I am trying to understand the proof of Theorem 16.14 of Probability Theory by A. Klenke (3rd version) about the Levy-Khinchin formula.
I would like to know how to prove this:
$$E[X]=\int x e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}(dx)}{n!}=\int xv(dx)$$
Where:
- $X$ is distributed as a Compound Poisson Distribution with intensity $v$
- $v$ is a $\sigma$-finite measure on $(0,\infty)$.
My attempt
What am I missing in the following steps?
$$\int x \textrm{CPoi}_{v}(dx)= \int x e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}(dx)}{n!} \\=e^{-v(\mathcal{R})}\left[\int x v^{*0}(dx)+\int x v(dx)+\int x \frac{1}{2!}(v*v)(dx)+\dots\right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+\iint(s+z)\frac{1}{2!}v(ds)v(dz)+\dots \right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+\frac{1}{2!}v(\mathcal{R})\int(s)v(ds)+\dots\right] \\=e^{-v(\mathcal{R})}\int x v(dx)\left[1+\frac{1}{2!}v(\mathcal{R})+\dots \right ]$$
Doubts
Is it right my way of calculating the integral involving the convolution?
I was thinking about using the Taylor expansion of $e^x$ with the terms in the square brackets but it doesn't seem to work.
Thanks for the help.
Solution
I think I found the error in the convolution integral. By exploiting the linearity of the integral, I did not see that for $n=2$ I have 2 identical terms, for $n=3$ I have 3 and so on. This is key to obtain the right factorial in the denominator. Thus I could apply the Taylor expansion above to get the desired result.
$$\int x \textrm{CPoi}_{v}(dx)= \int x e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}(dx)}{n!} \\=e^{-v(\mathcal{R})}\left[\int x v^{*0}(dx)+\int x v(dx)+\int x \frac{1}{2!}(v*v)(dx)+\dots\right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+\iint(s+z)\frac{1}{2!}v(ds)v(dz)+\dots \right] \\=e^{-v(\mathcal{R})}\left[0+\int x v(dx)+v(\mathcal{R})\int(s)v(ds)+\dots\right] \\=e^{-v(\mathcal{R})}\int x v(dx)\left[1+v(\mathcal{R})+\dots \right ]$$