1. ## To prove martingale

I have problem in the following question

Let $\displaystyle \eta_1,...,\eta_n$ be a sequence of independent identically distributed random variables with $\displaystyle E\eta_i=0$.
Show that the sequence $\displaystyle \xi=(\xi_k)$ with

$\displaystyle \xi_k = \frac{exp \lambda.\left(\eta_1+...+\eta_k \right)}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)^{k}}$
is a martingale.

I need to prove $\displaystyle E(\xi_{k+1}|F_k)=\xi_k$

$\displaystyle \xi_{k+1} = \frac{exp \lambda.\left(\eta_1+...+\eta_{k+1} \right)}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)^{k+1}}$

It can be decomposed as

$\displaystyle \xi_{k+1} = \frac{exp \lambda . \left(\eta_1+...+\eta_k \right)}{\left(E\ \textrm{exp}\lambda \eta_1 \right)^{k}}.\frac {\textrm{exp} \lambda\eta_{k+1}}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)}$

$\displaystyle \xi_{k+1}=\xi_k.\frac {\textrm{exp} \lambda\eta_{k+1}}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)}$

$\displaystyle E(\xi_{k+1}|F_k)=\xi_k.E(\frac {\textrm{exp} \lambda\eta_{k+1}}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)}|F_k)$

Anybody could help to bring
$\displaystyle E(\frac {\textrm{exp} \lambda\eta_{k+1}}{\left(E\ \textrm{exp}\ \lambda \eta_1 \right)}|F_k) =1$

2. Hello,

Since $\displaystyle E[\exp(\lambda \eta_1)]$ is a constant, we can pull it out from the conditional expectation and we get $\displaystyle \frac{1}{E[\exp(\lambda \eta_1)]}\cdot E\left[\exp(\lambda \eta_{k+1})\mid \mathcal F_k\right]$

Since $\displaystyle (\eta_k)$ is a sequence of independent rv's, $\displaystyle \eta_{k+1}$ and $\displaystyle \mathcal F_k$ are independent, and so is $\displaystyle \exp(\lambda \eta_{k+1})$ (since it's a measurable function of $\displaystyle \eta_{k+1}$.

Thus we have $\displaystyle \frac{1}{E[\exp(\lambda \eta_1)]} \cdot E[\exp(\lambda \eta_{k+1})]$ and since the $\displaystyle \eta_k$ have the same distribution, these two expectations are equal...