Levy processes - infinitely divisible distribution

Question

I am reading the following text but can't understand the last sentence (source: Andreas E. Kyprianou "Fluctuations of Levy Processes with Applications"):

From the definition of a Levy process, we see that, for any $t>0$, $X_t$ is a random variable belonging to the class of infinitely divisible distributions. This follows from the fact that, for any $n=1,2,\ldots,$ $$ X_t = X_{t/n}+(X_{2t/n}-X_{t/n})+\cdots+(X_t-X_{(n-1)t/n}), \label{a}\tag{1.1} $$together with the facts that $X$ has stationary independent increments and that $X_0=0$. Suppose, now, that we define for all $\theta\in\mathbb{R}$, $t\ge 0$, $$ \psi_t(\theta)=-\log \mathbb{E}\left(e^{i\theta X_t}\right) $$Then using \ref{a} twice, we have, for any two positive integers $m,n$ that $$ m\psi_1(\theta) =\psi_m(\theta)=n\psi_{m/n}(\theta). $$Hence, for any rational $t>0$, $$ \psi_t(\theta)=t\psi_1(\theta) \label{b}\tag{1.2} $$If $t$ is an irrational number, then we can choose a decreaing sequence of rationals $\{t_n: n\ge 1\}$ such that $t_n \downarrow t$ as $n$ tends to infinity. Almost sure right-continuity of $X$ implies right-continuity of $\exp\big(-\psi_t(\theta)\big)$ (by dominated convergence)and hence \ref{b} holds for all $t\ge 0$.

Could someone explain in more detail the application of the dominated convergence theorem?

Answer 1

Welcome to math.stackexchange!

$$\lim_{n\rightarrow\infty}\exp\Big(-\psi_{t+\frac{1}{n}}(\theta)\Big)=\lim_{n\rightarrow\infty}\mathbb{E}\Big[e^{i\theta X_{t+\frac{1}{n}}}\Big]=\mathbb{E}\Big[\lim_{n\rightarrow\infty}e^{i\theta X_{t+\frac{1}{n}}}\Big]=\mathbb{E}[e^{i\theta X_t}]=\exp\Big(-\psi_{t}(\theta)\Big)$$

EDIT: Show that (1.2) holds for all $t\ge 0$:

Let $t\in\mathbb{R}_+\backslash\mathbb{Q}$. Consider a sequence $(t_n)_{n\in\mathbb{N}}\subset\mathbb{Q}$ with $t_n\downarrow t$ for $n\rightarrow \infty$. The function

$$t\mapsto\psi_t(\theta)=-\log(\exp(-\psi_t(\theta)))$$

is right-continuous, since $t\mapsto\psi_t(\theta)$ is right-continuous and $x\mapsto-\log(x)$ is continuous. (See second part of the accepted answer in the topic Composition of a Cadlag function with a Continuous Function)

This gives $$\psi_t(\theta)=\lim_{n\rightarrow\infty}\psi_{t_n}(\theta)=\lim_{n\rightarrow\infty}t_n\psi_1(\theta)=t\psi_1(\theta)$$

Answer 2

Based on the above solution, I made the same reasoning for the characteristic function From infinitely divisible distribution of $X_m$ we can write $$X_m=X_1+(X_2-X_1)+...+(X_m-X_{m-1})$$ $$X_m=X_{\frac{m}{n}}+(X_{2\frac{m}{n}}-X_{\frac{m}{n}})+...+(X_m-X_{(m-1)\frac{m}{n}})$$

So $\phi_1(u)^m=\phi_m(u)=\phi_{\frac{m}{n}}(u)^n$ where $\phi_m(u)=\mathbb{E}(e^{iuX_m})$ This imples that for any rational $t> 0$ we have $\phi_t(u)=\phi_1(u)^t$

Now because for all $u\in \mathbb{R}$ and $t\ge 0$ $|\phi_t(u)|\le 1$ (property of characteristic function) we can use Dominated Convergence Theorem and rght continuity of $X$: $$\lim_{n \to \infty}\phi_{t+\frac{1}{n}}(u)=\lim_{n \to \infty}\mathbb{E}(e^{iuX_{t+\frac{1}{n}}})=\mathbb{E}(\lim_{n \to \infty}e^{iuX_{t+\frac{1}{n}}})=\mathbb{E}(e^{iuX_{t}})=\phi_t(u)$$

so $\phi_t(u)$ is right continous with respect to $t$. Finally $$\phi_t(u)=\lim_{n \to \infty}\phi_{t_n}(u)=\lim_{n \to \infty}t_n\phi_1(u)=t\phi_1(u)$$ so $(1.2)$ hold for all $t\ge0$