Q:

Find characteristic function of given Random Variable

Let $X_1, ..., X_n$ be iid ~Poi($\frac {\lambda}{\sqrt n}$)
Let $Y_n=\frac{(\sum_{k=1}^n X_k)-\lambda \sqrt n}{\sqrt \lambda n^{\frac 1 4}}$
Find the characteristic function of $Y_n$

It was a question in an older exam so I think there should be an easy/fast way to solve this. I see that $Y_n$ is standardised but not how that will help me here.
I started calculating
$\Phi_{Y_n}(t)=\mathbb E[exp(itY_n)]=\mathbb E[exp(it\frac{(\sum_{k=1}^n X_k)-\lambda \sqrt n}{\sqrt \lambda n^{\frac 1 4}})]=exp(-it \sqrt\lambda n^{\frac 1 4 })  \mathbb E[exp(\frac {it\sum_{k=1}^n X_k)}{\sqrt \lambda n^{\frac 1 4}})]$
But from there it got messier and I don't think this is the wanted solution.
Any help is appreciated.

A:

Useful facts:

If $X$ is Poisson with parameter $\alpha$ then $\varphi_X(t)=E(e^{itX})=\exp(-\alpha(1-e^{it}))$.
If $Y=aX-b$ then $\varphi_Y(t)=e^{-itb}\varphi_X(at)$.
If $(X_i)$ is i.i.d. Poisson with parameter $\beta$ then $X_1+\cdots+X_n$ is Poisson with parameter $n\beta$.

You might try to use 3. with $\beta=\lambda/\sqrt{n}$, then 1. with $\alpha=n\beta=\lambda\sqrt{n}$, then 2. with $b=1/a=\sqrt{\alpha}$.

A:

As the Random variables $X_i$ are i.i.d. you can use that the characteristic function of the sum of independent randomvariables is the product of the characteristic functions of the random variables.
To continue your computation
\begin{eqnarray}
\Phi_{Y_n}(t)&=&exp(-it \sqrt\lambda n^{\frac 1 4 })  \mathbb E\left[exp\left(\frac {it\sum_{k=1}^n X_k)}{\sqrt \lambda n^{\frac 1 4}}\right)\right]
\\
&=&exp(-it \sqrt\lambda n^{\frac 1 4 }) \mathbb \prod_{k=1}^n E\left[exp\left(\frac {it X_k}{\sqrt \lambda n^{\frac 1 4}}\right)\right]
\\
&=&exp(-it \sqrt\lambda n^{\frac 1 4 }) \mathbb \prod_{k=1}^n E\left[exp\left(\frac {it}{\sqrt \lambda n^{\frac 1 4}}X_k\right)\right].
\end{eqnarray}
Now you can plug in the characteristic function of the Poisson distribution $Poi(\theta)$ $$\Phi_{Poi(\theta)}(u)=\exp\left(\theta(e^{iu}-1)\right)$$
with the respective $\theta$ and $u$.

