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?
Best Answer
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$$\lim_{n\rightarrow\infty}\exp\Big(-\psi_{t+\frac{1}{n}}(\theta)\Big)=\lim_{n\rightarrow\infty}\mathbb{E}\Big[e^{i\theta X_{t+\frac{1}{n}}}\Big]=\mathbb{E}\Big[\lim_{n\rightarrow\infty}e^{i\theta X_{t+\frac{1}{n}}}\Big]=\mathbb{E}[e^{i\theta X_t}]=\exp\Big(-\psi_{t}(\theta)\Big)$$
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 $$\psi_t(\theta)=\lim_{n\rightarrow\infty}\psi_{t_n}(\theta)=\lim_{n\rightarrow\infty}t_n\psi_1(\theta)=t\psi_1(\theta)$$