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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.

2 Answers 2

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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 $u$.

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    @Vincent-w Thank you very much but there is still one thing I don't understand. We have $\Phi_{Y_n}(t) =exp(-it \sqrt\lambda n^{\frac 1 4 }) \mathbb \prod_{k=1}^n E\left[exp\left(\frac {it}{\sqrt \lambda n^{\frac 1 4}}X_k\right)\right] = exp(-it \sqrt\lambda n^{\frac 1 4 }) *exp(-\lambda \sqrt n) *exp(\lambda \sqrt n*exp(\frac {it}{\sqrt \lambda n^{\frac 1 4}}))$ But as $Y_n$ is standardised it should converge to the standarddistribution for $n \to \infty$. But $\Phi_{Y_n}(t)$ converges to $0$ instead of $e^{\frac {-t^2}{2}}?$ – 2017-02-09
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    @PeterGarder You have to bare in mind, that the variance of the considered random variables is $\frac{\lambda}{\sqrt{n}}$ which converges to zero if $n\rightarrow\infty$ the result is a degenerate Normal variable. You can see that if you plug in $t=0$ and take the limit (I also think there is a $\sqrt{n}$ missing in the denominator of $Y_n$ if you want to apply the CLT). – 2017-02-09
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Useful facts:

  1. If $X$ is Poisson with parameter $\alpha$ then $\varphi_X(t)=E(e^{itX})=\exp(-\alpha(1-e^{it}))$.
  2. If $Y=aX-b$ then $\varphi_Y(t)=e^{-itb}\varphi_X(at)$.
  3. 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 $b=1/a=\sqrt{\alpha}$.

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    Ooooh... a silent downvote. So sweet. – 2017-02-11