TITLE: Showing a process is a martingale
QUESTION [7 upvotes]: 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.

REPLY [1 votes]: 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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