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 $\sigma=\sigma(t)>0$?


1 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 $$X_t=X_0\exp\left(\left( b-\frac{1}{2}\sigma^2\right)t +\sigma B_t\right).$$

  • $\begingroup$ @fsp-b I'm sorry. Must've read your question too quickly. Does this help? For any $s\leq t$, $$\ln(X_t)|\mathcal{F}_s=\ln(X_s)+\int_s^t\left( b(u)-\frac{1}{2}\sigma^2(u)\right)\text{d}u +\int_s^t \sigma(u)\text{d}B_u,$$ which also follows from the standard conditional expectation arguments. Thus, it's just a small adjustment to get the conditional distribution. You can write $x$ for $X_s$ if you want to. $\endgroup$
    – Kevin
    Commented Sep 22, 2020 at 15:19
  • $\begingroup$ Firstly, $\mathcal{F}_s=\sigma(X_u|u\in[0,s])$, i.e. the natural filtration (all information up to time $s$ and not just the information at time $s$). On the other hand, $X_t$ is Markovian, so it doesn't really matter. I didn't mean the conditional expectation (remember that the expectation of the second integral would be zero). It's more like an expression to derive the conditional distribution. Perhaps I was a bit less formal but that expression was meant to be more useful. You can write $\ln(X_t)$ instead of $\ln(X_t)|\mathcal{F}_s$ if that confuses you. $\endgroup$
    – Kevin
    Commented Sep 22, 2020 at 15:31
  • $\begingroup$ Thanks, that clears it up :) $\endgroup$
    – fsp-b
    Commented Sep 22, 2020 at 15:33
  • $\begingroup$ Why $𝔼[𝑋𝑡]=exp(𝔼[ln(𝑋𝑡)]+1/2𝕍ar[ln(𝑋𝑡)])$ ? $\endgroup$
    – Mike HWANG
    Commented Jan 22, 2022 at 2:29
  • $\begingroup$ Think about the expectation of a log normal random variable @MikeHWANG $\endgroup$
    – Kevin
    Commented Jan 22, 2022 at 8:31

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