I have to derive the Geometric Brownian motion (with not constant drift and volatility), and to find the mean and variance of the solution.
$\quad \left\{\begin{aligned}
& d X_t = \mu(t) X_t d t + \sigma(t) X_t d W_t \\
& X_0 = \xi
\end{aligned}\right.$
The solution can be obtained in a classical manner by Ito's Lemma:
$X_t = \xi e^{\int_0^t \left(\mu(s) – \frac{\sigma^2(s)}{2}\right) d s + \int_0^t \sigma(s) d W_s}$
And we can find the mean and variance:
$\mathbb{E}[X_t] = \xi e^{\int_0^t \left(\mu(s) – \frac{\sigma^2(s)}{2}\right) ds} \mathbb{E}\left[e^{\int_0^t \sigma(s) dW_s}\right]$
$Var(X_t) = \xi^2 e^{\int_0^t \left(2\mu(s) – \sigma^2(s)\right) d s} \left(\mathbb{E}\left[e^{2 \int_0^t \sigma(s) d W_s}\right] – \mathbb{E}\left[e^{\int_0^t \sigma(s) d W_s}\right]^2\right)\\$
These expression are not really simple, as they are when $\mu$ and $\sigma$ are constant.
Does someone know if we have a general expression for the expectation of the exponential of an Itô's Integral? (in this case of a deterministic function). I.e:
$\mathbb{E}\left[e^{\int_0^t \sigma(s) dW_s}\right]$
Thanks.
Best Answer
Given that $\sigma(s)$ is a deterministic function, then the process $(X(t))_{t \geq 0}$, where $$X(t) = \int_{0}^{t} \sigma(s) dW_s,$$ is a Gaussian process with zero mean and covariance function $\rho(s,t) = \displaystyle \int_{0}^{\min(s,t)} \sigma(u)^2 du$. A proof of this theorem can be found in Schreve's stochastic calculus for Finance II. Hence, $$\mathbb{E}\left[\exp\left(\int_{0}^{t} \sigma(s) dW_s\right)\right],$$ is the moment generating function of a Normally distributed random variable with zero mean and variance $\rho(t,t)$ evaluated in $1$.