# Expectation of functions with Brownian Motion embedded

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!

• Any chance you could use LaTex commands for your formulas pls? Jan 16 at 16:43
• Another user kindly did so for me! Thank you for the reminder and thank you for editing Jan 16 at 16:58
• 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. Jan 17 at 14:57
• Thank you, as mentioned below, all answers have helped my understanding! Jan 17 at 19:16

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

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

Then,

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

is the characteristic function of the Normal distribution.

• Thank you!! I should have spotted this, combined with the below answer it helps with the framework a lot 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.

• Thank you so much, the step by step (especially the GBM) is really helpful Jan 17 at 19:15