# 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.

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 $$n$$.
• Thank you very much for your help! I was able to realize some mistakes in question 2 to find that $\alpha \in (0,1)$ and thanks to the development of the integral I was able to complete the whole exercise. May 30, 2021 at 23:32