I have the following definition of a Lévy process:

An adapted process $X=(X_t)$ with $X_0=0$ a.s. is a Lévy process if

  1. it has independent increments of the past, i.e. $X_t-X_s$ is independent of $\mathcal{F}_s$
  2. it has stationary increments, i.e. $X_t-X-s$ has the same distribution as $X_{t-s}$
  3. $X_t$ is continuous in probability, i.e. $\lim_{t\to s}X_t=X_s $ in probability.

Let $f_t(u):=E[e^{iuX_t}]$, then it is easily seen $f_{t+s}(u)=f_t(u)f_s(u)$. Now I want to show the continuity of $f_t(u)$. So here is, what I've done:

Let $t_n\to t$ and we need to prove $f_{t_n}(u)\to f_t(u)$. By $3$. we can choose a subsequence (also denoted by $t_n$) such that $X_{t_n}\to X_t$ a.s. Applying dominated convergence and using continuity of $\exp(x)$ we get

$$\lim_n f_{t_n}(u)=E[\lim_ne^{iuX_{t_n}}]=f_t(u)$$

is this correct? Now I know the theorem that any continuous function $h$ which solves $h(x+y)=h(x)h(y)$ can be written of the form $h(x)=e^{a x}$ for some unique $a$. How can I use this result to conclude that $f_t(u)=e^{-t\phi(u)}$ for some function $\phi(u)$?

  • $\begingroup$ Have you not just shown that $t\mapsto f_t(u)$ is continuous and satisfies $f_{t+s}(u)=f_t(u)f_s(u)$ for any fixed $u$? $\endgroup$ – Stefan Hansen Jan 31 '14 at 18:21
  • $\begingroup$ @StefanHansen So my proof of continuity of $f_t(u)$ (in $t$) is right? The problem is, that $f$ depends on $u$ as well. I have a functional equation $f(t,u)f(s,u)=f(t+s,u)$. How do I know that this is of the form $e^{-t\psi (u)}$ for another function $\psi$? $\endgroup$ – user20869 Feb 1 '14 at 7:25
  • $\begingroup$ Not sure that your proof of continuity holds. As for the other thing, let $u$ be fixed. Then $t\mapsto f(t,u)$ solves the functional equation $f(t+s,u)=f(t,u)f(s,u)$ and hence there exists a unique $a_u\in\mathbb{R}$ (this of course depends on $u$) such that $f(t,u)=e^{ta_u}$ for all $t\geq 0$. Now put $\phi(u)=-a_u$. $\endgroup$ – Stefan Hansen Feb 1 '14 at 7:31
  • $\begingroup$ @StefanHansen Thx for your comment. Why do you have doubts about the continuity proof? $\endgroup$ – user20869 Feb 1 '14 at 7:42
  • $\begingroup$ Since you haven't proven convergence of the sequence $(t_n)$ but only of a subsequence of it. $\endgroup$ – Stefan Hansen Feb 1 '14 at 8:35

For any $n$ which is a positive integer, by stationary independent increment. Let $f_n = (f_1)^n$. Moreover for $m,n$ integers, we get $(f_1)^m = f_m = (f_{m/n})^n$

This means $f_{m/n} = (f_1)^\frac{m}{n}$ or $f_t = (f_1)^t$ for $t\in\mathbb{Q}$.

Then for an irrational $t$, take $t_n\downarrow t$, where $t_n\in\mathbb{Q}$, by almost sure right continuity of $X_t$ and the dominated convergence theroem.

$(f_1)^t = \lim\limits_{n\rightarrow\infty} (f_1)^{t_n} = \lim\limits_{n\rightarrow\infty} Ee^{iuX_{t_n}} = E \lim\limits_{n\rightarrow\infty} e^{iuX_{t_n}}= Ee^{iuX_t} = f_t$.

From this, $f_{s+t}=f_sf_t$ follows.

PS it is might actually better to define $\psi_t = - \log f_t$ instead and do it directly: i.e. for $m,n$ integers,

$m\psi_1 = \phi_m = n\psi_{m/n}$ ect.

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  • $\begingroup$ thanks for your answer. Writing $f_n$ you omitting $u$, so $f_n=E[e^{iuX_n}]$? My problem is: I was able to prove $f_{t+s}(u)=f_t(u)f_s(u)$. However the general theorem I mentioned can here not be applied, since we our function $f_t(u)$ depends also on $u$. $\endgroup$ – user20869 Feb 1 '14 at 7:24
  • $\begingroup$ @hulik i made a few typos but which bit is exactly wrong? $\endgroup$ – Lost1 Feb 1 '14 at 13:22
  • $\begingroup$ What exactly do you want? Prove this is continuous in u? $\endgroup$ – Lost1 Feb 1 '14 at 13:25
  • $\begingroup$ I have the following question: Note, we do not assume and now so far that $X$ is RCLL. It should be possible due to my script to prove continuity of $f_t(u)$ for every $u$ by using continuity in probability of $X$. But it is not that easy, see Stefan Hansen comment above $\endgroup$ – user20869 Feb 1 '14 at 16:30

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