Let $X$ be a Levy process. Let $$f(t):=E[e^{i u^{tr}X_t}]$$ I want to show that $f(t+h)=f(t)f(h)$ holds and I know that $M_t:=e^{i u^{tr}X_t}/f(t)$ is a martingale.

I tried to replace $X_{t+h}=X_t+X_{t+h}-X_t$, then I have by independence of increments $$f(t+h)=E[e^{i u^{tr}(X_t+X_{t+h}-X_t)}]=E[e^{i u^{tr}(X_t+X_{t+h})}e^{-i u^{tr}X_t}]=E[e^{i u^{tr}(X_t+X_{t+h})}]E[e^{-i u^{tr}X_t}]$$

But I don't see how to get $f(t)f(h)$

  • $\begingroup$ @HartoSaarinen $L_t$ is not important here, it is part of the whole proof. So I removed it. Thanks for pointing out. $\endgroup$ – Matriz Feb 9 '18 at 19:01

Levy processes also have stationary increments, which means that $X_{t+h}-X_t =^d X_{t+h-t}=X_{h}$ Since these have the same distribution their characteristic functions are the same so that $$E[e^{iu^{tr}(X_{t+h}-X_t)}]=E[e^{iu^{tr}(X_h)}].$$

  • $\begingroup$ Thank you! That was all I needed! $\endgroup$ – Matriz Feb 9 '18 at 19:27

You are pretty close. Hints to get you trough:

Write $X_{t+h}= (X_{t+h}-X_t)+X_t$. Then use the independent increments property (as you already did) and after that stationary increments property.

  • $\begingroup$ Thank you for the hints. $\endgroup$ – Matriz Feb 9 '18 at 19:30

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