Translating Exposure-Response Laws Into Stochastic Effect Models

From an Exposure-Response Law to a Response SDE

A concentration-response law describes how an effect is associated with drug concentration. It does not, by itself, specify how a response changes over time or how much unexplained variation it has. To construct a stochastic differential equation (SDE), specify the response state, the role of the exposure-response law in the drift, the process-noise term, and the initial response.

A key modelling decision is whether the law gives a target response at each concentration or a rate of response change. These quantities have different units and lead to different SDEs. Treating a target as a rate, or a rate as a target, changes the model rather than merely changing its notation.

Definition: Let $C_t$ be drug concentration and $R_t$ a response state. A response SDE has the form $$ dR_t=b(t,R_t,C_t)\,dt+\sigma(t,R_t,C_t)\,dW_t,\qquad R_0=r_0. $$ Here $b$ is the drift, specifying the conditional direction and rate of change, $\sigma$ is the diffusion coefficient, specifying the scale of process noise, $W_t$ is Brownian motion, and $r_0$ is the initial response. The units of $b$ are response per time; those of $\sigma$ are response per square root of time.

An exposure-response law is often written $g(C)$, with units of response. If it represents the response approached under sustained concentration, a useful translation is $$ dR_t=\kappa\{g(C_t)-R_t\}\,dt+\sigma(t,R_t,C_t)\,dW_t, $$ where $\kappa>0$ has units of inverse time. The concentration-response curve supplies the target; $\kappa$ specifies how quickly the response moves toward it. This distinction is important: putting $g(C_t)$ directly in the drift would give it the wrong units unless $g$ had instead been defined as a rate.

A standard target curve is the inhibitory Emax law $$ g(C)=E_0-\frac{E_{\max}C}{EC_{50}+C},\qquad C\geq 0, $$ where $E_0$ is the baseline target, $E_{\max}$ is the maximum decrease from baseline, and $EC_{50}>0$ is the concentration giving half of that maximum decrease. A stimulatory version uses a plus sign. A Hill curve replaces $C/(EC_{50}+C)$ by $C^h/(EC_{50}^h+C^h)$, with $h>0$. The choice of curve describes the concentration-response relationship; it does not determine the noise model or initial state.

Assumptions and a Basic Derivation

For the results below, take $C_t$ to be a deterministic, continuous concentration profile on a finite interval $[0,T]$. Suppose $g(C_t)$ is continuous and bounded there, $\kappa>0$, $\sigma$ is constant, and $R_0=r_0$ is fixed. Brownian motion is defined on a filtered probability space, and the equation is interpreted in the Itô sense. These assumptions give an explicit solution; more general response-dependent coefficients require standard existence and uniqueness conditions, such as measurability in time, local Lipschitz continuity in the response state, and a linear-growth bound.

Theorem: Under the assumptions above, the target-relaxation SDE $$ dR_t=\kappa\{g(C_t)-R_t\}\,dt+\sigma\,dW_t,\qquad R_0=r_0 $$ has the solution $$ R_t=e^{-\kappa t}r_0+\kappa\int_0^t e^{-\kappa(t-s)}g(C_s)\,ds +\sigma\int_0^t e^{-\kappa(t-s)}\,dW_s. $$ Consequently, $$ E[R_t]=e^{-\kappa t}r_0+\kappa\int_0^t e^{-\kappa(t-s)}g(C_s)\,ds,\qquad \operatorname{Var}(R_t)=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t}). $$

Proof. Apply Itô's product rule to $e^{\kappa t}R_t$. Since $e^{\kappa t}$ has finite variation, its quadratic covariation with $R_t$ is zero. Therefore $$ d(e^{\kappa t}R_t)=\kappa e^{\kappa t}g(C_t)\,dt+\sigma e^{\kappa t}\,dW_t. $$ Integrating from $0$ to $t$, then multiplying by $e^{-\kappa t}$, gives the stated solution. The deterministic terms have the displayed expectation, and the Itô integral has mean zero because its deterministic integrand is square-integrable on $[0,t]$. By Itô isometry, its variance is $$ \sigma^2\int_0^t e^{-2\kappa(t-s)}\,ds =\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t}). $$ The remaining terms are deterministic, so this is also the variance of $R_t$. $\square$

This result shows that with a changing concentration, the expected response is a weighted history of the target curve, not generally $g(C_t)$ itself. Recent target values receive more weight than distant ones. If concentration is constant at $c$, the mean simplifies to $$ E[R_t]=g(c)+(r_0-g(c))e^{-\kappa t}. $$ The response approaches the target in expectation, while process noise leaves nonzero variability at positive times.

Result: With fixed concentration $c$ and no process noise, the target-relaxation equation has the unique equilibrium $R=g(c)$, and that equilibrium is stable. Indeed, $dR_t/dt=\kappa\{g(c)-R_t\}$ has solution $R_t=g(c)+(R_0-g(c))e^{-\kappa t}$. Since $\kappa>0$, the exponential term tends to zero, so every initial state converges to $g(c)$. The uniqueness follows because the drift is zero only when $R=g(c)$. $\square$

Worked Example: An Inhibitory Emax Target

Suppose the target response follows $g(C)=10-6C/(2+C)$, where response is measured in units and concentration in mg/L. Let $\kappa=0.5$ per hour, $\sigma=0.3$ response units per square root of hour, concentration be held at $C=4$ mg/L, and the initial response be $R_0=10$.

First calculate the target: $$ g(4)=10-\frac{6(4)}{2+4}=10-4=6. $$ The SDE is $dR_t=0.5(6-R_t)\,dt+0.3\,dW_t$. At the initial state, the drift is $0.5(6-10)=-2$ response units per hour: the expected local movement is downward.

After two hours, the mean and variance are $$ E[R_2]=6+(10-6)e^{-0.5(2)}=6+4e^{-1}\approx 7.472, $$ $$ \operatorname{Var}(R_2)=\frac{0.3^2}{2(0.5)}(1-e^{-2(0.5)(2)}) =0.09(1-e^{-2})\approx 0.0778. $$ Thus the standard deviation is approximately $\sqrt{0.0778}=0.279$ response units. The mean has moved toward 6 but has not reached it; the variance describes stochastic spread around that mean.

Choosing the Noise and the Initial State

For additive noise, $\sigma$ is constant and the response has the same instantaneous noise scale at every concentration and response level. A concentration-dependent choice, such as $\sigma(C_t)$, can represent different process variability at different exposures. A response-dependent coefficient may represent variability that changes with the response itself, but it must be checked for existence, uniqueness, and scientifically plausible behavior. In all cases, $\sigma^2$ is a variance rate, not a variance at a single observation time.

The initial condition is part of the model. If $R_t$ is a measured response, $R_0$ may be set to the pre-dose measurement or represented as an uncertain state when that is justified by the data. If the model defines $R_t$ as a change from baseline, $R_0=0$ may be appropriate. These are different definitions of the state, not interchangeable defaults. For a random initial response independent of $W_t$, the solution formula still applies with random $R_0$; the unconditional variance also includes the propagated initial-state variance $e^{-2\kappa t}\operatorname{Var}(R_0)$.

Worked Example: A Hill Target and Concentration-Dependent Noise

Let the target be $g(C)=20-12C^2/(3^2+C^2)$. Take $C=3$ mg/L, current response $R_t=11$, $\kappa=0.4$ per hour, and $\sigma(C)=0.2+0.05C$ response units per square root of hour.

At this concentration, the Hill fraction is $3^2/(3^2+3^2)=9/18=0.5$, so $g(3)=20-12(0.5)=14$. The SDE at this exposure is $$ dR_t=0.4(14-R_t)\,dt+0.35\,dW_t, $$ because $\sigma(3)=0.2+0.05(3)=0.35$. At $R_t=11$, the drift is $0.4(3)=1.2$ response units per hour, while the instantaneous variance rate is $0.35^2=0.1225$ response-units squared per hour.

The positive drift does not mean every response path must increase: the Brownian increment can move the response in either direction. Nor does the Hill curve specify $0.35$; that number comes from the separately chosen noise model. This example makes the two modelling contributions explicit.

When the Law Is a Rate Rather Than a Target

Some exposure-response relationships describe a drug-dependent rate of response change. Write that rate as $q(C)$, with units of response per time. A corresponding additive-noise model is $$ dR_t=q(C_t)\,dt+\sigma\,dW_t. $$ This equation does not pull the response toward $q(C_t)$. It accumulates the rate over time, and unless another drift term is specified, the response has no fixed equilibrium under constant exposure.

Worked Example: A Drug-Dependent Response Rate

Suppose the rate law is $q(C)=0.4+0.3C/(2+C)$ response units per hour, with $C$ in mg/L. Hold concentration at $C=2$ mg/L for three hours, set $R_0=5$, and use $\sigma=0.2$ response units per square root of hour.

The rate is $$ q(2)=0.4+\frac{0.3(2)}{2+2}=0.4+0.15=0.55 $$ response units per hour. Thus $dR_t=0.55\,dt+0.2\,dW_t$ during the exposure interval. Integrating gives $R_3=5+0.55(3)+0.2W_3$. Since $E[W_3]=0$ and $\operatorname{Var}(W_3)=3$, $$ E[R_3]=6.65,\qquad \operatorname{Var}(R_3)=0.2^2(3)=0.12. $$ The expected increase is $1.65$ units. Unlike a target-relaxation model, this rate model does not predict a return toward a concentration-specific target after the rate is applied.

The target and rate formulations are not competing notations for the same curve. The target formulation requires a response-valued curve and a relaxation timescale. The rate formulation requires a rate-valued curve and accumulates its effect. Use the interpretation, measurements, and units of the stated exposure-response law to decide which mathematical object it supplies.

Diagnostics, Alternatives, and Limitations

Several checks can reveal an incorrectly translated model. First, check units in every drift term and in the diffusion coefficient. Second, check the sign and size of the drift at clinically relevant pairs $(R,C)$; an inhibitory target should not accidentally produce an upward drift when the response is above that target. Third, compare the specified initial state with the definition and timing of the response measurements. Finally, inspect simulated paths and one-step innovations: persistent systematic departures from the target pattern, implausible variability across concentration, or repeated boundary violations can indicate a mismatch in the curve, drift, noise, or state definition.

A deterministic response model omits process noise and is a useful alternative when unexplained path-to-path variation is negligible or not supported by the data. A stochastic effect model adds process noise to the response state. Alternatively, one may place uncertainty in the concentration process or in parameters, rather than adding response-process noise; these represent different sources of variation and should not be interpreted as interchangeable. The earlier material on residual diagnostics and model comparison provides tools for checking predictive adequacy once a model has been specified.

The target-relaxation construction has limitations. It assumes the response has a meaningful target and a relaxation rate; a single constant $\kappa$ may be inadequate when response timing changes with concentration or time. Additive Gaussian noise can generate values outside a scientifically possible range. The bounded fractional-effect construction is a separate modelling approach. Also, a concentration-response curve estimated from steady-state or cross-sectional data does not alone identify $\kappa$ or $\sigma$; time-resolved response data and an explicit observation model are needed to learn those features.

Common pitfall: Writing $dR_t=g(C_t)\,dt+\sigma\,dW_t$ for a curve $g$ measured in response units treats a response level as a rate. The units of $g(C_t)\,dt$ are response multiplied by time, not response. If $g$ is the desired target, a dimensionally consistent example is $\kappa\{g(C_t)-R_t\}\,dt$. If the law genuinely specifies a rate, define it as such and use its rate units explicitly.

Check Your Understanding

Work each problem by identifying what the concentration-response law represents before writing the SDE. Solutions are included for checking.

A target law is $g(C)=8+4C/(C+2)$. Find the target at $C=2$, and give the drift at $R=10$ if $\kappa=0.25$ per hour.
For a target-relaxation SDE with constant target 6, $\kappa=0.5$ per hour, and $R_0=10$, find the mean response after two hours.
A rate law gives $q(C)=0.3$ response units per hour at the current concentration. With $R_0=2$, constant exposure for four hours, and no process noise, what is $R_4$?
Explain why a curve in response units cannot be inserted directly as the drift in an SDE for response.

Solutions. (1) $g(2)=8+4(2)/(2+2)=10$. The drift is $0.25(10-10)=0$ response units per hour. (2) $E[R_2]=6+(10-6)e^{-0.5(2)}=6+4e^{-1}\approx7.472$. (3) The rate accumulates: $R_4=2+0.3(4)=3.2$. (4) The drift must have units of response per time so that its product with $dt$ has response units. A response-valued curve needs a rate conversion, such as multiplication by a relaxation rate and subtraction of the current response to form a target-relaxation drift.