Transition Distribution of Geometric Brownian Motion with Variable Coefficients
Summary
The document derives the distribution of a scalar geometric Brownian motion when its drift and volatility are deterministic functions of time. Applying Itô’s lemma to the logarithm converts the process into an integral of a deterministic drift term and a stochastic integral. The latter is Gaussian, with variance given by the integrated squared volatility, so the level of the process is lognormally distributed.
It also gives expressions for the mean and variance of the log process and the expected level, and notes that constant coefficients recover the standard geometric Brownian motion. The displayed density formula in the answer is not correctly parameterized: a lognormal density uses the mean and variance of the logarithm, rather than the mean and variance of the level. The derivation supports the distributional result, but readers should verify the density expression before using it.
Key ideas
- Taking the logarithm and applying Itô’s lemma yields an additive process with a time-varying drift.
- The stochastic integral with deterministic volatility is Gaussian, with variance equal to integrated squared volatility.
- The process level is lognormally distributed, with parameters determined by integrated drift and volatility.
- The density printed in the answer incorrectly uses moments of the level in its logarithmic terms.
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# Transition density of geometric Brownian motion with time-dependent drift and volatility
# Transition density of geometric Brownian motion with time-dependent drift and volatility
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 by Kevin (score 1, accepted)
https://quant.stackexchange.com/a/58143
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 $$X_t=X_0\exp\left(\left( b-\frac{1}{2}\sigma^2\right)t +\sigma B_t\right).$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.