I'm trying to solve the following problem, and am having problems with the expectation operator:
Let $(X_n)_{n\geq1}$ be independent such that $E(X_i)=m_i$, $var(X_i)=\sigma_i^2$, $i\geq1$. Let $S_n=\sum_{i=1}^n X_i$ and $\mathcal{F}=\sigma(X_i,1\leq1\leq n)$. Assume there is a real number $\lambda$ such that $e^{\lambda X_i} \in L^1$ for any $i\geq1$. Find a sequence $(a_n^{(\lambda)})_{n \geq 1}$ such that $e^{\lambda S_n - a_n^{(\lambda)}}$ is a $(\mathcal{F}_n)_{n \geq 1}$-martingale.
Here is my attempted solution:
Let $z_n=e^{\lambda S_n - a_n^{(\lambda)}}$
$E(z_{n+1}|F_n)=E(z_{n}|F_n)E(\dfrac{z_{n+1}}{z_{n}}|F_n)$.
We need to show $E(\dfrac{z_{n+1}}{z_{n}}|F_n)=1$.
$\dfrac{z_{n+1}}{z_{n}}=\dfrac{e^{\lambda(S_n+X_{n+1}) - a_{n+1}^{(\lambda)}}}{e^{\lambda S_n - a_n^{(\lambda)}}}$
$=e^{\lambda X_{n+1} - (a_{n+1}^{(\lambda)}-a_{n}^{(\lambda)})}$
But the expectation of that last expression is not simply equal to the expression with $X_{n+1}$ replaced by $m_{n+1}$. If it were it would be simple to define the sequence as $a_{n+1}^{(\lambda)}-a_{n}^{(\lambda)}=\lambda m_{n+1}$. Please can someone suggest where to go from here? Jensen's inequality doesn't seem to help as I need actually to define the sequence. Thanks!
Since $X_{n+1}$ and $F_n$ are independent, we have
$$\mathbb{E}(g(X_{n+1}) \mid F_n) = \mathbb{E}(g(X_{n+1}))$$
(for all measurable $g$ such that $g(X_{n+1}) \in L^1$). Thus, in your case,
$$\mathbb{E} \left( \exp\big( \lambda \cdot X_{n+1} - (a_{n+1}^{(\lambda)}-a_n^{(\lambda)}) \big) \mid F_n \right) = \exp \big(- (a_{n+1}^{(\lambda)}-a_n^{(\lambda)}) \big) \cdot \mathbb{E}e^{\lambda \cdot X_{n+1}} \stackrel{!}{=} 1$$
so
$$a_n^{(\lambda)} := \sum_{j=1}^n \ln (\mathbb{E}e^{\lambda \cdot X_j}) $$
does the job (and is well-defined).