Composing Transitions Across Unobserved Intermediate Times

Why Compose Transitions?

A discretely observed diffusion may be measured at times $t_0$ and $t_2$, even though its state evolves continuously between them. The transition density over that full interval is not generally obtained by pretending the diffusion made one simple step. The Markov property instead says that we can condition on the state at an intermediate time $t_1$, then integrate over every possible intermediate state. This is useful when an interval is long, when a short-step approximation is more accurate, or when an analytic transition is available only over shorter intervals.

The intermediate state in this calculation is unobserved. It is integrated out; it is not treated as additional data. The resulting density can be used for a likelihood contribution between observed endpoints. The formula also provides the foundation for approximating a long-interval transition by combining shorter transition kernels.

Definition: Let $X_t$ be a time-inhomogeneous Markov process on a state space $E$, with transition density $p(s,x;t,y)$ with respect to a common reference measure $\mu$ on $E$. This means $P(X_t\in A\mid X_s=x)=\int_A p(s,x;t,y)\,\mu(dy)$ for measurable sets $A$. For $s<u<t$, the composition formula is $$ p(s,x;t,y)=\int_E p(s,x;u,z)\,p(u,z;t,y)\,\mu(dz). $$ The equality holds for $\mu$-almost every $y$, subject to the usual choice of density versions. For a diffusion with a Lebesgue density, $\mu(dz)=dz$; for a different reference measure, the integral must use that measure.

Why the Composition Formula Holds

The relevant assumptions are that the process is Markov at time $u$, and that the transition kernels from $s$ to $u$ and from $u$ to $t$ have densities with respect to the same reference measure. The proof uses conditional probability and integration of nonnegative functions, so Tonelli’s theorem permits the order of integration without requiring a separate integrability bound.

Condition on the intermediate state.
For a measurable set $A$, the Markov property gives $P(X_t\in A\mid X_s=x)=\int_E P(X_t\in A\mid X_u=z)\,P(X_u\in dz\mid X_s=x)$.

Insert the transition densities.
The two conditional measures have densities, so the expression is $\int_E \int_A p(u,z;t,y)\,\mu(dy)\,p(s,x;u,z)\,\mu(dz)$.

Interchange the integrals.
The integrand is nonnegative. Tonelli’s theorem gives $\int_A [\int_E p(s,x;u,z)p(u,z;t,y)\,\mu(dz)]\,\mu(dy)$.

Identify the density.
This expression gives the probability of every measurable $A$. Therefore the bracketed function is a version of the transition density from $s$ to $t$, establishing the formula for $\mu$-almost every $y$. $\square$

For a discretely observed path $x_0,\ldots,x_n$ at times $t_0,\ldots,t_n$, the Markov likelihood is formed from transitions between observed times. If one observed gap is itself represented through an unobserved intermediate time, its transition factor can be written as the composition integral above. With a Lebesgue density, the integral is over state values $z$; with counting measure, it is a sum. The reference measure is part of the density definition, not an optional detail.

Gaussian Kernels and Exact Convolution

When the short-interval kernels are Gaussian and their conditional means are affine in the state, the composition integral can often be evaluated exactly. Suppose $Z\mid X_s=x\sim N(m_1,v_1)$ and $Y\mid Z=z\sim N(az+b,v_2)$, with the second transition’s random error independent of the first. Then $Y=aZ+b+\varepsilon_2$, where $\varepsilon_2\sim N(0,v_2)$. A linear combination of independent normal variables is normal, so the composed kernel has mean $am_1+b$ and variance $a^2v_1+v_2$.

Derived result: Under the Gaussian and affine-mean assumptions just stated, integrating out $Z$ gives $$ \int_{\mathbb R} \phi(z;m_1,v_1)\,\phi(y;az+b,v_2)\,dz =\phi(y;am_1+b,\ a^2v_1+v_2), $$ where $\phi(\,\cdot\,;m,v)$ is the $N(m,v)$ density. Indeed, write $Z=m_1+\varepsilon_1$, with $\varepsilon_1\sim N(0,v_1)$. Then $Y=am_1+b+a\varepsilon_1+\varepsilon_2$; independence gives mean $am_1+b$ and variance $a^2v_1+v_2$. Its normal density is precisely the integral on the left, by conditioning on $Z$. This proves the identity. $\square$

Worked Example: Composing Two Ornstein–Uhlenbeck Transitions

Consider $dX_t=-0.5X_t\,dt+dW_t$, starting from $X_0=1$, and compose transitions over $0$ to $1$ and $1$ to $3$. For an interval of length $h$, this process has conditional mean $e^{-0.5h}x$ and variance $1-e^{-h}$.

For the first interval, $Z=X_1\mid X_0=1\sim N(e^{-0.5},1-e^{-1})=N(0.606531,0.632121)$. For the second interval, $X_3\mid X_1=z\sim N(e^{-1}z,1-e^{-2})$, so $a=e^{-1}=0.367879$, $b=0$, and $v_2=0.864665$.

The composed mean is $a m_1=e^{-1}e^{-0.5}=e^{-1.5}=0.223130$. Its variance is $a^2v_1+v_2=e^{-2}(1-e^{-1})+(1-e^{-2})=1-e^{-3}=0.950213$. Thus $X_3\mid X_0=1\sim N(0.223130,0.950213)$. At $y=0.2$, the composed density is $\phi(0.2;0.223130,0.950213)=\exp[-(0.2-0.223130)^2/(2(0.950213))]/\sqrt{2\pi(0.950213)}\approx 0.4091$. The direct three-unit transition has the same mean and variance, which checks the composition.

This exact simplification depends on the kernel family being closed under the composition. A Gaussian approximation to each short transition does not imply that the true diffusion’s long-interval density is Gaussian. Composing the approximations gives a well-defined approximation, but it does not remove their modeling error.

Numerical Quadrature for Non-Gaussian Compositions

When the integral has no closed form, deterministic quadrature replaces it with a weighted sum. For example, if the first short-step kernel is Gaussian, write the composition as an expectation over $Z\sim N(m,v)$. A Gauss–Hermite rule approximates that expectation using selected nodes and weights. The calculation is usually efficient in one state dimension; in several dimensions, tensor-product rules can require many evaluations as the number of dimensions grows.

Worked Example: A Gauss–Hermite Approximation

Take two normalized kernels $p_1(z)=\phi(z;0,1)$ and $p_2(y\mid z)=\phi(y;z^2,1)$. At $y=0$, their composition is $$ \int_{\mathbb R}\phi(z;0,1)\phi(0;z^2,1)\,dz =\frac{1}{\sqrt{2\pi}}E\!\left[\exp(-Z^4/2)\right],\qquad Z\sim N(0,1). $$ The three-node Gauss–Hermite rule for a standard normal has nodes $0,\pm\sqrt{3}$ and weights $2/3,1/6,1/6$. The expectation is therefore approximated by $(2/3)\exp(0)+(1/6)\exp(-(\sqrt{3})^4/2)+(1/6)\exp(-(\sqrt{3})^4/2)$. Since $(\sqrt{3})^4=9$, this equals $2/3+(1/3)e^{-4.5}\approx0.670370$. Multiplying by $1/\sqrt{2\pi}\approx0.398942$ gives the approximate composed density $0.2675$.

As a sensitivity check, the five-node standard-normal rule has center weight $8/15$, and paired nodes $\pm\sqrt{5-\sqrt{10}}$ and $\pm\sqrt{5+\sqrt{10}}$, with respective per-node weights $(7+2\sqrt{10})/60$ and $(7-2\sqrt{10})/60$. Substitution gives an expectation of about $0.6154$, hence a density of about $0.2456$. The noticeable change shows that the three-node result is not yet stable at this tolerance. Increasing the order or using adaptive quadrature is appropriate; the difference between two orders is a diagnostic, not a guaranteed error bound.

A practical quadrature calculation should preserve the density’s normalization and nonnegativity to numerical precision. Check stability as the order or integration range changes, and verify that nodes cover regions where both transition factors are appreciable. A truncated integration range can miss tail mass; a coarse rule can miss a narrow peak. In multiple dimensions, the cost of a tensor grid can become prohibitive, so the computational method must be chosen with the state dimension and required accuracy in mind.

Positive States and Coordinate Changes

For a positive concentration or amount, a log transformation can make Gaussian calculation convenient, but densities change under transformations. If $Y_t=\log X_t$, then a density with respect to $dx$ is not numerically the same as a density with respect to $dy$. When composing in log coordinates, the intermediate-state Jacobian must be handled consistently.

Worked Example: Composing Geometric Brownian Transitions

Let $dX_t=0.1X_t\,dt+0.3X_t\,dW_t$, with $X_0=10$, and consider a three-unit gap split at time $1$. The log state $Y_t=\log X_t$ has drift $0.1-0.3^2/2=0.055$ and diffusion coefficient $0.3$. Its composed transition has mean $\log(10)+0.055(3)$ and variance $0.3^2(3)=0.27$.

For the endpoint $y=12$, $\log(12/10)=0.182322$, so its difference from the log-transition mean is $0.182322-0.165=0.017322$. The density with respect to $dx$ is the lognormal density $\exp[-(0.017322)^2/(2(0.27))]/[12\sqrt{2\pi(0.27)}]\approx0.06394$.

To see the intermediate-state measure explicitly, put $v=\log z$. The first $X$-scale density contributes a factor $1/z$, and $dz=z\,dv$ cancels it; the second transition contributes the endpoint factor $1/y$. Thus the original composition over $z>0$ equals $1/y$ times the Gaussian composition integral over $v\in\mathbb R$. Omitting the final factor $1/y$ would report a density on the log scale, not on the concentration scale.

Diagnostics, Alternatives, and Limitations

Exact Gaussian convolution is preferable when its assumptions hold: it avoids quadrature error and gives a direct calculation. Numerical quadrature is more flexible when the integrand is smooth and the state dimension is modest. Other numerical integration strategies may be needed for irregular or high-dimensional integrands, but they do not change the underlying composition formula. In every case, distinguish approximation error in the short-step kernels from error in evaluating the composition integral.

Check normalization: integrate the computed long-interval density over the endpoint state, or check that its total mass is close to one.
Check quadrature stability: increase the order and, if truncating the domain, expand the range. Large changes indicate unresolved numerical error.
Check support and reference measure: a positive-state model should not acquire mass at negative states, and a transformed density must include its Jacobian.
Check the approximation itself: agreement across quadrature orders only checks integration of the chosen kernels. It does not establish that those kernels accurately describe the diffusion.

Composition can be costly when many intermediate states or state dimensions are involved. It can also be inaccurate when a short-step approximation is used far outside the time-step or state range for which it is reliable. Splitting a long interval more finely may reduce local approximation error, but adds integrations and does not guarantee that accumulated error decreases monotonically. The appropriate check is to compare results under finer subdivisions and tighter integration settings, while monitoring the resulting likelihood.

Common pitfall: Writing $\int p(s,x;u,z)p(u,z;t,y)\,dz$ after defining densities with respect to an arbitrary measure is inconsistent. The correction is to integrate with the common reference measure $\mu(dz)$; use $dz$ only when the state density is with respect to Lebesgue measure.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

What assumptions allow the transition-composition formula, and what measure belongs in its integral?
If the first Gaussian transition has variance $0.4$, and the second has conditional mean $2z+1$ and variance $0.5$, find the composed variance.
Why does agreement between two quadrature orders not prove that a short-step transition approximation is accurate for the diffusion?
For a positive state $X$ and $Y=\log X$, what factor converts a density on the log scale to a density with respect to $dx$?

Solutions. (1) The process must be Markov at the intermediate time, and both transition kernels must have densities with respect to a common reference measure; integrate with that measure, such as $dz$ for Lebesgue densities. (2) The variance is $2^2(0.4)+0.5=2.1$. (3) The orders assess numerical integration of the selected kernels; both can integrate an inaccurate kernel approximation consistently. (4) The density with respect to $dx$ is the log-scale density evaluated at $\log x$, multiplied by $1/x$.