I want to prove why the mean and variance of a $\operatorname{Poisson}(\lambda)$, is different when the time index approaches infinite (it's approximated by the mean and variance of a Normal).
For example:
$$
N_k = N_1 + (N_2 – N_1) + (N_3 – N_2) + … + (N_k – N_{k-1})
$$
Using CLT:
$\frac{N_k – k\lambda}{\sqrt{k\lambda}}$ is normally distributed (in the limit).
I want to answer why is that a Poisson R.V. characterized by $\mathbb{E}[X] = \lambda$ and $\operatorname{Var}[X] = \lambda$.
Because when it approaches a normal distribution, $\mathbb{E}[Z] = \mu$ and $\operatorname{Var}[Z] = \sigma^2$.
Best Answer
The Poisson distribution does not approach the normal distribution, the centered Poisson distribution does. More precisely, if $X_\lambda$ is Poisson with parameter $\lambda$, then $Y_\lambda$ converges in distribution to a standard normal random variable $Z$, where $Y_\lambda=(X_\lambda-\lambda)/\sqrt{\lambda}$. In particular, for every $\lambda$, $E[Y_\lambda]=E[Z]=0$ and $\mathrm{var}(Y_\lambda)=\mathrm{var}(Z)=1$ (in your language, $\mu=0$ and $\sigma^2=1$).