Hello everyone if have the following task:
Prove that the process $Y_t=(B_t+t)\exp(-B_t-t/2)$ is a martingale.
I know that the stochastic exponential $X_t=\exp(B_t-t/2)$ is a martingale, where $B_t$ is the standard Brownian Motion.
What I have got so far is the following:
\begin{alignat*}{2}
\mathbb{E}\left[Y_t|\mathscr{F}_s\right] & = \mathbb{E}[(B_t+t)\exp(-B_t-t/2)|\mathscr{F}_s] \\
& = \mathbb{E}[B_t\exp(B_t-t/2)|\mathscr{F}_s]+t\mathbb{E}[\exp(B_t-t/2|\mathscr{F}_s)]
\end{alignat*}
Now the second term is equal to $t\exp(B_s-s/2)$.
For the first term I calculated: $$ \mathbb{E}[B_t\exp(B_t-t/2)|\mathscr{F}_s]=\exp(-t/2)\mathbb{E}[B_t\exp(B_t-B_s)\exp(B_s)|\mathscr{F}_s] $$
We can take the term $\exp(-B_s)$ out of the integral and have left $\mathbb{E}[B_t\exp(B_t-B_s)|\mathscr{F}_s]$.
Now I think we can split the expectation as $B_t$ is independent of $(B_t-B_s)$. Hence we get \begin{alignat*}{2} \mathbb{E}[B_t\exp(B_t-B_s)|\mathscr{F}_s] & = B_s\mathbb{E}[\exp(B_t-B_s)|\mathscr{F}_s]\\ & = B_s\exp(1/2(t-s))\\ \end{alignat*}
So all in all $\mathbb{E}[Y_t|\mathscr{F}_s]=B_s\exp(B_s-s/2)+t\exp(B_s-s/2)\neq Y_s$
My question is where I went wrong. Any hints on how to solve the problem would be greatly appreciated.