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!

  • $\begingroup$ The second one is right; the intuitive point is that by conditioning on $\mathcal{F}_s$, you know the value of $W_s$ which gives you better information on what $e^{W_t}$ will ultimately turn out to be. $\endgroup$
    – Ian
    Commented Jul 3, 2019 at 15:04
  • $\begingroup$ @Ian Thanks, but what is wrong with the former? $\endgroup$ Commented Jul 3, 2019 at 15:06
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    $\begingroup$ $W_t$ is not independent with $\mathcal F_s$. Wiener process doesn't have independent values. That's why $e^{W_t}$ also is not independent with $\mathcal F_s$. $\endgroup$ Commented Jul 3, 2019 at 15:12
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    $\begingroup$ @GalvinNg that's right. It should be very careful, cause independent increments means the future increments are independent of the past values. But we cannot say that value is independent with past increments (and especially with past values). $\endgroup$ Commented Jul 3, 2019 at 15:39
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    $\begingroup$ Note that it actually suffices to show that $\mathbb{E}(e^{W_t})$ is not constant (in time). $\endgroup$
    – saz
    Commented Jul 3, 2019 at 16:40

1 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.

  • $\begingroup$ Thanks for replying. I understand the second argument. I worked out $E[e^{W_t - W_s}|\mathcal F_s] = e^{{t-s}/2}$ as an exercise earlier. Also, I believe you meant $E[e^{W_t}|\mathcal F_s] = E[e^{W_t}]$ in the second line right? $\endgroup$ Commented Jul 5, 2019 at 11:00
  • $\begingroup$ No, that is incorrect! The conditional expectation of $e^{W_t}$ given $\mathcal F_s$ is not deterministic number... it depends on $W_s$. In particular, it does not equal $\mathbb E[e^{W_t}]$ $\endgroup$
    – pre-kidney
    Commented Jul 6, 2019 at 8:35
  • $\begingroup$ I know that $E[e^{W_t}|\mathcal F_s] \neq E[e^{W_t}]$ but I was asking if you meant "It is not true (when $s < t$) that $E[e^{W_t}|\mathcal F_s] = E[e^{W_t}]$" instead of "It is not true (when $s < t$) that $E[e^{W_t}|\mathcal F_s] = e^{W_t}$" as is your second line Basically, did you make a typo? Regardless, I understand and appreciate your advice! Thanks $\endgroup$ Commented Jul 6, 2019 at 9:10

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