1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ @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
    Sep 22 '20 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
    Sep 22 '20 at 15:31
  • $\begingroup$ Thanks, that clears it up :) $\endgroup$
    – fsp-b
    Sep 22 '20 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.