Let $\zeta(t)$ be a process with independent increments and $M(t)=E(\exp(\zeta(t))) < \infty $, show that $M(t)^{-1}\exp(\zeta(t))$ is a martingale.

So what I need to show is

$$E(M(t)^{-1}\exp(\zeta(t))|F_s)= M(s)^{-1}\exp(\zeta(s))$$

What I've tried so far: \begin{align} E(M(t)^{-1}\exp(\zeta(t))|F_s) &= M(s)^{-1}\exp(\zeta(s)) \\ &= E(M(t)^{-1}\exp(\zeta(t)+\zeta(s))|F_s).\exp(-\zeta(s)) \end{align}

but I don't know whether this step helps in any way?

(This is not a homework nor an assignment, just found old exams, which I'm trying to solve.)

  • 1
    $\begingroup$ Welcome to the site. I tried to tweak your latex to improve readability. Please ensure it still says what you want it to. $\endgroup$ – gung - Reinstate Monica Jun 28 '16 at 1:53
  • $\begingroup$ @gung yes, it is still says what I want to do, and thanks for the edit $\endgroup$ – Parinn Jun 28 '16 at 8:19

Let $X_t = M_t^{-1}\mathbb e^{\xi_t}$. Then $$\mathbb E[|X_t|] = \mathbb E\left[\frac{e^{\xi_t}}{\mathbb E\left[e^{\xi_t}\right]}\right] = 1 $$ so that $X_t$ is integrable, and for $s<t$ we have \begin{align} \mathbb E[X_t\mid\mathcal F_s] &= \mathbb E\left[ \frac{e^{\xi_t}}{\mathbb E\left[e^{\xi_t}\right]}\,\big\vert\, \mathcal F_s\right]\\ &= \mathbb E\left[e^{\xi_t-\xi_s}e^{\xi_s}\mid\mathcal F_s\right]\mathbb E\left[e^{\xi_t}\right]^{-1}\\ &=\mathbb E\left[e^{\xi_t-\xi_s}\right]e^{\xi_s}\mathbb E\left[ e^{\xi_t-\xi_s}e^{\xi_s}\right]^{-1}\\ &= \mathbb E\left[e^{\xi_s} \right]^{-1}e^{\xi_s}\\ &= X_s, \end{align} which implies that $X_t$ is a martingale.


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