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? $\endgroup$ Jan 16 at 16:43
  • $\begingroup$ Another user kindly did so for me! Thank you for the reminder and thank you for editing $\endgroup$ Jan 16 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. $\endgroup$ Jan 17 at 14:57
  • $\begingroup$ Thank you, as mentioned below, all answers have helped my understanding! $\endgroup$ Jan 17 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 $\endgroup$ Jan 17 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 $\endgroup$ Jan 17 at 19:15

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