# Martingale measure

I'm stuck with an exercise here. So there's a random walk on the integers with probability 0.5 of going left or right, i.e. $P[X_i=1]=P[X_i=-1]=1/2$ and $S_n=X_1+...+X_n$. for a real number $c$, set

$M_n=e^{cS_n}(\frac{2}{e^c+e^{-c}})$ and show that $M_n$ is a martingale.

What does that mean? Do I have to show that we have a martingale measure, i.e. some risk-neutral probabilities that sum up to 1? Would that be $\sum_{n \in \mathbb{Z}} M_n=1$? I don't see how to apply the definition on this, if I'm even taking the right definition here. Can someone please explain me what I exactly have to do?

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No. You have to show that $M_n$ is a martingale, so it satisfies three properties: $M_n$ is $\mathcal{F}_n$-measurable; it is integrable and $E[M_{n+1}|\mathcal{F}_n]=M_n$, where (I suppose) $\mathcal{F}_n$ is the natural filtration. –  Kolmo Mar 21 '12 at 18:17
Try this‌​. (And you are probably asked to show that $(M_n)_{n\geqslant0}$, not $M_n$, is a martingale.) –  Did Mar 21 '12 at 20:02
Ah okay, then I think I know what to do. Cool! –  Marie. P. Mar 21 '12 at 20:45

Instead of $$M_n = e^{c \cdot S_n} \cdot \left( \frac{2}{e^c+e^{-c}} \right)$$ it should read $$M_n = e^{c \cdot S_n} \cdot \left( \frac{2}{e^c+e^{-c}} \right)^{\color{red}{n}}$$ Otherwise, $(M_n)_n$ is not a martingale (with respect to the natural filtration).
Denote by $\mathcal{F}_n$ the natural filtration, i.e. $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$.
• Obviously, $M_n$ is $\mathcal{F}_n$ measurable for all $n \in \mathbb{N}$ and $M_n \in L^1$.
• For $n \in \mathbb{N}$, we have \begin{align} \mathbb{E}(M_n \mid \mathcal{F}_{n-1}) &= \left(\frac{2}{e^c+e^{-c}} \right)^n \cdot e^{c \cdot S_{n-1}} \cdot \mathbb{E}(e^{c \cdot X_n} \mid \mathcal{F}_{n-1}) \\ &=\left(\frac{2}{e^c+e^{-c}} \right)^n \cdot e^{c \cdot S_{n-1}} \cdot \mathbb{E}(e^{c \cdot X_n}) \tag{1} \end{align} where we used the independence of the random variables in the last step. Moreover, $$\mathbb{E}(e^{c \cdot X_n}) = \frac{1}{2} \cdot \left( e^c + e^{-c} \right) \tag{2}$$ since $X_n \sim \frac{1}{2} (\delta_1+\delta_{-1})$. Thus, $$\mathbb{E}(M_n \mid \mathcal{F}_{n-1}) \stackrel{(1)}{=}\left(\frac{2}{e^c+e^{-c}} \right)^n \cdot e^{c \cdot S_{n-1}} \cdot \mathbb{E}(e^{c \cdot X_n}) \stackrel{(2)}{=} \left(\frac{2}{e^c+e^{-c}} \right)^{n-1} \cdot e^{c \cdot S_{n-1}} = M_{n-1}$$