Bolus Doses as Jump Maps for Probability Densities

Why a dose changes the density by a map

Between bolus doses, a concentration state may evolve continuously and have a density described by a forward equation. At the instant of a bolus, the state instead changes according to a deterministic update. The density after dosing is therefore not obtained by evolving the forward equation for a positive amount of time: it is the distribution of the pre-dose state after applying the dose map.

The familiar additive dose in concentration coordinates is a translation, whose Jacobian is one. That special case can conceal an important point: if the state update is nonlinear, probability must be adjusted for how the map stretches or compresses intervals. A density is probability per unit of state, so it changes inversely with that stretching.

Definition: Let $X^-$ be the state immediately before a bolus, with density $p^-(x)$ on a state domain $I$. A deterministic bolus jump map $T:I\to J$ specifies the post-dose state by $X^+=T(X^-)$. The post-dose distribution is the pushforward of the pre-dose distribution: for every measurable set $A\subseteq J$, $P(X^+\in A)=P(X^-\in T^{-1}(A))$.

The superscripts $-$ and $+$ mean the left and right limits at the dosing time. The bolus is assumed instantaneous on the time scale of the diffusion, and the state update is deterministic conditional on the pre-dose state. A random dose or a random dose volume requires an additional probability mixture; it is not represented by a single deterministic map.

The change-of-variables rule

First suppose the state is one-dimensional. Assume $X^-$ has a density on an interval $I$, and $T$ is continuously differentiable, one-to-one, and has $T'(x)\ne 0$ on $I$. The image $J=T(I)$ is then an interval, and the inverse $T^{-1}$ is differentiable there.

Theorem: Under these assumptions, $X^+=T(X^-)$ has density $$ p^+(z)=p^-\!\left(T^{-1}(z)\right) \left|\frac{d}{dz}T^{-1}(z)\right| =\frac{p^-\!\left(T^{-1}(z)\right)} {\left|T'\!\left(T^{-1}(z)\right)\right|},\qquad z\in J, $$ and $p^+(z)=0$ outside $J$.

To justify the rule, take any measurable $A\subseteq J$. By the definition of the transformed state and the density of $X^-$,

P(X^+\in A)=\int_{T^{-1}(A)}p^-(x)\,dx.

Since $T$ is one-to-one and differentiable with nonzero derivative, substitute $z=T(x)$, so $x=T^{-1}(z)$ and $dx=|(T^{-1})'(z)|\,dz$. The integral becomes

P(X^+\in A)=\int_A p^-\!\left(T^{-1}(z)\right) \left|(T^{-1})'(z)\right|\,dz.

This is the defining integral of the stated density. It also proves normalization: setting $A=J$ gives $P(X^+\in J)=P(X^-\in I)=1$. Finally, differentiating $T(T^{-1}(z))=z$ gives $T'(T^{-1}(z))(T^{-1})'(z)=1$, which yields the second form. $\square$

In $d$ dimensions, the corresponding formula applies when $T$ is a continuously differentiable one-to-one map between open state domains and its Jacobian matrix is nonsingular. With $|\det DT(x)|$ denoting the absolute determinant, the density is $p^+(z)=p^-(T^{-1}(z))/|\det DT(T^{-1}(z))|$. The determinant measures local volume expansion; it is not generally enough to multiply by one componentwise derivative.

Worked examples of bolus maps

Worked Example: An Additive Concentration Dose

Suppose pre-dose concentration $C^-$ has an exponential density with rate $2$ on $c>0$, so $p^-(c)=2e^{-2c}$ for $c>0$. A fixed bolus produces an immediate concentration increment $d=1.5$, in the same concentration units. Find the post-dose density.

The jump map is $T(c)=c+1.5$, with inverse $T^{-1}(z)=z-1.5$ and derivative $T'(c)=1$. Its image of the pre-dose support is $z>1.5$. Thus

p^+(z)= \begin{cases} 2e^{-2(z-1.5)},&z>1.5,\\ 0,&z\le 1.5. \end{cases}

The density is zero below the dose increment because no positive pre-dose concentration maps there. Its integral is $\int_{1.5}^{\infty}2e^{-2(z-1.5)}\,dz =\int_0^\infty 2e^{-2u}\,du=1$, where $u=z-1.5$. The unchanged height and shape result from the unit Jacobian of a translation.

Worked Example: A Nonlinear Concentration Update

Consider a positive concentration state $C^-$ with density $p^-(c)=e^{-c}$ for $c>0$. As an illustrative state-dependent dose update, let $T(c)=c+0.5c^2$. This increasing map represents a nonlinear jump rule; it is not the usual fixed-volume additive bolus. Find the post-jump density and verify its normalization.

For $c>0$, $T'(c)=1+c>0$, so the map is one-to-one. Solving $z=c+0.5c^2$ gives $c=\sqrt{1+2z}-1$ on $z>0$. Also $T'(c)=1+c=\sqrt{1+2z}$ at this inverse. Therefore

p^+(z)= \frac{\exp\!\left(1-\sqrt{1+2z}\right)}{\sqrt{1+2z}}, \qquad z>0,

and $p^+(z)=0$ for $z\le 0$. To check the integral, substitute $z=c+0.5c^2$, for which $dz=(1+c)\,dc$ and $\sqrt{1+2z}=1+c$. Then

\int_0^\infty p^+(z)\,dz =\int_0^\infty \frac{e^{-c}}{1+c}(1+c)\,dc =\int_0^\infty e^{-c}\,dc=1.

The denominator is essential: at a given $z$, the map expands the state interval by $1+c$, reducing probability per unit of post-jump concentration by that factor.

Worked Example: A Fixed Dose in Log-Concentration Coordinates

Let $Y^-=\log C^-$ have distribution $N(0,0.25)$, and let the physical concentration receive a fixed increment $d=1$. The post-dose log concentration is $Y^+=\log(e^{Y^-}+1)$. Find its density and evaluate it at $z=\log 2$.

The map $T(y)=\log(e^y+1)$ is strictly increasing, with derivative $T'(y)=e^y/(e^y+1)$. Its image is $z>0$, and its inverse is $y=\log(e^z-1)$. Differentiating that inverse gives $dy/dz=e^z/(e^z-1)$. The pre-dose normal density is $p^-(y)=\frac{1}{0.5\sqrt{2\pi}}\exp(-y^2/(2(0.25)))$. Hence

p^+(z)= \frac{e^z}{e^z-1}\, \frac{1}{0.5\sqrt{2\pi}} \exp\!\left( -\frac{[\log(e^z-1)]^2}{0.5} \right),\qquad z>0.

At $z=\log 2$, the inverse is $y=\log(2-1)=0$, and $dy/dz=2$. The pre-dose density at zero is $1/(0.5\sqrt{2\pi})$, so $p^+(\log 2)=2/(0.5\sqrt{2\pi})=4/\sqrt{2\pi}$, approximately $1.596$. This is a density value per unit of log concentration, not a probability at the single point.

Non-one-to-one maps and sequential updates

A nonlinear jump need not be one-to-one. If each regular value $z$ has finitely many inverse images $x_1,\ldots,x_m$, with $T'(x_i)\ne 0$, each inverse branch contributes probability to the density at $z$. Provided the critical points have pre-dose probability zero and the map is differentiable and locally invertible away from them, the density is

p^+(z)=\sum_{x:T(x)=z}\frac{p^-(x)}{|T'(x)|}.

This follows by dividing the domain into intervals on which $T$ is one-to-one, applying the one-to-one change-of-variables rule on each interval, and adding the resulting integrals. The sum is required because disjoint pre-dose regions can land at the same post-dose state. At critical values the formula is understood almost everywhere; a singularity in the density there does not by itself imply an atom.

For example, $T(x)=x^2$ on the real line sends both $x=\sqrt z$ and $x=-\sqrt z$ to $z>0$. If $X^-\sim N(0,1)$, then

p^+(z)= \frac{p^-(\sqrt z)}{2\sqrt z} +\frac{p^-(-\sqrt z)}{2\sqrt z} =\frac{e^{-z/2}}{\sqrt{2\pi z}},\qquad z>0.

Keeping only one root would lose half the probability mass. More generally, if a map is constant on a set with positive pre-dose probability, that probability becomes an atom and cannot be represented by an ordinary density alone.

Result: For a sequence of deterministic jumps $T_1,\ldots,T_r$, the composed map is $S=T_r\circ\cdots\circ T_1$. If the maps are differentiable and one-to-one with nonzero derivatives on the relevant domains, the final density is $p^-(S^{-1}(z))|(S^{-1})'(z)|$. Equivalently, each intermediate density update can be applied in order.

For completeness, the derivative chain rule gives $S'(x)=T_r'(x_{r-1})\cdots T_1'(x)$, where $x_j=T_j(x_{j-1})$ and $x_0=x$. The inverse derivative is $1/S'(S^{-1}(z))$. Applying the single-map formula to $S$ therefore gives the stated final density. Applying that formula successively multiplies the same inverse-derivative factors, so the sequential and composed updates agree. $\square$

Connecting the jump to diffusion between doses

Suppose a diffusion evolves between fixed dose times $t_k$, and let $p(t,x)$ denote its density away from those instants. The forward equation governs the continuous-time evolution on each open interval between doses. At $t_k$, the density has a jump condition: transform $p(t_k-,x)$ through the dose map to obtain $p(t_k+,z)$. For an invertible scalar map this is

p(t_k+,z)= \frac{p(t_k-,T_k^{-1}(z))} {|T_k'(T_k^{-1}(z))|}.

The forward equation is not applied across the instantaneous jump as though it were continuous drift. A model can alternatively encode a dose in an amount coordinate $A$ by $A^+=A^-+D$. In that coordinate the jump is a translation; if the modeled state is a nonlinear function of amount, transforming the resulting distribution into that state still requires its Jacobian.

Diagnostics, alternatives, and limitations

Several checks catch most errors in a density update. First, the support after dosing must be the image of the pre-dose support. Second, integrate the proposed density over that image; the result must be one, except where explicitly modeled probability is assigned to atoms or other singular components. Third, compare simulations: draw pre-dose states, apply the jump map to each draw, and compare the resulting histogram or empirical distribution with the transformed density. A mismatch often indicates a missing absolute value, a missing inverse branch, or an incorrect inverse.

A common mistake is to write $p^+(z)=p^-(z)$ because the dose is “deterministic,” or to substitute $T(z)$ for $T^{-1}(z)$. Determinism says how each state moves; it does not say density values stay fixed. The density at $z$ comes from the pre-dose state or states that map to $z$, and it must be divided by the local expansion factor. For an additive dose the factor happens to be one, which can hide this error.

Directly transforming Monte Carlo samples is a useful alternative when the map is complicated or has multiple branches, but a histogram is an approximation and can obscure sharp density features. Analytic change of variables gives the density directly when its assumptions hold. An amount-coordinate update may simplify the dose map, while the resulting distribution in concentration coordinates still needs a coordinate transformation. These approaches answer related but not identical computational needs.

The ordinary-density formula has limits. It requires an absolutely continuous pre-dose law and a map that does not collapse positive probability onto a lower-dimensional set. If the dose amount is random, condition on the dose and average the resulting densities (including any singular components). If volume, bioavailability, or another state is itself updated, use the full state vector and its Jacobian rather than treating that component as fixed. Finally, a mathematically valid jump map does not establish that its pharmacokinetic assumptions are scientifically appropriate.

Check Your Understanding

Answer each question, then compare with the short solutions below.

If $X^-$ has density $p^-$ on $x>0$ and $X^+=X^-+d$ for $d>0$, state the post-dose density and support.
For $T(x)=3x$ on $x>0$, express $p^+(z)$ in terms of $p^-$, and identify the Jacobian factor.
Why must the density for $T(x)=x^2$ with a real-valued pre-dose state include two terms for $z>0$?
A map sends every pre-dose state in an interval of positive probability to one fixed post-dose value. Can the full post-dose law have an ordinary density? Explain.
What is the key density check after applying an invertible dose map?

Solutions. (1) The inverse is $z-d$, the derivative is one, and $p^+(z)=p^-(z-d)$ for $z>d$, zero otherwise. (2) The inverse is $z/3$ and its derivative is $1/3$, so $p^+(z)=p^-(z/3)/3$ for $z>0$. (3) Both $\sqrt z$ and $-\sqrt z$ map to $z$, so both pre-dose regions contribute probability. (4) No. That interval's positive probability becomes an atom at the fixed value, which is not represented by an ordinary density alone. (5) Integrate the transformed density over the image of the pre-dose support; it must equal one.