Transition density of geometric Brownian motion with time-dependent drift and volatility

Question

Can you provide a reference to the transition density of the scalar geometric Brownian Motion with time-dependent drift and volatility, i.e. the scalar process $X = (X_t)_{t\geq 0}$ defined by the SDE

$\mathrm{d}X_t = X_t\cdot(b(t)\,\mathrm{d}t + \sigma(t)\,\mathrm{d}B_t)$

for (sufficiently smooth) functions $b= b(t)\in\mathbb{R}$ and $\sigma=\sigma(t)>0$?

Answer 1

You can do what we always do and take logs and Itô's Lemma:

$$\text{d}\ln(X_t)= \left( b(t)-\frac{1}{2}\sigma^2(t)\right)\text{d}t+\sigma(t)\text{d}B_t.$$ Then, by definition, $$\ln(X_t)=\ln(X_0)+\int_0^t\left( b(s)-\frac{1}{2}\sigma^2(s)\right)\text{d}s +\int_0^t \sigma(s)\text{d}B_s$$ or $$X_t=X_0\exp\left(\int_0^t\left( b(s)-\frac{1}{2}\sigma^2(s)\right)\text{d}s +\int_0^t \sigma(s)\text{d}B_s\right).$$

Because $\int_0^t f(s)\text{d}B_s$ is Gaussian (with zero mean, see here) if $f$ is deterministic (as in your case), your process remains log-normally distributed, just with time-dependent drift and volatility. Note that

\begin{align*} \mathbb{E}[\ln(X_t)] &= \ln(X_0)+\int_0^t\left( b(s)-\frac{1}{2}\sigma^2(s)\right)\text{d}s,\\ \mathbb{V}\text{ar}[\ln(X_t)] &= \int_0^t \sigma^2(s)\text{d}s. \end{align*} As always, $\mathbb{E}[X_t]=\exp\left(\mathbb{E}[\ln(X_t)]+\frac{1}{2}\mathbb{V}\text{ar}[\ln(X_t)]\right)=X_0\exp\left(\int_0^t b(s)\text{d}s\right)$. The variance of $X_t$ is found similarly. If you know the first two moments, you can write down the density of $X_t$, that is

$$f_{X_t}(x) = \frac{1}{x}\frac{1}{\sqrt{2\pi\mathbb{V}\text{ar}[X_t]}}\exp\left(-\frac{\left(\ln(x)-\mathbb{E}[X_t]\right)^2}{2\mathbb{V}\text{ar}[X_t]}\right).$$

If $b(t)\equiv b$ and $\sigma(t)\equiv\sigma$ are constants, you recover the standard $$X_t=X_0\exp\left(\left( b-\frac{1}{2}\sigma^2\right)t +\sigma B_t\right).$$