Composing Transition Densities

Why Transition Densities Must Compose

A transition density describes the distribution of a process at a later time when its present state is specified. If we observe the process first at an intermediate time, then its later distribution is obtained in two stages: integrate over every possible intermediate state, weighting the second transition by the probability of reaching that state in the first stage. This is the Chapman–Kolmogorov principle in density form. The integral is not an optional approximation: it is exactly what removes the unobserved intermediate state.

We work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For an Itô integral \(\int_0^t H_u\,dB_u\), the integrand \(H_u\) is assumed adapted and progressively measurable, with \(\mathbb{E}\int_0^T H_u^2\,du<\infty\) on each finite time interval under consideration. The examples below use such integrals only to describe the Ornstein–Uhlenbeck process; the density compositions themselves are ordinary integrals with respect to Lebesgue measure.

Let \(P_{s,t}(x,A)\) denote the transition probability of a process from state \(x\) at time \(s\) to a Borel set \(A\) at time \(t\), where \(s<t\). A transition density \(p_{s,t}(x,y)\) is a nonnegative measurable function such that \(P_{s,t}(x,A)=\int_A p_{s,t}(x,y)\,dy\) for every Borel set \(A\). The density is defined only up to changes on a set of \(y\)-values of Lebesgue measure zero.

The phrase “every starting state” matters. An identity established only almost surely for the random state \(X_s\) under one initial law does not determine kernel values at states that this law never visits. In that setting one can conclude the kernel identity only for almost every state under the distribution of \(X_s\), unless an additional assumption specifies the kernels at all states. Here we use the Markov-family version, so the Chapman–Kolmogorov identity is assumed, and hence available, for every \(x\).

From Kernels to Densities

Proof. Fix \(s<r<t\), a starting state \(x\), and a Borel set \(A\). The kernels have densities, so Chapman–Kolmogorov gives

Both densities are nonnegative and measurable. Tonelli’s theorem therefore permits interchanging the order of integration, including when an integral is initially allowed to be infinite:

Taking \(A=\mathbb{R}\) shows \(\int q(y)\,dy=P_{s,t}(x,\mathbb{R})=1\), so \(q\) is a probability density. Two densities of the same measure agree Lebesgue-almost everywhere: if their difference were positive or negative on a set of positive measure, their integrals over an appropriate measurable subset would differ. If the two density versions are continuous and differ at some point, continuity makes their difference have the same strict sign on a nonempty interval, contradicting their almost-everywhere equality. This proves both claims, with the starting state \(x\) fixed but arbitrary. \(\square\)

The formula integrates over the full state space, not merely over states that seem likely or convenient. It also composes kernels across unequal time intervals: the first and second densities need not have the same functional form when the process is time-inhomogeneous.

A Gaussian Convolution Calculation

Proof. The density being integrated is the product of the \(N(m_1,v_1)\) density at \(z\) and the \(N(z+m_2,v_2)\) density at \(y\). In its exponent, complete the square as follows:

The first term on the right is a square in \(z\). Integrating its exponential over \(z\) gives the normalizing factor that turns the product of prefactors \(1/(2\pi\sqrt{v_1v_2})\) into \(1/\sqrt{2\pi(v_1+v_2)}\). The remaining exponential is \(\exp(-(y-m_1-m_2)^2/(2(v_1+v_2)))\). This is the \(N(m_1+m_2,v_1+v_2)\) density, as claimed. \(\square\)

Brownian Motion: The Heat-Kernel Composition

For Brownian motion started at \(x\) at time \(s\), the increment \(B_t-B_s\) has distribution \(N(0,t-s)\). Thus its transition density is \(p_{s,t}(x,y)=\frac{1}{\sqrt{2\pi(t-s)}}\exp(-(y-x)^2/(2(t-s)))\). The intermediate state after time \(r\) has mean \(x\) and variance \(r-s\); the remaining increment has mean zero and variance \(t-r\). The Gaussian convolution calculation therefore predicts mean \(x\) and variance \((r-s)+(t-r)=t-s\).

Ornstein–Uhlenbeck Motion: Decay and Accumulated Variance

Consider the Ornstein–Uhlenbeck stochastic differential equation \(dX_t=-\kappa X_t\,dt+\eta\,dB_t\), with \(\kappa>0\), \(\eta>0\), and initial state \(X_s=x\). Here \(-\kappa x\) is the drift coefficient and \(\eta\) is the diffusion coefficient. Its explicit solution is

The integrand \(H_u=\eta e^{-\kappa(t-u)}\mathbf{1}_{[s,t]}(u)\) is deterministic, hence progressively measurable, and is square-integrable on every finite interval: \(\mathbb{E}\int_s^t H_u^2\,du=\eta^2\int_s^t e^{-2\kappa(t-u)}du<\infty\). The stochastic integral is therefore defined in the Itô sense. Its deterministic-integrand Brownian distribution is centered normal with variance equal to the integral of \(H_u^2\). Consequently the transition is normal with mean \(a_hx\) and variance \(v_h\), where \(h=t-s\), \(a_h=e^{-\kappa h}\), and \(v_h=\frac{\eta^2}{2\kappa}(1-e^{-2\kappa h})\).

Proof. Write \(a_i=e^{-\kappa h_i}\) and \(v_i=\frac{\eta^2}{2\kappa}(1-a_i^2)\) for \(i=1,2\). The intermediate state has density \(N(a_1x,v_1)\), while the density from intermediate state \(z\) to the final state is \(N(a_2z,v_2)\). Since \(a_2>0\), scaling the intermediate normal variable by \(a_2\) gives a normal variable with mean \(a_2a_1x\) and variance \(a_2^2v_1\). Adding the second independent centered normal increment gives mean \(a_2a_1x\) and variance \(a_2^2v_1+v_2\). Substitution yields

Because \(a_1a_2=e^{-\kappa(h_1+h_2)}\), the mean and variance are precisely those in the proposition. The density identity follows from the Gaussian convolution calculation (or the change of variable \(w=a_2z\) followed by that calculation). \(\square\)

A Common Mistake: Freezing the Intermediate State

A tempting but incorrect calculation is to multiply the two transition densities at one chosen intermediate state \(z_0\) and call that the composed density. For Brownian motion started at \(x=0\) and run for two unit intervals, choosing \(z_0=0\) would give \(p_{0,1}(0,0)p_{1,2}(0,y)\). This is a product of two density values, not the density of the final state: it has not accounted for the continuum of possible intermediate states, and it does not have the required units of a density in \(y\).

The correction is to integrate over all intermediate states:

This is the \(N(0,2)\) density, which integrates to one. The integration variable \(z\) is precisely the state that is unobserved between the two transitions.