Let $B_t$ be a brownian motion. Find $a$ and $b$ so that $W_t=exp(aB_t+bt)$ is martingale relative to the filtration $F_t^B$.
My attempt: For $s \leq t$ we have $E(W_t | F_s)=E(e^{b(t-s)}e^{a(B_t-B_s)}W_s | F_s)$. We know that $B_t-B_s$ is independent from $B_s$ and that $B_t-B_s$ has zero mean. It seems like we are almost there but I can't put the rest of the pieces together.
Best Answer
What you need to know is $Ee^{c(B_v-B_u)}=e^{c^{2} (v-u)/2} $ for $u<v$.
In a martingale expectations remain a constant. So we need $Ee^{aB_t+bt}=1$ for all $t$ (since $B_0=0$). This gives $e^{a^{2}t/2}e^{bt}=1$ or $b=-a^{2}/2$.
To show that $e^{aB_t-\frac {a^{2}} 2 t}$ is martinagle do the following:
$E(e^{aB_t-\frac {a^{2}} 2 t}|\mathcal F_s)=E(e^{a(B_t-B_s)|\mathcal F_s) E(e^{aB_s-\frac {a^{2}} 2 t}}|\mathcal F_s)$. Of course, $E(e^{aB_s-\frac {a^{2}} 2 t}|\mathcal F_s)=e^{aB_s-\frac {a^{2}} 2 t}$. Also, $E(e^{a(B_t-B_s)}|\mathcal F_s)=E(e^{a(B_t-B_s)})=e^{a^{2}(t-s)/2} $. I will let you finish from here.