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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.

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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 $1$.

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  • $\begingroup$ Fast and precise, thanks! $\endgroup$ – Edouard Berthe May 11 '16 at 15:09
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I would like to add something to the excellent answer of Siron: although it is good to know the Ito Integral of a deterministic function is Gaussian, we can solve my problem without that property.

Let

$\quad g \colon [0,T] \longmapsto \mathbb{R}$

a deterministic function. Then if we define:

$\quad Y_t = \int_0^t g(s) \mathrm{d} W_s \iff \mathrm{d} Y_t = g(t) \mathrm{d}W_t\\ \quad f \colon (t, x) \mapsto e^x \\ \quad Z_t = f(t, Y_t) = e^{Y_t}$

Then by Ito's Lemma: $\def\d{\mathrm{d}}$ $\d Z_t = \left(0 + 0 \times Z_t + \frac{1}{2}g^2(t)Z_t\right)\d t + g(t)Z_t\d W_t = \frac{1}{2}g^2(t)Z_t\d t + g(t)Z_t\d W_t$

I.e:

$\quad Z_t = Z_0 + \int_0^t \frac{1}{2}g^2(t)Z_t\d t + \int_0^t g(t)Z_t\d W_t$

So:

$\def\E{\mathbb{E}}$ $\E[Z_t] = \E\left[Z_0 + \int_0^t \frac{1}{2}g^2(t)Z_t\d t\right] = \E[Z_0] + \int_0^t \frac{1}{2}g^2(s)\E[Z_t]\d t$

Because expectation of Ito Integral is null + Fubini.

So $h \colon t \mapsto \E[Z_t]$ checks: $\quad h'(t) = \frac{1}{2}g^2(t)h(t)$.

Therefore: $\quad h(t) = h(0)e^{\int_0^t \frac{1}{2}g^2(s)\d s} = e^{\int_0^t \frac{1}{2}g^2(s)\d s}$

This is what we want!

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