Let $X_t$ be a random variable $F_t-measurable$, and $\epsilon_t$ a standardized normal random variable independent of $F_t$. ($t\in R$)
Express $log(E[exp(xX_t\epsilon_t) | F_t ])$ in function of $X_t$ and $x$.
My thought were to rewrite it using the moment generating function of the normal distribution, i.e.,
$log(E[exp(xX_t\epsilon_t) | F_t ]) = log(E[exp(xX_t\epsilon_t)]) =log(M_Z(xX_t))$
And this is equal to $0.5(xX_t)^2$.
But how can I go from the conditional expectation to the unconditional one? I don't really know if I'm allowed to do that. Is there a better way to do this ?
Use that $$ \exp(tX_t\epsilon_t) = \sum_k \frac 1{k!} t^kX_t^k \epsilon_t^k$$ Now $X_t^k$ is $F_k$-measurable, hence $\def\E{\mathbf E}\E[X_t^k\epsilon_t^k \mid F_t] = X_t^k\E[X_t^k\mid F_t]$ and $\epsilon_t^k$ is $F_t$-independent, hence $\E[\epsilon_t^k\mid F_t] = \epsilon_t^k$. Therefore $$ \E[\exp(tX_t\epsilon_t)\mid F_t] = \sum_{k} \frac {t^k}{k!}X_t^k \E[\epsilon_t^k] = M_{Z}(tX_t) = \exp\left(\frac 12 t^2X_t^2\right)$$