Deriving the Geometric Brownian Motion Density with the Fokker–Planck Equation

# Density of Geometric BM via Fokker-Planck


# Density of Geometric BM via Fokker-Planck












Attempting to derive density of a GBM (which we know is log-normal) the long way, using the Fokker Planck-equation. Can't figure out where I went wrong - would appreciate a few sets of extra eyes!

GBM: $dS_t = \mu S_t dt + \sigma S_t dW_t$

Let $p(t,T,x,y) = \mathbb{P}(S(T) = y|S(t) = x)$ be the transition density for $S_t$, for which we know \begin{align} p(0,T,x,y) = \frac{1}{y\sigma \sqrt{2 \pi T}} \exp\left(-\frac{1}{2}\left(\frac{\log(y) - \left(\log(x) + (\mu - \frac{\sigma^2}{2})T\right)}{\sigma \sqrt{T}}\right)^2 \right). \end{align}

I'd like to derive this result using the fact that $p$ (viewed as $p(T,y)$) satisfies the Fokker Planck PDE

\begin{align*} \frac{\partial p}{\partial T} & = -\frac{\partial}{\partial y}(\mu yp) + \frac{\partial^2}{\partial y^2}\left(\frac{1}{2}\sigma^2 y^2 p\right) \\ & = -\mu \left(p + y\frac{\partial p}{\partial y}\right) + \frac{1}{2}\sigma^2 \left(2p + 4y\frac{\partial p}{\partial y} + y^2 \frac{\partial^2 p}{\partial y^2}\right) \\ & = \frac{\sigma^2 y^2}{2} \frac{\partial^2 p}{\partial y^2} + (2 \sigma^2 - \mu)y \frac{\partial p}{\partial y} + (\sigma^2 - \mu)p. \end{align*}

To solve this PDE, I first transform it to a constant coefficient problem using the change of variables $w = \log y$, so $\frac{\partial p}{\partial y} = \frac{\partial p}{\partial w}\frac{1}{y}$ and $\frac{\partial^2 p}{\partial y^2} = \frac{\partial^2 p}{\partial w^2}\frac{1}{y^2} - \frac{\partial p}{\partial w} \frac{1}{y^2}$, giving

\begin{align*} \frac{\partial p}{\partial T} & = \frac{\sigma^2}{2} \left(\frac{\partial^2 p}{\partial w^2} - \frac{\partial p}{\partial w}\right) + (2 \sigma^2 - \mu) \frac{\partial p}{\partial w}+ (\sigma^2 - \mu)p \\ & = \frac{\sigma^2}{2} \frac{\partial^2 p}{\partial w^2} + \left(\frac{3\sigma^2}{2} - \mu\right) \frac{\partial p}{\partial w} + (\sigma^2 - \mu)p. \end{align*}

Take Fourier transform to convert this to an ODE: \begin{align*} \frac{\partial \hat{p}}{\partial T} & = -\frac{\sigma^2 \omega^2}{2} \hat{p} + i\omega\left(\frac{3\sigma^2}{2} - \mu\right) \hat{p} + (\sigma^2 - \mu)\hat{p} \\ \hat{p} & = \hat{p}_0\exp\left((\sigma^2 - \mu)T\right)\exp\left(-\frac{\sigma^2 \omega^2}{2}T + i\omega\left(\frac{3\sigma^2}{2} - \mu\right)T \right) \end{align*} where the transform I'm using is $$ \mathcal{F}[f](\omega) = \int_{-\infty}^\infty e^{-i \omega x} f(x) dx. $$ Since $$ \mathcal{F}\left[\frac{1}{s \sqrt{2\pi}} \exp\left(\frac{1}{2}\left(\frac{w - m}{s}\right)^2\right)\right](\omega) = \exp(-i\omega m - \omega^2s^2/2), $$ with $m = -\left(\frac{3\sigma^2}{2} - \mu\right)T$ and $s = \sigma \sqrt{T}$ we see that \begin{align*} \hat{p} & = \hat{p}_0\exp\left((\sigma^2 - \mu)T\right)\mathcal{F}\left[\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{w -\left(\mu - \frac{3\sigma^2}{2}\right)T}{\sigma \sqrt{T}}\right)^2 \right)\right]. \end{align*} Now, since $\mathcal{F}[f \ast g] = \mathcal{F}[f] \mathcal{F}[g]$ and $p_0 = p(0,w) = \delta(w - w_0)$, \begin{align*} p(T,w) & = \exp\left((\sigma^2 - \mu)T\right)\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{w - w_0 -\left(\mu - \frac{3\sigma^2}{2}\right)T}{\sigma \sqrt{T}}\right)^2 \right). \end{align*} Finally, changing back $w \to \log y$ we get the supposed solution \begin{align*} p(T,y) & = \exp\left((\sigma^2 - \mu)T\right)\frac{1}{\sigma \sqrt{2\pi T}}\exp\left(-\frac{1}{2}\left(\frac{\log(y) - \left(\log(y_0) + \left(\mu - \frac{3\sigma^2}{2}\right)T\right)}{\sigma \sqrt{T}}\right)^2 \right), \end{align*} which is not the lognormal density as above.

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/17891

Hi bcf: This is a good question. As you pointed out below, \begin{align*} p_0 &= \delta(y-y_0)\\ &=\delta(e^w-y_0). \end{align*} Then, \begin{align*} p_0 * g &= \int_{-\infty}^{\infty}\delta(e^z-y_0) g(w-z) dz\\ &=\int_{0}^{\infty}\delta(u-y_0) g(w-\ln u) \frac{1}{u}du\\ &= \frac{1}{y_0}g(w-\ln y_0). \end{align*} Consequently, your last equality becomes \begin{align*} p(T,y) &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{3\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg) \end{align*} Let $Z= \ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big]$. Then, \begin{align*} p(T,y) &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{3\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg)\\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}} + \sigma \sqrt{T}\bigg)^2 \bigg)\\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}} + \sigma \sqrt{T}\bigg)^2 \bigg)\\ &= \frac{1}{y_0}\exp\big((\sigma^2 - \mu)T\big)\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}}\bigg)^2 - Z - \frac{\sigma ^2}{2}T \bigg)\\ &= \frac{1}{y_0}\frac{1}{\sigma \sqrt{2\pi T}}\exp\bigg(-\frac{1}{2}\bigg(\frac{Z}{\sigma \sqrt{T}}\bigg)^2 + \ln y_0 -\ln y \bigg)\\ &= \frac{1}{y \sigma \sqrt{2\pi T}} \exp\bigg(-\frac{1}{2}\bigg(\frac{\ln y - \big[\ln y_0 + \big(\mu - \frac{\sigma^2}{2}\big)T\big]}{\sigma \sqrt{T}}\bigg)^2 \bigg). \end{align*} This is the transition density you provided above.