Show CLT for Poisson random variables, using no generating function

Question

Question is as following:

$X\sim Po(\lambda)$

$$\frac{X-\lambda}{\sqrt{\lambda}} \,{\buildrel d \over \rightarrow}\, N(0,1)$$

as $\lambda \rightarrow \infty$.

Obs. One is asked not to show the convergence with generating functions.

I begin by setting $Y=\frac{X-\lambda}{\sqrt{\lambda}}$ and thus:

$P(Y\leq y)=P(\frac{X-\lambda}{\sqrt{\lambda}} \leq y)$.

After some arrangements:

$P(X \leq y\sqrt{\lambda}+\lambda)$.

The plan after this is to use the cdf of the $Po(\lambda)$ and send $\lambda \rightarrow \infty$. The problem is that the cdf is not given and is expected to be calculated which stops me in my tracks. I tried using $1-P(X>y\sqrt{\lambda}+\lambda)$ but the sum is something that I can not handle.

I also tried the simpler way to derive the cdf by simply doing:

$P(X\leq x) = \sum^x_{k=0}e^{\lambda}\frac{\lambda^k}{k!}$.

But it still stops me...I tried mixing around with the sum such that I could be able to use some known sums but I can't seem to handle it..

Any form of help is appreciated.

Answer 1

Edit: after the editing of the question, the following does not match the request -- indeed, it is based (solely) on characteristic functions...

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Using characteristic functions, for $Y=\frac{X-\lambda}{\sqrt{\lambda}}$ and any $t\in\mathbb{R}$: $$ \mathbb{E} e^{it Y} = e^{-it\sqrt{\lambda}}\mathbb{E} e^{i\frac{t}{\sqrt{\lambda}}X} = e^{-it\sqrt{\lambda}} e^{\lambda(e^{it/\sqrt{\lambda}}-1)} $$ using the expression of the characteristic function of a Poisson random variable (applied to $u=\frac{t}{\sqrt{\lambda}}$). Now, for any fixed $t$, we have $$\begin{align} -ita+a^2(e^{i\frac{t}{a}}-1) &= -ita+ + a^2\left(i\frac{t}{a}+\frac{i^2t^2}{2a^2} + o\!\left(\frac{1}{a^2}\right)\right) \\ &= -ita+ ita+\frac{i^2t^2}{2} + o(1) = -\frac{t^2}{2} + o(1) \\ &\xrightarrow[a\to\infty]{} -\frac{t^2}{2} \end{align}$$ so that $e^{-it\sqrt{\lambda}+\lambda(e^{it/\sqrt{\lambda}}-1)} \xrightarrow[\lambda\to\infty]{} e^{-\frac{t^2}{2}}$ (applying the above with $a=\sqrt{\lambda}$).

But recalling that $\phi\colon t \mapsto e^{-\frac{t^2}{2}}$ is the characteristic function of a normal random variable $\mathcal{G}(0,1)$, this means that the characteristic function $\phi_Y$ converges pointwise to $\phi$, or equivalently that $$Y\xrightarrow[\lambda\to\infty]{d} \mathcal{G}(0,1)$$ as claimed. $\square$

Answer 2

The admonestation by the OP's TA to "read the book" that was mentioned in a comment (and that the OP instantly dismissed as being due to some kind of psychological disorder their TA would be suffering from) seems rather sound and might be based on the following basic remark.

For every integer $\lambda$, $X_\lambda$ is distributed as the sum of $\lambda$ i.i.d. random variables $\xi_k$ Poisson with parameter $1$. Since $E(\xi_k)=\mathrm{var}(\xi_k)=1$, for integer values of $\lambda$, the desired result directly follows from the standard CLT.

To recover the result for noninteger values of $\lambda$ as well, note that one can realize $(X_\lambda)$ in such a way that $X_\lambda\leqslant X_\nu$ almost surely for every $\lambda\leqslant \nu$. Then, bound the random variable $Y_\lambda=(X_\lambda-\lambda)/\sqrt{\lambda}$ almost surely, using $Y_n$ and $Y_{n+1}$ for $n=\lfloor\lambda\rfloor$, and apply the previous case.