Identifying State-Dependent Diffusion Functions

Why Conditional Variance Reveals State Dependence

In a pharmacokinetic stochastic differential equation, random variation may change with the current state. For example, fluctuations may be larger at higher concentrations than at lower concentrations. A constant diffusion coefficient cannot represent that pattern; a state-dependent coefficient can. The key identification idea is that, over a sufficiently short interval, conditional increment variance per unit time approaches the squared diffusion coefficient at the current state.

This is a local statement. It concerns increments starting from a specified state and an interval whose length tends to zero. It does not say that the variance of an observed increment over an ordinary sampling interval equals the diffusion coefficient squared times that interval. The distinction matters both for interpreting data and for comparing candidate models.

Definition: Consider the scalar Itô diffusion $$ dX_s=b(X_s,s)\,ds+\sigma(X_s,s)\,dW_s, $$ where $W_s$ is standard Brownian motion, $b$ is the drift function, and $\sigma$ is the diffusion function. For a process started at $X_t=x$, define the increment $\Delta_h X=X_{t+h}-X_t$. The conditional local variance rate at $(x,t)$, when the limit exists, is $$ q(x,t)=\lim_{h\downarrow 0}\frac{\operatorname{Var}(\Delta_h X\mid X_t=x)}{h}. $$

In the usual regular setting, $q(x,t)=\sigma^2(x,t)$. The conditional variance identifies the square of the diffusion function, not its sign. It is conventional to take $\sigma$ to be nonnegative; even then, the result is about a local variance rate, not a direct equality for finite-time increments.

The Small-Time Variance Result

The following assumptions give a simple sufficient setting for the local result. Suppose $b$ and $\sigma$ are continuous in time and globally Lipschitz in state, uniformly over a time interval containing $t$. Suppose also that both functions are bounded there, and that the SDE has a solution with finite second moments. Start that solution at the deterministic state $X_t=x$. These conditions are stronger than necessary, but make the argument and its error control explicit.

Theorem: Under the conditions just stated, $$ \lim_{h\downarrow 0}\frac{\operatorname{Var}(X_{t+h}-X_t\mid X_t=x)}{h} =\sigma^2(x,t). $$

Proof. Write $b_0=b(x,t)$ and $\sigma_0=\sigma(x,t)$. By the integral form of the SDE, $$ X_{t+h}-x=b_0h+\sigma_0(W_{t+h}-W_t)+R_h, $$ where $$ R_h=\int_t^{t+h}\{b(X_s,s)-b_0\}\,ds+ \int_t^{t+h}\{\sigma(X_s,s)-\sigma_0\}\,dW_s. $$ The boundedness assumptions imply, for $s$ sufficiently close to $t$, that $E[|X_s-x|^2]\leq C(s-t)$ for a finite constant $C$. This follows by applying the Cauchy–Schwarz inequality to the drift integral, Itô isometry to the stochastic integral, and boundedness of $b$ and $\sigma$. Lipschitz continuity in state and continuity in time then imply that the drift and diffusion integrands in $R_h$ converge to zero in mean square as $s\downarrow t$. More specifically, local uniform continuity in time and the bound on $E[|X_s-x|^2]$ give $$ E[|R_h|^2]=o(h). $$ For example, the squared drift-integral contribution is bounded by $h$ times the integral of the mean squared drift difference, and the stochastic-integral contribution equals the integral of the mean squared diffusion difference by Itô isometry. Both divided by $h$ tend to zero.

The deterministic term $b_0h$ does not affect variance. The variance of $\sigma_0(W_{t+h}-W_t)$ is $\sigma_0^2h$. By Cauchy–Schwarz, the covariance between this Brownian term and $R_h$, divided by $h$, tends to zero: its absolute value is at most $|\sigma_0|\sqrt{h}\sqrt{E[|R_h|^2]}=o(h)$. Also, $\operatorname{Var}(R_h)\leq E[|R_h|^2]=o(h)$. Therefore $$ \operatorname{Var}(X_{t+h}-x)=\sigma_0^2h+o(h), $$ which proves the stated limit. $\square$

The same reasoning explains why drift is not the leading term in this limit. Over an interval of length $h$, a locally fixed drift contributes approximately $b(x,t)h$, whereas Brownian variation has variance $\sigma^2(x,t)h$. A deterministic first-order drift shift does not itself add conditional variance. State changes during the interval can affect the variance at finite $h$, but under the theorem’s assumptions those effects vanish after division by $h$ as $h$ tends to zero.

From a Local Variance Rate to Shape and Scale

Suppose the diffusion function is parameterized as $\sigma(x,t)=c\,g(x,t;\eta)$, where $c>0$ is an overall scale and $\eta$ controls the shape. Then the local variance rate is $q(x,t)=c^2g^2(x,t;\eta)$. If $g$ is positive and a reference state $x_0$ is available, the ratio $$ \frac{q(x,t)}{q(x_0,t)}= \frac{g^2(x,t;\eta)}{g^2(x_0,t;\eta)} $$ cancels $c$. Thus the pattern of variance rates across states can identify shape without first identifying the overall scale, provided the normalized shape is itself identifiable from the states considered. A reference normalization such as $g(x_0,t;\eta)=1$ can make the scale parameter’s interpretation explicit.

This separation is not automatic. If different values of $\eta$ produce the same normalized ratios over the states that the process visits, those shape parameters cannot be distinguished from those data. If observations cover only a narrow range of states, substantially different functions may have nearly identical values over that range. And if $g$ is allowed to include an arbitrary multiplicative constant, that constant can be traded against $c$; a normalization or another constraint is needed to give the two parameters distinct meanings.

Theorem: Suppose two regular scalar diffusions have the same conditional transition distributions from every state $x$ in a set $S$, at time $t$, for every sufficiently small positive $h$. If each satisfies the small-time variance result, then their squared diffusion functions agree on $S$ at time $t$.

Proof. Equality of the conditional transition distributions implies equality of their conditional increment variances for each such $x$ and $h$. Divide those equal variances by $h$ and take the limit as $h\downarrow0$. By the small-time variance result, the two limits are the respective squared diffusion functions. The limits are equal, so the squared diffusion functions agree at each $x\in S$. $\square$

This is a structural identification statement under ideal conditional transition information and the stated regularity conditions. It is not a claim that a finite data set, a single observed path, or measurements at a few time points determine an unrestricted function exactly. In practice, finite sampling, measurement error, and limited state coverage affect what can be estimated.

Worked Example: A State-Dependent Variance Rate

Consider $dX_s=-0.3X_s\,ds+\sqrt{0.04+0.01X_s^2}\,dW_s$. Starting at $x=2$, calculate the small-time mean increment and variance for $h=0.1$.

The drift at $x=2$ is $b(2)=-0.3(2)=-0.6$. Its leading contribution to the mean increment is $b(2)h=-0.6(0.1)=-0.06$.

The squared diffusion function at $x=2$ is $$ \sigma^2(2)=0.04+0.01(2^2)=0.08. $$ Hence the local variance approximation is $\operatorname{Var}(\Delta_hX\mid X_t=2)\approx 0.08(0.1)=0.008$. The corresponding standard deviation is approximately $\sqrt{0.008}\approx0.0894$. These are small-time approximations, not exact finite-interval moments.

At $x=0$, the variance rate is $0.04$, while at $x=2$ it is $0.08$. The model therefore predicts a larger local variance rate at the higher state. The drift changes the expected direction of movement but does not replace this state-dependent variance pattern.

Worked Example: Recovering Shape Separately from Scale

Suppose $\sigma(x)=c(1+\eta x)$ for $x\geq0$, with $c>0$ and $\eta\geq0$. The local variance rates are estimated as $q(0)=0.16$ and $q(2)=0.36$. Find $c$ and $\eta$ under this model.

At $x=0$, $q(0)=c^2(1+\eta\cdot0)^2=c^2$. Since $c>0$, $c=\sqrt{0.16}=0.4$.

At $x=2$, the ratio cancels the scale: $$ \frac{q(2)}{q(0)}=\frac{(1+2\eta)^2}{1^2} =\frac{0.36}{0.16}=2.25. $$ Because $1+2\eta$ is positive, taking the positive square root gives $1+2\eta=1.5$. Thus $\eta=0.25$. Checking, $c^2(1+2\eta)^2=0.16(1.5)^2=0.36$, as required. Here the reference-state rate identifies scale, and the ratio identifies shape.

Worked Example: Checking a Candidate Variance Shape

Suppose local variance-rate estimates at three states are $q(0)=0.04$, $q(2)=0.08$, and $q(4)=0.20$. Check whether they agree with a model $q(x)=a+bx^2$, and recover the parameters if they do.

At $x=0$, $a=q(0)=0.04$. At $x=2$, $0.08=0.04+4b$, so $b=0.04/4=0.01$. The resulting model predicts $q(4)=0.04+0.01(16)=0.20$, exactly the third supplied rate. Thus the three values agree with $q(x)=0.04+0.01x^2$, corresponding to the nonnegative diffusion function $\sigma(x)=\sqrt{0.04+0.01x^2}$.

The third state is a check, not a guarantee that the model is correct at all states. With noisy estimates, exact agreement is not expected; the question becomes whether discrepancies are consistent with estimation variability and whether the chosen shape is adequate across the observed state range.

Estimation, Diagnostics, and Alternatives

With repeated short-interval increments starting near a common state $x$, a natural empirical quantity is the sample variance of those increments divided by the interval length. With many observations and sufficiently small intervals, it targets $q(x,t)$. In longitudinal pharmacokinetic data, however, repeated increments rarely start at exactly the same state. One may group nearby states or fit a parameterized variance-rate function, but grouping introduces a choice of bandwidth and model fitting imposes a shape assumption. These choices should be reported and checked rather than treated as part of the small-time theorem.

Check the interval regime. If estimates change substantially as the interval length changes, finite-interval effects may be important; the local approximation may not yet be adequate.
Check state coverage. A fitted shape supported only by a narrow range of observed states should not be extrapolated confidently to states the data do not visit.
Check the conditioning. Pooling increments from widely different starting states can hide state dependence or create a misleading average variance rate.
Check the observation process. Measurement error can contribute to observed increment variability. The variance of noisy observed differences need not equal the process variance considered here.

A common mistake is to set $\operatorname{Var}(\Delta_hX\mid X_t=x)=\sigma^2(x,t)h$ for every sampling interval and then treat that equality as exact. It is only the leading small-time relation under the stated regularity conditions. The correction is to interpret the ratio as a limit, assess whether the sampling intervals are sufficiently short for the approximation, and account for observation error or finite-interval transition behavior when they are not.

Another distinction is between this local-variance approach and identifying diffusion parameters from a full transition-density likelihood. A transition-density analysis uses the distribution of finite-time transitions and can use more than their variance, but it depends on a suitable density or approximation. Conditional variance rates isolate the local diffusion contribution and provide an interpretable way to examine state dependence; they do not by themselves validate the full transition model or resolve every parameter confounding.

The method has clear limits. It identifies the squared diffusion function only where the relevant conditional behavior is available, and an unrestricted function cannot generally be recovered from finitely many noisy observations. It also does not identify the sign of a diffusion coefficient from conditional variances alone. In applications, a nonnegative parameterization, adequate coverage of states, and explicit treatment of measurement error are important parts of interpreting the result.

Check Your Understanding

Answer each question, then compare with the solutions.

If a diffusion has local variance rate $q(x,t)=0.09$ at a given state and time, what is the leading conditional increment variance for $h=0.02$?
For $\sigma(x)=c(1+\eta x)$, suppose $q(0)=0.25$ and $q(1)=0.36$, with $c>0$ and $\eta\geq0$. Find $c$ and $\eta$.
Why does equality of conditional transition distributions at arbitrarily small intervals imply equality of squared diffusion functions under the theorem’s assumptions?
A researcher estimates increment variances at only one state. Can those estimates generally distinguish an unknown overall scale from an unknown state-dependent shape? Explain.

Solutions. (1) The leading variance is $q h=0.09(0.02)=0.0018$. (2) At zero, $c=\sqrt{0.25}=0.5$. The rate ratio is $0.36/0.25=1.44=(1+\eta)^2$, so $1+\eta=1.2$ and $\eta=0.2$. Checking gives $0.25(1.2)^2=0.36$. (3) Equal conditional distributions have equal conditional variances; dividing by $h$ and taking the small-time limit gives the squared diffusion functions. (4) Generally not: one state supplies only one local variance rate, which confounds scale with the shape’s value there unless an additional normalization or constraint is imposed.