Trying to solve a problem set with:

Let $W_t$ be a Brownian Motion and $X_t = e^{izW_t}$ where $z$ is real, $i = \sqrt{-1}$.

I need to find $\mathbb{E}\left(X_t\right)$... I am a bit stuck.

I have got the SDE $X_t$ satisfies as: $$ dX_t = \left(iz\:dW_t + 0.5 \cdot (iz)^2 \:dt\right) X_t $$ From here I'm trying to get the expectation....

I set:

$m(t) = \mathbb{E}\left(X_t\right)$ and work through to end up with $dM(t) = \mathbb{E}\left(dX(t)\right)$

Then $iz\:dW_t$ in $dX_t$ expectation is 0 as $dW_t$ is $N(0,1)$ distributed....

I end up with the ODE : $$\frac{dm}{dt} = \frac{1}{2} \cdot m \cdot (iz)^2$$

Am I missing something? I am self teaching and new to this!

  • $\begingroup$ Any chance you could use LaTex commands for your formulas pls? $\endgroup$ Commented Jan 16, 2022 at 16:43
  • $\begingroup$ Another user kindly did so for me! Thank you for the reminder and thank you for editing $\endgroup$ Commented Jan 16, 2022 at 16:58
  • $\begingroup$ As per the answer below: if you can use the properties of the Characteristic function without proving these properties, you can just state that $$\varphi(z)\equiv\mathrm{E}\left(e^{izW_t}\right)$$ is how the Characteristic function of a normally distributed random variable is defined: then, referencing the wiki link, you can jump directly to the result. If you need to evaluate the expectation explicitly, have a look at this post. $\endgroup$ Commented Jan 17, 2022 at 14:57
  • $\begingroup$ Thank you, as mentioned below, all answers have helped my understanding! $\endgroup$ Commented Jan 17, 2022 at 19:16

2 Answers 2


As you have already noted, $W_t$ is normally distributed with

$$ W_t\sim \mathrm{N}(0,t) $$


$$ \varphi(z)\equiv\mathrm{E}\left(e^{izW_t}\right)=e^{-\frac{1}{2}z^2t} $$

is the characteristic function of the Normal distribution.

  • $\begingroup$ Thank you!! I should have spotted this, combined with the below answer it helps with the framework a lot $\endgroup$ Commented Jan 17, 2022 at 19:15

Alternatively, you can solve the ODE and recover the result:

With regards to the SDE for $X_t$, I got the same ODE as described above. We can derive the solution, $m(t)$, to the ODE as follows:

\begin{align} \frac{dm(t)}{dt} &= \frac{1}{2} (iz)^2\cdot m(t)\\ &\Updownarrow\\ \frac{2\cdot\frac{dm(t)}{dt}}{m(t)} &= -z^2\\ &\Updownarrow\\ \int \frac{2\cdot\frac{dm(t)}{dt}}{m(t)} \: dt&= \int-z^2\: dt \\ &\Updownarrow\\ 2\ln\left(m(t)\right) &= -z^2t+C_1\\ &\Updownarrow\\ m(t)&=e^{\frac{-z^2t}{2}}e^{\frac{C_1}{2}}, \end{align} where in the third and fourth equation, we have integrated with respect to $t$ on both sides and then collected the constants on one side, called $C_1$. The initial value of the ODE is just the initial value of the SDE, $m(0) := \mathbb{E}\left[X_0\right]=X_0 = e^{\frac{C_1}{2}}$. In conclusion, you will end up with the solution:

$$m(t) := \mathbb{E}\left[X_t\right] = X_0e^{-\frac{1}{2}z^2t}$$

In this regard, I believe that you're missing the initial value, $X_0$, in the first part of your statement for $X_t$.

The solution can be verified by observing that the SDE of $X_t$ is the well-known Geometric Brownian Motion (GBM) with solution: \begin{align} X_t &= X_0 e^{\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t}\\ &= X_0 e^{\left(\frac{1}{2}(iz)^2-\frac{(iz)^2}{2}\right)t + iz W_t}\\ &= X_0 e^{iz W_t} \end{align}

where in your scenario, $\mu = \frac{1}{2}(iz)^2$ and $\sigma = iz$. Here, we know that the expected value of a GBM is given by:

\begin{align} \mathbb{E}\left[X_t\right] &= X_0e^{\mu t}\\ &=X_0e^{-\frac{1}{2}z^2 t}, \end{align}

giving us the same result as the solution of the ODE.

  • $\begingroup$ Thank you so much, the step by step (especially the GBM) is really helpful $\endgroup$ Commented Jan 17, 2022 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.