Let $\newcommand{\F}{\mathcal F} S_n=S_{n-1} +X_n $ where $S_0=0$ , and $X_k$ are iid, and let $\phi(t)=\mathbb{E}e^{itX_1}$ be the characteristic function of $X_k$.

Consider a process $Y_n=e^{itS_n-n\log(\phi(t))}$. Show that the process $(Y_n, \F_n)$ is a martingale, where $\F_n=\sigma(X_1,...,X_n)$.

I am not too sure about how to calculate the conditional expectation $\mathbb{E}[Y_n \mid \F_{n-1}]$.

$\mathbb{E}[Y_n \mid \F_{n-1}]=\mathbb{E}[e^{itS_n-n\log\phi(t)}]=\mathbb{E}[e^{it(S_{n-1}+X_n)-n\log\phi(t)} \mid \F_{n-1}]=e^{itS_{n-1}}\mathbb{E}[e^{itX_n-n\log\phi(t)} \mid \F_{n-1}]$ $=e^{itS_{n-1}}\mathbb{E} [e^{itX_n} - \phi (t)^n \mid \F_{n-1}]$

At this point, I am stuck.

  • $\begingroup$ This is probably homework (with a typo). What do you know? What have you tried? Where are you stuck? $\endgroup$ – Did Oct 22 '11 at 16:47
  • $\begingroup$ I am stuck after the last step $\endgroup$ – Steven Ap Oct 22 '11 at 17:05
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    $\begingroup$ Good. You got your exponentials wrong at the last step but nevermind: one term in your next-to-last conditional expectation is not random (the one with $\phi(t)$) so you can get rid of it and you are left with the conditional expectation of a function of $X_n$ conditionally on $F_{n-1}$. Well, this is the time to remember the independence structure of your process, I would say... Does that ring a bell? $\endgroup$ – Did Oct 22 '11 at 17:25
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    $\begingroup$ $e^{-n\log\phi(t)}$ is not $-\phi (t)^n$. $\endgroup$ – user940 Oct 22 '11 at 18:59
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    $\begingroup$ @Steven: I've merged your accounts. Please register your account to avoid creating duplicate accounts in the future. $\endgroup$ – Qiaochu Yuan Jan 21 '12 at 19:07

Note that $\log(\phi(t))$ is not well-defined for $t \in \mathbb{R}$ such that $\phi(t) \leq 0$. Consider for example $X_1 \sim U_{[-1,1]}$, then $\phi(t)= \frac{\sin t}{t}$, thus $\phi(t) \leq 0$ for $t \in [\pi,2\pi]$. To avoid these inconveniences, define $Y_n$ by $$Y_n := e^{\imath \, t \cdot S_n} \cdot \phi(t)^{-n}$$ where $t \in \mathbb{R}$ such that $\phi(t) \not= 0$. For all $t \in \mathbb{R}$ such that $\phi(t) > 0$, this coincides with your definition of $Y_n$ since $$\exp(-n \cdot \log \phi(t)) = \phi(t)^{-n}$$ if $\phi(t)>0$.

To prove $(Y_n)_n$ a martingale we use the independence of the random variables $(X_n)_n$: $$\mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) = e^{\imath \, t \cdot S_{n-1}} \cdot \phi(t)^{-n} \cdot \mathbb{E}(e^{\imath \, t \cdot X_n} \mid \mathcal{F}_{n-1}) = e^{\imath \, t \cdot S_{n-1}} \cdot \phi(t)^{-n} \cdot \mathbb{E}(e^{\imath \, t \cdot X_n})$$ Since the random variables $(X_n)_n$ are identically distributed we have $$\mathbb{E}e^{\imath \, t \cdot X_n} = \mathbb{E}e^{\imath \, t \cdot X_1} = \phi(t)$$ Thus $$\mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) = e^{\imath \, t \cdot S_{n-1}} \cdot \phi(t)^{-n} \cdot \phi(t)=Y_{n-1}$$


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