Calculate $E[X_t]$ and Var$(X_t)$ Let $(X_t)_{t \ge 0}$ be the process given by $X_t= e^{-B_t-t/2}$ for all $t\ge 0$.
Calculate $E[X_t]$ and $Var(X_t)$.
For the expectation, one would have
$E[X_t]=E[e^{-B_t}]e^{-t/2}=1$ due to the moment generating function?
My other idea was to consider the Ito form $dX_t= -X_t dB_t $ . So one would get $X_t-1= \int_0^t -X_ss dB_s$. Following that logic one has: $$E[X_t]=E[\int_0^t-X_sdB_s]+1= 1$$
For $Var(X_t)$, its just  $Var(X_t)=E[X_t^2]-1=e^{-t}E[e^{-2B_t}]-1=e^t-1$ ?
 A: There is no need to use Itô's formula here, as everything can be done using the mgf as you mentioned. Your calculation for $\mathbb{E}[X_t]$ is correct. Then, for $Z 
 \sim N(0,1)$:
\begin{equation}
\mathbb{E}[X_{t}^{2}] = e^{-t} \mathbb{E}\left[ e^{2 \sqrt{t}Z }\right] = e^{-t}e^{2t} = e^t.
\end{equation}
So $\text{Var}[X_t] = \mathbb{E}[X_{t}^{2}] - \mathbb{E}[X_t] = e^t -  1$ as you mentioned.
It's usually good practice to avoid using Itô's formula when more elementary tools are available. The reason for this is that local martingale terms in the semimartingale decomposition may not be true martingales, and  the most common reason for this is lack of integrability. Hence, sometimes the expectation
\begin{equation}
\mathbb{E}\left[ \int_{0}^{t} X_s dB_s \right]
\end{equation}
may  not exist. In  this case you are fine, since you can check that the stochastic exponential is a true martingale, for example as presented here, so you can readily apply Itô's formula and the Itô isometry. But in more general cases, it is something worth worrying about.
