Let $v$ a finite measure on $\mathcal{R}$ and define the Compound Poisson distribution as
$$\text{CPoi}_v=e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}}{n!}$$
Suppose now that $v$ is concentrated on $(-1/k,-1/(k+1)] \cup [1/(k+1),1/k)$ for $k=1,2,\dots$.
I know that
$$E[X_k]=\int_{(-1/k,-1/(k+1)] \cup [1/(k+1),1/k)} x v(dx)$$
I have to prove that
$$\text{Var}[X_k]=\int_{(-1/k,-1/(k+1)] \cup [1/(k+1),1/k)} x^2 v(dx)$$
I am stuck since $E[X_k]$ is not zero from what I know. What am I doing wrong?
Thank you.
Forget about $k$. You have $Y_1,\ldots,Y_n,\ldots$ iid with distribution $\nu=v/\lambda$ with $\lambda=v(\mathcal{R})$ and $N$ a Poisson random variable of mean $\lambda$ independent of the $Y_i$. Finally you consider $X=Y_1+\ldots+Y_N$ (with the convention $X=0$ if $N=0.$) You are asked to find the variance of $X$ in terms of $\nu$ or $v.$ Use the two formulas $$E(N)=\lambda, \ E(N(N-1))=\lambda^2$$ that you can prove by derivating twice $$z\mapsto E(z^N)=e^{\lambda(z-1)}$$ and doing $z=1$ after. Now $$E(X)=\sum _{n=0}^{\infty}E(X|N=n)\Pr(N=n)=\sum _{n=0}^{\infty}nE(Y_1)\Pr(N=n)=\lambda E(Y_1).$$
$$E(X^2)=\sum _{n=0}^{\infty}E(X^2|N=n)\Pr(N=n)=\sum _{n=0}^{\infty}(nE(Y_1^2)+n(n-1)(E(Y_1))^2)\Pr(N=n)=\lambda E(Y^2_1)+\lambda^2(E(Y_1))^2.$$ As a consequence the variance of $X^2$ is $$E(X^2)-(E(X))^2=\lambda E(Y_1)^2=\lambda\int_{-\infty}^{\infty}y^2\nu(dy)=\int_{-\infty}^{\infty}y^2v(dy)$$ as desired.