I want to compute the characteristic function of the standard geometric brownian motion. Note that I know that it is easy when you exploit the distributional properties of the process, but I'm trying to come up with some exercises by myself in order to apply the same approach to broader classes of stochastic processes.
Consider the usual equation of the GBM:
$$
dX_t = \mu X_t dt + \sigma X_t dW_t
$$
Then, by Feynman-Kac, we have to solve the PDE:
$$
\frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x \frac{\partial^2 f}{\partial x^2} = 0
$$
In order to find the characteristic function, we have to take into account the terminal condition
$$
f(x,T) = \exp(i \omega x)
$$
Substituting $f(x,T)$ inside the PDE yields
$$
\mu x i \omega \exp(i \omega x) – \frac{1}{2}\sigma^2 x \omega^2 \exp(i \omega x)=0
$$
But then, how do I proceed? The above doesn't look at all like the characteristic function of a random variable with lognormal distribution).
Obviously considering the logarithm of $\frac{dX_t}{X_t}$ simplifies things a lot, however I really want to came up with the characteristic function of $X_t$, not its logarithm (see p. 41-42 in Pricing Bermudan and American Options Using the FFT Method).
I hope that somebody can help me, or even discuss things a little bit. By the way, excuse me for my poor way of handling PDEs. 🙂
Best Answer
First of all there is a typo in the partial differential equation(PDE) that you state. Let us firstly recall that PDE describes the time evolution of the one-point probability density function of the stochastic process in question. That PDE is termed the backward Chapman-Kolmogorov equation; you can look up the Wikipedia page to learn how to get from the Langevin equation (the stochastic differential equation) to the Chapman-Kolmogorov equation. Now the correct version is : \begin{equation} \frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2x^2 \frac{\partial^2 f}{\partial x^2} = 0 \end{equation} Now solving PDEs is in general a hard task but in this particular case the substitution $y=\log(x)$ turns the PDE into one with constant coefficients. Indeed we have: \begin{eqnarray} \frac{\partial f}{\partial y} &=& x \frac{\partial f}{\partial x}\\ \frac{\partial^2 f}{\partial y^2} &=& x \frac{\partial f}{\partial x} + x^2 \frac{\partial ^2 f}{\partial x^2} \end{eqnarray} and therefore our PDE takes the form: \begin{equation} \frac{\partial f}{\partial t} + \left(\mu-\frac{1}{2} \sigma^2\right) \frac{\partial f}{\partial y} + \frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2} = 0 \end{equation} which is basically saying that the logarithm of a geometrical Brownian motion with parameters $(\mu,\sigma^2)$ is a Gaussian stochastic process with drift $\mu - 1/2 \sigma^2$ and variance $\sigma^2$. Interestingly enough we would have arrived at the same conclusion by using the original Langevin equation and Ito's lemma only. Now going back to the PDE in question the standard way of solving it is to take Fourier transforms which is actually something you seek to obtain. We define: \begin{equation} \tilde{f}(k,t) := \frac{1}{2\pi} \int\limits_{\mathbb R} e^{\imath k y} f(y,t) d y \end{equation} and we transform the PDE above into a ODE: \begin{equation} \frac{d \tilde{f}}{d t} + \left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right] \cdot \tilde{f} = 0 \end{equation} which, given an initial condition at time $t=T$, is solved by: \begin{equation} \tilde{f}(k,t) = \tilde{f}(k,T) e^{\left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right](T-t)} \end{equation} This is the characteristic function that you were seeking. In applications like option pricing the initial condition $\tilde{f}(k,T)$ is equal to a Fourier transform of the option's payoff at maturity and the fair price of an option at some time before maturity is obtained by taking an inverse Fourier transform of the characteristic function.
Now, the really interesting question is to generalize that approach to more generic stochastic processes for example to L\'{e}vy stable processes. This leads to a fractional PDE. Those are much more difficult to handle yet it is worthwhile to pursue this approach because in here the Ito's lemma does not work.