Poisson random measure and $\alpha$-stable processes Let $N$ be a Poisson random measure in $(0,\infty)^2$ with intensity $\eta$ given by
$$\eta(ds, dx) = \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}} ds \, dx.$$

*

*Find the values $\alpha$ for which $\eta$ is a $\sigma$-finite measure.


*Let $X_t = N(f_t)$ with
$$f_t(s,x) = \mathbb{I}_{\{s\leq t\}}\cdot x.$$
Find the values $\alpha$ for which $X_t<\infty$ for all $t\geq 0$ a-s.


*Compute $\mathbb{E}\left[e^{-\lambda X_t}\right]$


*Prove $X_t \overset{d}{=} t^{1/\alpha}X_1$.
Here is the problem, I have read a bit about stable Levy $\alpha$-processes and I can realize that the $\eta$ measure is related to a $\alpha$-stable process, which means that $\alpha$ must be between 0 and 2 for the $\eta$ measure to be $\sigma$-finite.
My problem with the first question is that I don't know how to prove that alpha must take those values mentioned. I understand that for a measure to be sigma-finite we must find a succession of countable sets with finite measure covering all $S = (0,\infty)^2$ and that is easy considering the intervals $A_n = \left(\dfrac{1}{n}, n\right)$. So my reasoning (which I think is not correct) is the following
$$\eta\left(A_n\times A_n\right) = \int_{1/n}^{n} \int_{1/n}^{n}\mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}} \, ds \, dx$$
but here we can conclude that this measure is finite for any value $\alpha \in \mathbb{R}$, which I believe is incorrect to assume from what I have already said about $\alpha -$stable processes.
For questions 2 and 3 I am using Campbell's theorem: in question two I had no problem finding that $\alpha \in (1,2]$ (this is assuming I had concluded that $\alpha \in (0,2]$ in question 1).
For question 3, again using the result of Campbell's theorem and assuming $\alpha \in (1,2]$, I have the following calculation:
\begin{align*}
        \mathbb{E}\left[e^{-\lambda X_t}\right] &= \exp\{-\int_S (1-e^{-\lambda f_t(s,x)})\ \eta\left(ds,dx\right)\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda f_t(s,x)}\right) \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}}x}\right) \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}} x}\right) \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}}x}\right)\ ds\right]\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \int_0^t \left(1-e^{-\lambda x}\right)\ ds\right]\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \left(1-e^{-\lambda x}\right) \int_0^t \ ds\right]\,dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}} \left(1-e^{-\lambda x}\right) \cdot t\  dx\}\\[0.2cm]
        &= \exp\{-C\cdot t \int_0^\infty \left(1-e^{-\lambda x} \right) \dfrac{1}{x^{\alpha+1}} \, dx\} 
    \end{align*}
However, once that point is reached, I am stuck, since I don't know how to calculate that remaining integral, I am almost sure that the development up to that point is correct, but I would appreciate some comment in case something is wrong or some alternative how to conclude the calculation.
Finally, for question 4, I think it is enough to find the expectation of the Laplace transform of $t^{1/\alpha} X_1$, and see that it matches the previous calculation, but again I run into a problem when developing the calculation
\begin{align*}
        \mathbb{E}\left[e^{-\lambda t^{1/\alpha} X_1}\right] &= \exp\{-\int_S \left(1-e^{-\lambda t^{1/\alpha} f_1(s,x)}\right)\ \eta\left(ds,dx\right)\}\\[0.2cm]
        &= \exp\{-\int_{0}^{\infty} \int_{0}^{\infty} \left(1-e^{-\lambda t^{1/\alpha} \mathbb{I}_{\{s\leq 1\}}x}\right)\ \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}}\ ds \, dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \int_0^1 \left(1-e^{-\lambda t^{1/\alpha} x}\right)\ \dfrac{C}{x^{\alpha + 1}}\ ds \, dx\}\\[0.2cm]
        &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha + 1}} \left[\int_0^1 \left(1-e^{-\lambda t^{1/\alpha} x}\right)\ \ ds\right] dx\}
    \end{align*}
Again, I would appreciate any comments or observations that may help me to proceed.
 A: 1. For $\alpha>0$, the integral,
$$
\int_0^\infty(1-e^{-\lambda x}){dx\over x^{1+\alpha}}
$$
converges if and only if $0<\alpha<1$, because the integrand is $\sim \lambda x^{-\alpha}$ for $x\to 0+$. This is precisely the range of $\alpha$s for which $X_t<\infty$ a.s.
For   $\alpha\in(0,1)$,  the change of variables $u=\lambda x$ results in
$$
\int_0^\infty(1-e^{-\lambda x}){dx\over x^{1+\alpha}}=\lambda^\alpha\int_0^\infty(1-e^{-u}){du\over u^{1+\alpha}}.
$$
Now write $1-e^{-u} =\int_0^u e^{-t} dt$ and use Fubini:
$$
\eqalign{
\int_0^\infty(1-e^{-u}){du\over u^{1+\alpha}}
&=\int_0^\infty\int_0^t e^{-t}dt {du\over u^{1+\alpha}}  \cr
&=\int_0^\infty\int_t^\infty {du\over u^{1+\alpha}} e^{-t} dt\cr
&=\int_0^\infty t^{-\alpha}e^{-t} dt\cr
&=\int_0^\infty t^{(1-\alpha)-1}e^{-t} dt\cr
&=\Gamma(1-\alpha).\cr
}
$$
2. The measure $\eta$ is $\sigma$-finite for each $\alpha>0$. Indeed
$$
\eta\left( (0,n)\times (1/n,n)\right)=\lambda n\cdot{C\over\alpha}(n^\alpha-n^{-\alpha})<\infty
$$
for each positive integer $n$.
