Any advice on the following will be appreciated.
I am trying to show that for $X_t = e^{W_t}$, $X_t$ is not a martingale, where $W_t$ is Brownian motion with respect to filtration $\mathcal F_t$
My approach is $\forall s < t$: \begin{align} E[X_t|Fs] & = E[e^{W_t}|\mathcal F_s] \\ &= E[e^{W_t}] \\ & = e^{t/2} \end{align}
- Is $e^{W_t}$ be independent to $\mathcal F_s$?
I used a pre-established result for the second to third line.
The following is another approach, which gives a different answer, \begin{align} E[X_t|\mathcal F_s] &= E[e^{W_t}|\mathcal F_s] \\ &= E[e^{W_t - W_s + W_s}|\mathcal F_s] \\ &= e^{W_s}E[e^{W_t - W_s}|\mathcal F_s] \\ &=e^{W_s}e^{\frac{t-s}{2}} \end{align}
I used the property $E[e^{W_t - W_s}|\mathcal F_s] = e^{\frac{t-s}{2}}$
Two different approaches that led to two different answers. I believe the former is wrong, but I cannot tell why it is wrong.
Thank you!