Why is $e^{W_t}$ not a martingale?

Question

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!

Answer 1

The mistake in the first approach occurred in the second equality. It is not true (when $s<t$) that $\mathbb E[e^{W_t}\mid \mathcal F_s]=e^{W_t}$. In fact, the simplest way I see to compute this conditional expectation (for $s<t$) is to use the fact that $W_t$ is a Markov process, and therefore so is $e^{W_t}$. Thus, $$ \mathbb E[e^{W_t}\mid \mathcal F_s]=\mathbb E[e^{W_t}\mid W_s]=e^{W_s}\mathbb E[e^{W_t-W_s}\mid W_s]. $$ Now use the independence of increments to conclude that $$ E[e^{W_t-W_s}\mid W_s]=\mathbb Ee^{W_t-W_s}=e^{(t-s)/2}. $$ Consequently, $$ \mathbb E[e^{W_t}\mid \mathcal F_s]=e^{(t-s)/2}W_s. $$

In summary, your second approach is correct, although it sounds like it is a transcription of an argument you heard somewhere and didn't fully understand. Thus in my answer I have attempted to walk through the same steps but in more detail. Please ask if there is a specific step that doesn't make sense or that you are treating as a "black box" without understanding why it is true.