I am supposed to find the following ($B_t$ is a Brownian motion, and $\mathcal{F}_s$ the generated filtration):
$\mathbb{E}[e^{B_t}|\mathcal{F}_s]=?$ I tried this: shifting by $s$ to the left to get $\mathbb{E}[e^{B_{t-s}}|\mathcal{F}_0]$ and then treat $B_{t-s}$ as an RV with normal distribution and $\mathbb{E}=0$ and $Var=t-s$, and then plug this in into $\mathbb{E}[e^X]=e^{\mu+\frac{\sigma^2}{2}}$ to get $\mathbb{E}=e^\frac{t-s}{2}$. But then I remembered that for a Brownian motion, we must have $B_0=0$, and simply shifting $B_t$ to the left by $s$ will not give a distribution that is centered at $0$. So what will I do now? Is it enough to add $W_s$ to the expected value, or was the whole shifting a bad idea?
Yours, Marie