Transition Densities of Brownian Motion and Geometric Brownian Motion

From an SDE to a Transition Density

A transition density describes the distribution of a process at a future time given its present state. For Brownian motion, independent Gaussian increments give the density directly. For geometric Brownian motion, the logarithm of the process evolves as Brownian motion with drift, so its transition law is obtained by transforming a Gaussian distribution. The conditioning matters: the current state may itself be random, and the formulas below then hold with that state as the parameter, almost surely.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Thus, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Every Ito integral below is taken for a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_u^2\,du<\infty\) on the time interval in question. In particular, the integral with integrand \(1\) is admissible on every finite interval because \(\mathbb{E}\int_0^T1\,du=T\).

Brownian Motion: Gaussian Transition Density

For Brownian motion started at zero, \(B_t=B_s+(B_t-B_s)\). The first term is known at time \(s\), and the second is an independent \(N(0,t-s)\) random variable. Thus the transition density from state \(x\in\mathbb{R}\) at time \(s\) is

Proof. Put \(h=t-s>0\) and \(Z=B_t-B_s\). By the Brownian increment property, \(Z\) is independent of \(\mathcal{F}_s\) and has density \((2\pi h)^{-1/2}\exp(-z^2/(2h))\). For bounded Borel \(\varphi\), independence gives \(\mathbb{E}[\varphi(B_s+Z)\mid\mathcal{F}_s]=\int_{\mathbb{R}}\varphi(B_s+z)(2\pi h)^{-1/2}e^{-z^2/(2h)}\,dz\). The substitution \(y=B_s+z\) gives the stated transition-density identity. A normal variable with mean zero and variance \(h\) has mean zero and second moment \(h\); applying these to \(B_t=B_s+Z\) gives the conditional mean and variance. The density integrates to one by the same substitution and normalization of the \(N(0,h)\) density. \(\square\)

The formula is a conditional statement, not conditioning on an event such as \(B_s=x\), which usually has probability zero. It says that the conditional distribution is represented by the kernel evaluated at the random state \(B_s\). The same kernel applies to Brownian motion started at any deterministic \(x\): its position after elapsed time \(h\) is \(x\) plus an independent \(N(0,h)\) increment.

Geometric Brownian Motion: From Logarithms to a Density

Consider geometric Brownian motion with deterministic \(S_0>0\), constant drift \(\mu\in\mathbb{R}\), and constant diffusion coefficient \(\sigma\in\mathbb{R}\):

Here \(\mu S_t\) is the drift coefficient and \(\sigma S_t\) is the diffusion coefficient. The Ito integral in this equation has integrand \(H_u=\sigma S_u\); on each finite horizon this is an admissible square-integrable integrand for the solution below, since \(\mathbb{E}\int_0^T \sigma^2S_u^2\,du<\infty\). The case \(\sigma\ne0\) gives a lognormal transition density. We derive the solution and verify it rather than assuming the density formula.

For any positive solution, apply Ito's formula to \(f(x)=\log x\), whose derivatives are \(f'(x)=1/x\) and \(f''(x)=-1/x^2\). The correction term is \(\tfrac12 f''(S_t)(\sigma S_t)^2\,dt=-\tfrac12\sigma^2dt\). Thus \(d\log S_t=(\mu-\tfrac12\sigma^2)dt+\sigma\,dB_t\), and integrating yields the candidate \(S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)\).

To verify the candidate, set \(g(t,x)=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma x)\). Its partial derivatives are \(g_t=(\mu-\tfrac12\sigma^2)g\), \(g_x=\sigma g\), and \(g_{xx}=\sigma^2g\). The time-space Ito formula, including its correction, gives \(d g(t,B_t)=[g_t+\tfrac12g_{xx}]dt+g_xdB_t=\mu g\,dt+\sigma g\,dB_t\), as required. Also, \(\mathbb{E}[S_u^2]=S_0^2\exp((2\mu+\sigma^2)u)\), so \(\mathbb{E}\int_0^T(\sigma S_u)^2du=\sigma^2S_0^2\int_0^T e^{(2\mu+\sigma^2)u}du<\infty\). This confirms the stated integrability on finite horizons. The explicit formula is positive, so the logarithm argument applies; it also shows any positive solution must equal this candidate.

Proof. Write \(h=t-s>0\). The explicit solution gives \(S_t=S_s\exp((\mu-\tfrac12\sigma^2)h+\sigma(B_t-B_s))\). Conditional on \(\mathcal{F}_s\), the increment \(B_t-B_s\) is \(N(0,h)\). Therefore \(\log S_t\) is conditionally normal with mean \(\log S_s+(\mu-\tfrac12\sigma^2)h\) and variance \(\sigma^2h\). Its normal density, transformed by \(z=\log y\), acquires the Jacobian factor \(1/y\) on \(y>0\); this gives the displayed density. The transformation maps \((0,\infty)\) onto \(\mathbb{R}\), so the density integrates to one, and \(S_t\) has no mass on \(y\le0\).

For the moments, let \(Z=B_t-B_s\). For every real \(a\), completing the square in the \(N(0,h)\) density gives \(\mathbb{E}[e^{aZ}]=e^{a^2h/2}\). Hence \(\mathbb{E}[S_t^q\mid\mathcal{F}_s]=S_s^qe^{q(\mu-\sigma^2/2)h}\mathbb{E}[e^{q\sigma Z}]=S_s^q e^{(q\mu+\frac12q(q-1)\sigma^2)h}\). Taking \(q=1\) gives the mean; taking \(q=2\) gives the second raw moment \(S_s^2e^{(2\mu+\sigma^2)h}\). Subtracting the square of the mean yields \(S_s^2e^{2\mu h}(e^{\sigma^2h}-1)\), the stated conditional variance. \(\square\)

Degenerate Cases and a Common Mistake

If \(\sigma=0\), the SDE reduces to \(dS_t=\mu S_t\,dt\) and its solution is \(S_t=S_s e^{\mu(t-s)}\). The conditional distribution is then a point mass, not a density with respect to Lebesgue measure. The lognormal density formula must not be used in this case, because its variance parameter \(\sigma^2(t-s)\) is zero. If \(S_0>0\), the process stays positive; in the nondegenerate case \(\sigma\ne0\), its transition density is supported on positive values only.

Common mistake: omit the Ito correction. A tempting but incorrect proposed solution is \(S_t=S_0e^{\mu t+\sigma B_t}\). For \(g(t,x)=S_0e^{\mu t+\sigma x}\), the derivatives are \(g_t=\mu g\), \(g_x=\sigma g\), and \(g_{xx}=\sigma^2g\). Ito's formula therefore gives \(dg=(\mu+\tfrac12\sigma^2)g\,dt+\sigma g\,dB_t\), which has the wrong drift. The correction in the exponent, \(-\tfrac12\sigma^2t\), cancels the extra \(\tfrac12\sigma^2g\,dt\) term.

All the conditional laws and moments here are under \(\mathbb{P}\). A change of measure would require identifying the new measure and the Brownian motion under it; it is not implicit in a transition-density calculation. Also, the conditional moment formula for geometric Brownian motion is useful beyond the mean and variance: for any real \(q\) it gives a finite conditional moment, because the exponential moments of a Gaussian variable are finite for every real parameter.