda\to\infty} (\lambda) \left( x\lim_{\lambda\to\infty}\left( \frac{\ln\lambda}{\lambda} \right) -1 \right) $$

$ x \gt \ln x $ for all $x\gt 0$ so the limit $\frac{\ln\lambda}{\lambda}$ is $0$.

$$ \lim_{\lambda\to\infty} (\lambda) \times \left( 0x - 1\right) = -1\times \lim_{\lambda\to\infty} (\lambda) = -\infty $$

So the first limit evaluates to $e^{-\infty}=0$.

$$ \lim_{\lambda\to\infty} \left(\frac{e^{-\frac{(x-\lambda)^2}{2\lambda}}}{\sqrt{\lambda}}\right) = \lim_{\lambda\to\infty} \left(\frac{1}{\sqrt{\lambda}}\right) e^{-\textstyle\lim_{\lambda\to\infty}\frac{(x-\lambda)^2}{2\lambda}} $$

$\sqrt{\infty}=\infty$ so $1/\sqrt{\lambda}$ is 0. Additionally, the term $(x-\lambda)^2$ approaches $\lambda^2$ as $\lambda\to\infty$.

$$ 0 e^{-\textstyle\frac{1}{2}\lim_{\lambda\to\infty}\frac{(x-\lambda)^2}{\lambda}} = 0e^{-\frac{1}{2}\infty} = 0\times 0=0 $$