# Characteristic function of a Levy process

When I was reading Protter's textbook "Stochastic Integration and Differential Equations", it is stated without proof that (page 20)

If we take the Fourier transform of of each $X_t$ (a levy process). we get a function $f(t, u) = f_t(u)$ given by $f_t(u) = E[e^{iuX_t}]$. Using the right continuity in probability we conclude $f_t(u) = \exp(-t\psi(u))$ for some continuous function $\psi(u)$ with $\psi(0) = 0$.

I would like to know why the continuity in probability can imply the existence of $\psi(u)$. Thanks in advance!

## 1 Answer

Fix $u$ and consider the function $g(t)=f_t(u)$. Then $g(t+s)=g(t)g(s)$ and $g$ is continuous. [ Because $X_n \to X$ in probability and $\{X_n\}$ uniforlmy bounded implies $EX_n \to EX$]. The only such functions are given by $g(t)=e^{ct}$. Using that fact that $g$ is bounded conclude that $c<0$. $c$ depends on $u$. Call it $-\psi (u)$. Can you take it from here?

• Thanks! That helps a lot! – VanDDF Jul 7 '18 at 4:56