# Characteristic function of $Z=X-Y$, where $X$ and $Y$ follows Po($\lambda$) and prove that $E[Z^2]=2\lambda$

$$\lambda$$ 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?

• Why do you need the characteristic function to solve this? There are simpler ways. It is lucky that the Poisson has the reproductive property, otherwise it might be extinct by now. Aug 22, 2020 at 6:44
• @wolofies I need it because characteristic function is (1) of the problem and $E[Z^2]$ is (2). Aug 22, 2020 at 6:56
• Of course, you do not need the characteristic function to evaluate this expectation.\begin{align}\mathsf E(Z^2)&=\mathsf E(X^2-2XY+Y^2)\\&~~\vdots\\&=2\mathsf{Var}(X)\\&=2\lambda\end{align} But, yeah, you are sometimes requested to do things the hard way ... Aug 22, 2020 at 6:57

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 $$-1$$.

• Can we always assure that $-E[e^{-itX}]=E[e^{-itX}]$ for $E[e^{it(X-Y)}]=|E[e^{itX}]|^{2}$? Aug 22, 2020 at 6:24
• @ohisamadaigaku Whenever $X,Y$ are i.i.d we have $Ee^{it(X-Y)}=Ee^{itX} Ee^{-itY}=Ee^{itX} Ee^{-itX}$ and $Ee^{-itX}$ is the complex conjugate of $Ee^{itX}$. Hence $Ee^{it(X-Y)}=|Ee^{itX}|^{2}$. Aug 22, 2020 at 6:28
• Thank you, I understand this part. Then another question is how you got $|Ee^{itX}|^{2}=e^{-2\lambda (1-cos (\lambda t))}$, emersion of $cos(\lambda t)$ Aug 22, 2020 at 7:25
• @ohisamadaigaku If $c$ is any complex number then $|e^{c}|=e^{\Re c}$ (by $\Re c$ by I mean the real part of $c$). Note that real part of $-\lambda (1-e^{it})$ is $-\lambda (1-cos t)$. Aug 22, 2020 at 7:33
• ok, so should the final answer be $e^{-2\lambda(1-cost)}$? Aug 22, 2020 at 7:45