Prove or disprove convergence in distribution of a poisson variable.

Question

Let

$$S \overset{d}{\sim} Poisson(\lambda).$$ I would like to determine $\frac{S-\lambda}{\sqrt{\lambda}}$ converges in distribution as $\lambda \rightarrow \infty.$ So my set up is:

$$\Pr\left[a \le \frac{S-\lambda}{\sqrt{\lambda}} \le b\right] = Pr[S \le \sqrt{\lambda} b + \lambda] - Pr[S \le \sqrt{\lambda} a + \lambda].$$

Where $$Pr[S \le \sqrt{\lambda} b + \lambda] = \sum_{k=0}^{\sqrt{\lambda} b + \lambda} \frac{e^{-\lambda}\lambda^k}{k!}.$$

But as $\lambda$ tends to infinity, it seems like I could simply examine the last term in the summation:

$$\frac{e^{-\lambda} \lambda^{\sqrt{\lambda} b + \lambda}}{(\sqrt{\lambda}b + \lambda)!} \sim \frac{e^{-\lambda} \lambda^{\sqrt{\lambda}b}\lambda^{\lambda}}{\sqrt{2\pi}(\sqrt{\lambda} b+ \lambda)^{1/2}(\sqrt{\lambda} b+ \lambda)^{\sqrt{\lambda}b}(\sqrt{\lambda} b+ \lambda)^{\lambda}e^{-\sqrt{\lambda}b}e^{-\lambda}}$$

By Stirling's formula. And we see the $e^{-\lambda}$ cancel. However, I am not sure what to make of this, it certainly looks nothing like the normal distribution since the only exponential term is $e^{\sqrt{\lambda}b}$, which does not converge at large value of the exponent as it would in a normal.

1. First I would like to confirm if the steps thus far are correct. Especially the argument that it suffice to consider the largest term in the summation.
2. Secondly I would like to know if there's another way to attack this problem altogether.

Answer 1

A more convenient way is to use moment generating functions.

MGF of Poisson$:= E[e^S]=e^{\lambda (e^t-1)}$.

Define a sequence of random variables $S_n$, where $S_n\sim Poi(n)$, then $S_i=\sum\limits_{j=1}^{i}X_j\;\;X_j\sim Poi(1)$

Also, $S_i-i = \sum\limits_{j=1}^{i}(X_j-1)$

We can now use a convenient property of moment generating functions $\psi(t):Y_N=\sum\limits_{i=1}^N a_iX\implies \psi_{Y_N}(t)=\prod\limits_{i=1}^N\psi_i(a_it)$, provided that the $X_i$ are independent (not necessarily identically distributed).

For our particular case, we will need a second MGF to represent the "degenerate" distribution $\delta_a(x):MGF[\delta_a(x)]=e^{ta}$.

Now, we can see that $\frac{S_i-i}{\sqrt{i}}=\sum\limits_{j=1}^i \frac{X_j}{\sqrt{i}} - \frac{\delta_{1,j}}{\sqrt{i}} \implies MGF\left[\frac{S_i-i}{\sqrt{i}}\right]=e^{i\left(e^{\frac{t}{\sqrt{i}}}-1\right)}\times e^{-t\sqrt{i}}=\exp\left[i\left(e^{\frac{t}{\sqrt{i}}}-1\right)-t\sqrt{i}\right]$

Now, $\lim\limits_{i \to \infty} \exp\left[i\left(e^{\frac{t}{\sqrt{i}}}-1\right)-t\sqrt{i}\right] = e^{\frac{t^2}{2}}$, which is the MGF of a standard normal distribution. $\square$