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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.

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    This is probably homework (with a typo). What do you know? What have you tried? Where are you stuck?2011-10-22
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    I am stuck after the last step2011-10-22
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    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?2011-10-22
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    if its independent, then i am left with the uncond expectation of function of random variable thanks2011-10-22
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    Was that the definition of $Y_t$ given in the assignment? It seems to introduce some unnecessary complications arising from how to interpret $\log$. These are minor, but seem like they could be easily avoided by a simple redefinition of $Y_t$.2011-10-22
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    $e^{-n\log\phi(t)}$ is *not* $-\phi (t)^n$.2011-10-22
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    Steven, you seem to say you saw the idea of the proof. Then I suggest you write it down with all due details and that you post it here, so that we can check it. If you do that carefully, you will see that some hypothesis are missing. (And yes, please do get rid of these awful exp(-log)...)2011-10-22
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    carrying from where i left off $e^{itS_{n-1}-n\log\phi(t)}\mathbb{E}[e^{itX_n}|F_{n-1}]=e^{itS_{n-1}-n\log\phi(t)}\mathbb{E}(e^{itX_n})$ $=e^{itS_{n-1}-n\log\phi(t)}\phi(t)=e^{itS_{n-1}-n\log\phi(t)}e^{\log\phi(t)}=e^{itS_{n-1}-(n-1)\log\phi(t)}=Y_{n-1}$2011-10-23
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    Right, you got the basic idea. Now, you should try to rewrite the whole thing without $\exp(-\log\phi(t))$, which may cause unneeded problems. Consider for example the case of centered Bernoulli random variables $X_k$ then $\phi(t)=\cos(t)$ hence $\log\phi(t)$ is $\log(-1)$ for $t=\pi$... To rewrite the proof from the start with this in mind is a learning exercise.2011-10-23
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    @Steven: I've merged your accounts. Please register your account to avoid creating duplicate accounts in the future.2012-01-21

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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 $$\mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) = e^{\imath \, t \cdot S_{n-1}} \cdot \phi(t)^{-n} \cdot \phi(t)=Y_{n-1}$$