$\lambda$ is a positive real number. Two random variables $X$ and $Y$ are independent each other and follows Poisson distribution with mean $\lambda$.
We define $Z = X-Y$.
I need to get a characteristic function of $Z$, $\varphi=E[e^{itZ}]$ and prove that $E[Z^2]=2\lambda$.
What I have tried
I have found that Poisson distribution has reproductive property so a parameter of $Z$, $\lambda'$ is $\lambda-\lambda=0$.
Then I got a characteristic function of $$ \sum_{i=0}^\infty \frac{e^{itz_i}}{z_i!}=e^{it}$$
But this will not give any functions with $\lambda$ when I want to have a moment of $Z$.
Where did I get wrong?
Best Answer
It is not true that $X-Y$ has Poisson distribution with parameter $\lambda -\lambda$. Obviously $X-Y$ takes negative integer values also, so it cannot have a Poisson distribution.
$Ee^{itX}=\sum e^{-\lambda} \frac {\lambda^{n} e^{itn}} {n!}=e^{-\lambda} e^{\lambda e^{it}}=e^{-\lambda (1-e^{it})}$.
Hence $$Ee^{it(X-Y)}=|Ee^{itX}|^{2}=e^{-2\lambda (1-cos ( t))}$$.
To find $EZ^{2}$ differentiate this twice, put $t=0$ and multiply by $-1$.