Prediction and Dissipation in Nonequilibrium Molecular Sensors: Conditionally Markovian Channels Driven by Memoryful Environments

Abstract

Biological sensors must often predict their input while operating under metabolic constraints. However, determining whether or not a particular sensor is evolved or designed to be accurate and efficient is challenging. This arises partly from the functional constraints being at cross purposes and partly since quantifying the prediction performance of even in silico sensors can require prohibitively long simulations, especially when highly complex environments drive sensors out of equilibrium. To circumvent these difficulties, we develop new expressions for the prediction accuracy and thermodynamic costs of the broad class of conditionally Markovian sensors subject to complex, correlated (unifilar hidden semi-Markov) environmental inputs in nonequilibrium steady state. Predictive metrics include the instantaneous memory and the total predictable information (the mutual information between present sensor state and input future), while dissipation metrics include power extracted from the environment and the nonpredictive information rate. Success in deriving these formulae relies on identifying the environment’s causal states, the input’s minimal sufficient statistics for prediction. Using these formulae, we study large random channels and the simplest nontrivial biological sensor model—that of a Hill molecule, characterized by the number of ligands that bind simultaneously—the sensor’s cooperativity. We find that the seemingly impoverished Hill molecule can capture an order of magnitude more predictable information than large random channels.

Appendices

Revisiting the Thermodynamics of Prediction

For completeness, we review the derivation of Eq. (6). Let ${x} _t$ represent the input at time t, $y _t$ represent the sensor state at time t, and $E( {x} , y )$ the system’s energy function for the reservoir of interest. We assume constant temperature. The system’s temperature-normalized nonequilibrium free energy $F_{\hbox {neq}}$ is given by:

$$\begin{aligned} \beta F_{neq}[p( y | {x} )] = \beta \langle E( {x} , y )\rangle _{p( y | {x} )} - {\text {H}}[ Y | {X} = {x} ]. \end{aligned}$$

(A1)

Even if this is not a valid expression for nonequilibrium free energy, the validity of Still et al. (2012)’s derivation only rests on this expression being a Lyapunov function for the dynamics. Intuitively, this corresponds to an assumption that the system reduces its nonequilibrium free energy when the sensor thermalizes to its attached thermal bath. [Accordingly, the $\beta$ in the above expression refers to the temperature of the sensor when the environment is fixed, indirectly circumventing the difficulty with defining a nonequilibrium temperature (Casas-Vázquez and Jou 2003).] If so, then:

$$\begin{aligned} \beta F_{\hbox {neq}}[p( y _t| {x} _{t+\Delta t})] \ge \beta F_{\hbox {neq}}[p( y _{t+\Delta t}| {x} _{t+\Delta t})], \end{aligned}$$

giving:

$$\begin{aligned} 0&\le \beta F_{\hbox {neq}}[p( y _t| {x} _{t+\Delta t})] -\beta F_{\hbox {neq}}[p( y _{t+\Delta t}| {x} _{t+\Delta t})] \\&\le \left( \beta \langle E( {x} _{t+\Delta t}, y _t)\rangle _{p( y _t| {x} _{t+\Delta t})}\right. \\&\left. \quad - {\text {H}}[ Y _t| {X} _{t+\Delta t}= {x} _{t+\Delta t}]\right) - (\beta \langle E( {x} _{t+\Delta t}, y _{t+\Delta t})\rangle _{p( y _{t+\Delta t}| {x} _{t+\Delta t})}\\&\quad - {\text {H}}[ Y _{t+\Delta t}| {X} _{t+\Delta t}= {x} _{t+\Delta t}]) \\&\le \beta \lim _{\Delta t\rightarrow 0} \left( \frac{\langle E( {x} _{t+\Delta t},y_t)\rangle _{p(y_t| {x} _{t+\Delta t})}}{\Delta t}- \frac{\langle E( {x} _{t+\Delta t},y_{t+\Delta t}) \rangle _{p(y_{t+\Delta t}| {x} _{t+\Delta t})}}{\Delta t}\right) \nonumber \\&\quad + \lim _{\Delta t\rightarrow 0} \left( \frac{{\text {H}}[ Y _{t+\Delta t}| {X} _{t+\Delta t}= {x} _{t+\Delta t}]}{\Delta t}- \frac{{\text {H}}[ Y _t| {X} _{t+\Delta t}= {x} _{t+\Delta t}]}{\Delta t} \right) . \end{aligned}$$

Finally, we average over possible environmental realizations, equivalent in nonequilibrium steady states (NESSs) to averages over time, to find:

$$\begin{aligned} 0&\le \beta \lim _{\Delta t\rightarrow 0} \frac{\langle E( {x} _{t+\Delta t},y_t)\rangle - \langle E( {x} _{t+\Delta t},y_{t+\Delta t}) \rangle }{\Delta t}\\&\quad + \lim _{\Delta t\rightarrow 0} \frac{{\text {H}}[ Y _{t+\Delta t}| {X} _{t+\Delta t}] - {\text {H}}[ Y _t| {X} _{t+\Delta t}]}{\Delta t}. \end{aligned}$$

The former term could be called $-\dot{Q}$—the negative of the sensor’s heat dissipation rate into the reservoir of interest—and the latter $\dot{{\text {I}}}_\text {lost}$—the rate of lost information. And so:

$$\begin{aligned} 0&\le -\beta \dot{Q} + \dot{{\text {I}}}_\text {lost}. \end{aligned}$$

(A2)

This is valid even outside of NESSs. In a NESS, however, we can invoke stationarity, concluding that:

$$\begin{aligned} \dot{Q}=-P, \end{aligned}$$

(A3)

where P is the part of the power extracted by the sensor from the environment that is dissipated as heat into the reservoir of interest. Furthermore, $\dot{{\text {I}}}_\text {lost}$ reduces to the negative of the nonpredictive information rate, since ${\text {H}}[ Y _{t+\Delta t}| {X} _{t+\Delta t}] = {\text {H}}[ Y _{t}| {X} _{t}]$, giving:

$$\begin{aligned} \dot{{\text {I}}}_\text {lost}&= \lim _{\Delta t\rightarrow 0} \frac{{\text {H}}[ Y _t| {X} _t]-{\text {H}}[ Y _t| {X} _{t+\Delta t}]}{\Delta t}. \end{aligned}$$

Calling on a standard information theory identity (Cover and Thomas 2006)—${\text {H}}[U|V] = {\text {H}}[U]-{\text {I}}[U;V]$—leads to:

$$\begin{aligned} \dot{{\text {I}}}_\text {lost}&= \lim _{\Delta t\rightarrow 0} \frac{{\text {I}}[ Y _t; {X} _{t+\Delta t}]-{\text {I}}[ Y _t; {X} _t]}{\Delta t}. \end{aligned}$$

We recognize this as the continuous-time version of the nonpredictive information rate $\dot{{\text {I}}}_\text {np}$; also called the learning rate. Hence, the nonpredictive information rate is the increase in unpredictability of sensor state $Y _t$ given a slightly delayed environmental state:

$$\begin{aligned} \dot{{\text {I}}}_\text {np}&= \lim _{\Delta t\rightarrow 0} \frac{{\text {H}}[ Y _t| {X} _{t+\Delta t}]-{\text {H}}[ Y _t| {X} _t]}{\Delta t}. \end{aligned}$$

(A4)

In a NESS, then:

$$\begin{aligned} \dot{{\text {I}}}_\text {lost}&= - \dot{{\text {I}}}_\text {np}. \end{aligned}$$

(A5)

Outside of NESSs, these terms are augmented by the time derivative $d {\text {H}}[ Y _t| {X} _t] / \hbox {d}t$ of the conditional entropy. This leads to the addition of an Landauer-erasure information (Landauer 1961) when integrated.

One of Still et al. (2012)’s main results follows directly from Eqs. (A2), (A3), and (A5):

$$\begin{aligned} \dot{{\text {I}}}_\text {np}= \lim _{\Delta t\rightarrow 0} \frac{{\text {I}}[ Y _t; {X} _t] - {\text {I}}[ Y _t; {X} _{t+\Delta t}]}{\Delta t} \le \beta P. \end{aligned}$$

(A6)

In contrast to Still et al. (2012)’s implication, this is true only in a NESS and Eq. (A2) should be used otherwise.

Differences in presentation between the derivation here and that of Still et al. (2012) come from the difference between discrete- and continuous-time formulations. To make this clear, we present a continuous-time formulation of the same result, following Horowitz and Esposito (2014). We start from $\beta F_{\hbox {neq}}[p(y_{t'}|x_t)]$ being a Lyapunov function in $t'$:

$$\begin{aligned} 0&\ge \beta \frac{\partial F_{\hbox {neq}}[p(y_{t'}|x_t)]}{\partial t'} \\&= \beta \frac{\partial }{\partial t'} \left\langle E(x_t,y_{t'})\right\rangle _{p(y_{t'}|x_t)}\Big |_{t'=t} - \frac{\partial }{\partial t'} {\text {H}}[Y_{t'}|X_t=x_t]|_{t'=t} \\&= \beta \left\langle \frac{\partial E(x_t,y_t)}{\partial y_t} \dot{y}_t \right\rangle _{p(y_{t'}|x_t)} -\frac{\partial }{\partial t'} {\text {H}}[Y_{t'}|X_t=x_t]|_{t'=t}. \end{aligned}$$

Next, as before, we average over protocols (or, equivalently in NESS, over time) to find:

$$\begin{aligned} 0&\ge \beta \left\langle \frac{\partial E(x_t,y_t)}{\partial y_t} \dot{y}_t \right\rangle _{p(x_t,y_{t})} -\frac{\partial }{\partial t'} {\text {H}}[Y_{t'}|X_t]|_{t'=t} \end{aligned}$$

We then recognize $\beta \left\langle \frac{\partial E(x_t,y_t)}{\partial y_t} \dot{y}_t\right\rangle _{p(x_t,y_{t})}$ as the temperature-normalized rate of heat dissipation $\beta \dot{Q}$, so that:

$$\begin{aligned} \beta \dot{Q} \le \frac{\partial }{\partial t'} {\text {H}}[Y_{t'}|X_t]|_{t'=t}. \end{aligned}$$

The quantity on the right-hand side is simply the rate $\dot{{\text {I}}}_{\hbox {lost}}$ of information loss, defined earlier. In NESS, $d \langle E\rangle / \hbox {d}t$ and $d {\text {H}}[Y_t|X_t] / \hbox {d}t$ vanish. As a result, $\beta \dot{Q}+\beta P = 0$ and:

$$\begin{aligned} \frac{\partial }{\partial t'} {\text {H}}[Y_{t'}|X_t]|_{t'=t} = -\frac{\partial }{\partial t'} {\text {H}}[Y_t|X_{t'}]|_{t'=t}, \end{aligned}$$

giving:

$$\begin{aligned} \beta P \ge \frac{\partial }{\partial t'} {\text {H}}[Y_t|X_{t'}]|_{t'=t}. \end{aligned}$$

(A7)

We recognize this as the continuous-time formulation of Eq. (A4). Again invoking stationarity, $d {\text {H}}[X_t] / \hbox {d}t$ vanishes and so:

$$\begin{aligned} \beta P \ge -\frac{\partial }{\partial t'} {\text {I}}[X_{t'};Y_t] \vert _{t'=t}, \end{aligned}$$

(A8)

the continuous-time formulation of Eq. (A6). We have, in Eqs. (A4), (A6), (A7), and (A8), four equivalent definitions for the nonpredictive information rate in the NESS limit.

Closed-form Expressions for Unifilar Hidden Semi-Markov Environments

To find $\rho ( \sigma ^+, y )$, we start with the following:

$$\begin{aligned} \Pr ( \mathcal {S} ^+_{t+\Delta t}&=(g, {x} ,\tau ), Y _{t+\Delta t}= y ) \nonumber \\&= \sum _{g', {x} ,',\tau ', y '} \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau ), Y _{t+\Delta t}= y | \mathcal {S} ^+_t\nonumber \\&=(g', {x} ',\tau '), Y _t= y ') \Pr ( \mathcal {S} ^+_t=(g', {x} ',\tau '), Y _t= y '). \end{aligned}$$

(B1)

We decompose the transition probability using the lack of feedback as:

$$\begin{aligned} \Pr ( \mathcal {S} ^+_{t+\Delta t}&=(g, {x} ,\tau ), Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g', {x} ',\tau '), Y _t= y ') \\&= \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )| \mathcal {S} ^+_t\\&=(g', {x} ',\tau ')) \Pr ( Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g', {x} ',\tau '), Y _t= y '). \end{aligned}$$

From the setup, we have:

$$\begin{aligned} \Pr ( Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g', {x} ',\tau '), Y _t= y ') = {\left\{ \begin{array}{ll} k_{ y '\rightarrow y }( {x} ')\Delta t &{} y \ne y ' \\ 1-k_{ y '\rightarrow y '}( {x} ')\Delta t &{} y = y ', \end{array}\right. } \end{aligned}$$

with corrections of $O(\Delta t^2)$.

Now split this into two cases. As long as $\tau > \Delta t$, so that ${x} = {x} '$, we have:

$$\begin{aligned} \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )| \mathcal {S} ^+_t=(g', {x} ',\tau ')) = \frac{\Phi _{g}(\tau )}{\Phi _{g}(\tau ')} \delta (\tau -(\tau '+\Delta t)) \delta _{ {x} , {x} '} \delta _{g,g'}. \end{aligned}$$

Then, Eq. (B1) reduces to:

$$\begin{aligned} \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )&, Y _{t+\Delta t}= y ) \nonumber \\&= \sum _{ y '} \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )| \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t))\nonumber \\&\quad \times \Pr ( Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ') \nonumber \\&\quad \times \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ') \nonumber \\&= \sum _{ y '\ne y }\Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )| \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t))\nonumber \\&\quad \times \Pr ( Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ') \nonumber \\&\quad \times \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ') \nonumber \\&\quad + \Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau )| \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t))\nonumber \\&\quad \times \Pr ( Y _{t+\Delta t}= y | \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ) \nonumber \\&\quad \times \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ) \nonumber \\&= \sum _{ y '\ne y } \frac{\Phi _{g}(\tau )}{\Phi _{g}(\tau -\Delta t)} k_{ y '\rightarrow y }( {x} )\nonumber \\&\quad \times \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ') \Delta t \nonumber \\&\quad + \frac{\Phi _{g}(\tau )}{\Phi _{g}(\tau -\Delta t)} \left( 1-k_{ y \rightarrow y }( {x} )\Delta t\right) \nonumber \\&\quad \times \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ) ~, \end{aligned}$$

(B2)

plus terms of $O(\Delta t^2)$. We Taylor expand $\Phi _{g}(\tau +\Delta t) = \Phi _{g}(\tau ) - \phi _{g}(\tau )\Delta t$ to find:

$$\begin{aligned} \frac{\Phi _{g}(\tau )}{\Phi _{g}(\tau -\Delta t)} = 1-\frac{\phi _{g}(\tau )}{\Phi _{g}(\tau )}\Delta t `, \end{aligned}$$

plus terms of $O(\Delta t^2)$. And, similarly, assuming differentiability, we write:

$$\begin{aligned}&\Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau -\Delta t), Y _t= y ')\\&\quad = \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ') - \frac{d}{\hbox {d}\tau } \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ') \Delta t, \end{aligned}$$

plus terms of $O(\Delta t^2)$. Substitution into Eq. (B2) then gives:

$$\begin{aligned}&\Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau ), Y _{t+\Delta t}= y )\\&= \left( \sum _{ y '\ne y } k_{ y '\rightarrow y }( {x} ) \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ') \right) \Delta t + \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ) \\&\qquad - \frac{d\Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y )}{\hbox {d}\tau }\Delta t - \frac{\phi _{g}(\tau )}{\Phi _{g}(\tau )} \Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ) \Delta t \\&\qquad - k_{ y \rightarrow y }( {x} )\Pr ( \mathcal {S} ^+_t=(g, {x} ,\tau ), Y _t= y ) \Delta t, \end{aligned}$$

plus terms of $O(\Delta t^2)$. For notational ease, we denote:

$$\begin{aligned} \rho ((g, {x} ,\tau ), y ) := \Pr ( \mathcal {S} ^+_t=( {x} ,\tau ), Y _t= y ), \end{aligned}$$

which is equal to $\Pr ( \mathcal {S} ^+_{t+\Delta t}=(g, {x} ,\tau ), Y _{t+\Delta t}= y )$ since we assumed the system is in a NESS. Then we have:

$$\begin{aligned} \rho ((g, {x} ,\tau ), y )&= \left( \sum _{ y '\ne y } k_{ y '\rightarrow y }( {x} ) \rho ((g, {x} ,\tau ), y ') \right) \Delta t + \rho ((g, {x} ,\tau ), y ) - \frac{\hbox {d} \rho ((g, {x} ,\tau ), y )}{\hbox {d}\tau }\Delta t \\&\qquad - \frac{\phi _{g}(\tau )}{\Phi _g(\tau )} \rho ((g, {x} ,\tau ), y ) \Delta t - k_{ y \rightarrow y }( {x} ) \rho ((g, {x} ,\tau ), y )\Delta t \end{aligned}$$

plus corrections of $O(\Delta t^2)$. We are left equating the coefficient of the $O(\Delta t)$ term to 0:

$$\begin{aligned} \frac{d\rho ((g, {x} ,\tau ), y )}{d \tau }&= \sum _{ y '\ne y } k_{ y '\rightarrow y }( {x} ) \rho ((g, {x} ,\tau ), y ') - \frac{\phi _g(\tau )}{\Phi _{g}(\tau )} \rho ((g, {x} ,\tau ), y )\nonumber \\&\quad - k_{ y \rightarrow y }( {x} ) \rho ((g, {x} ,\tau ), y ). \end{aligned}$$

(B3)

Our task is simplified if we separate:

$$\begin{aligned} \rho ((g, {x} ,\tau ), y ) = p( y |g, {x} ,\tau ) \rho (g, {x} ,\tau ) \end{aligned}$$

and if we recall that:

$$\begin{aligned} \rho (g, {x} ,\tau ) = \mu _{g} \Phi _{g}(\tau ) p(g) p( {x} |g). \end{aligned}$$

These give:

$$\begin{aligned} \frac{\hbox {d}\rho (g, {x} ,\tau )}{\hbox {d}\tau } = -\mu _{g} \phi _{g}(\tau )p( {x} ) p( {x} |g). \end{aligned}$$

(B4)

Plugging Eq. (B4) into Eq. (B3) yields:

$$\begin{aligned} \frac{\hbox {d}p( y | {x} ,\tau )}{\hbox {d}\tau } \rho (g, {x} ,\tau )&- \mu _{g} \phi _{g}(\tau )p(g)p( {x} |g) p( y |g, {x} ,\tau ) \\&= \sum _{ y '\ne y } k_{ y '\rightarrow y }( {x} ) \rho (g, {x} ,\tau ) p( y '|g, {x} ,\tau )\\&\quad - \frac{\phi _{g}(\tau )}{\Phi _{g}(\tau )} \rho (g, {x} ,\tau ) p( y |g, {x} ,\tau )\\&\quad - k_{ y \rightarrow y }( {x} ) \rho (g, {x} ,\tau ) p( y |g, {x} ,\tau ), \end{aligned}$$

where we note that:

$$\begin{aligned} \mu _{g} \phi _{g}(\tau )p(g)p( {x} |g) p( y |g, {x} ,\tau ) = \frac{\phi _{g}(\tau )}{\Phi _{g}(\tau )} \rho (g, {x} ,\tau ) p( y |g, {x} ,\tau ). \end{aligned}$$

Hence, we are left with:

$$\begin{aligned} \frac{\hbox {d}p( y |g, {x} ,\tau )}{\hbox {d}\tau }&= \sum _{ y '\ne y } k_{ y '\rightarrow y }( {x} ) p( y '|g, {x} ,\tau ) - k_{ y \rightarrow y }( {x} ) p( y |g, {x} ,\tau ). \end{aligned}$$

We can summarize this ordinary differential equation in matrix-vector notation as follows. Let $\mathbf {v}(g, {x} ,\tau )$ be the vector:

$$\begin{aligned} \mathbf {v}(g, {x} ,\tau ) := \begin{pmatrix} p( y _1|g, {x} ,\tau ) \\ \vdots \\ p( y _{|\mathcal {Y}|}|g, {x} ,\tau ). \end{pmatrix} \end{aligned}$$

We have:

$$\begin{aligned} \frac{d\mathbf {v}}{d\tau } = M( {x} )\mathbf {v}, \end{aligned}$$

with solution:

$$\begin{aligned} \mathbf {v}(g, {x} ,\tau ) = e^{M( {x} )\tau } \mathbf {v}(g, {x} ,0). \end{aligned}$$

(B5)

The structure of $M( {x} )$ guarantees that probability is conserved, as long as $1^{\top }\mathbf {v}(g, {x} ,0)=1$ for all ${x} \in \mathcal {A}$.

Our next task is to find expressions for $\mathbf {v}(g, {x} ,0)$. We do this by considering Eq. (B1) in the limit that $\tau <\Delta t$. More straightforwardly, we consider the equation:

$$\begin{aligned} \rho ((g, {x} ,0), y )&= \sum _{g', {x} '} \int _0^{\infty } \hbox {d}\tau ~\frac{\phi _{g'}(\tau )}{\Phi _{g'}(\tau )} \delta _{g,\epsilon ^+(g', {x} ')} p( {x} |g) \rho ((g', {x} ',\tau ), y ), \end{aligned}$$

(B6)

which is based on the following logic. For probability to flow into $\rho ((g, {x} ,0), y )$ from $\rho ((g', {x} ',\tau ), y ')$, we need the dwell time for symbol ${x} '$ to be exactly $\tau$ and for $y '= y$. (The latter comes from the unlikelihood of switching both channel state and input symbol at the same time.) Again decomposing:

$$\begin{aligned} \rho ((g', {x} ',\tau ), y )&= p( y |g', {x} ',\tau ) \rho (g', {x} ',\tau ) \nonumber \\&= \mu _{g'} \Phi _{g'}(\tau ) p(g')p( {x} '|g') p( y |g', {x} ',\tau ) \end{aligned}$$

(B7)

and, thus, as a special case:

$$\begin{aligned} \rho ((g, {x} ,0), y ) = p( y |g, {x} ,0) p(g) p( {x} |g) \mu _{g}. \end{aligned}$$

(B8)

Plugging both Eqs. (B7) and (B8) into Eq. (B6), we find:

$$\begin{aligned} \mu _{g} p(g) p( {x} |g) p( y |g, {x} ,0)&= \sum _{g', {x} '} \int _0^{\infty } \mu _{g'} p(g') p( {x} '|g') \phi _{g'}(\tau ) \delta _{g,\epsilon ^+(g', {x} ')} p( {x} |g) p( y |g', {x} ',\tau ) \hbox {d}\tau \\ \mu _{g} p(g) p( y |g, {x} ,0)&= \sum _{g', {x} '} \int _0^{\infty } \mu _{g'} p(g') p( {x} '|g') \phi _{g'}(\tau ) \delta _{g,\epsilon ^+(g', {x} ')} p( y |g', {x} ',\tau ) \hbox {d}\tau . \end{aligned}$$

Using Eq. (B5), we see that $p( y |g', {x} ',\tau ) = \left( e^{M( {x} ')\tau } \mathbf {v}(g', {x} ',0)\right) _{ y }$ and $p( y |g, {x} ,0) = \left( \mathbf {v}(g, {x} ,0)\right) _{ y }$. So, we have:

$$\begin{aligned} \mu _{g} p(g) \mathbf {v}(g, {x} ,0) = \sum _{g', {x} '} \mu _{g'} \delta _{g,\epsilon ^+(g', {x} ')} p(g') p( {x} '|g') \left( \int _0^{\infty } \phi _{g'}(\tau ) e^{M( {x} ')\tau } d\tau \right) \mathbf {v}(g', {x} ',0). \end{aligned}$$

If we form the composite vector:

$$\begin{aligned} \mathbf {U}&= \begin{pmatrix} \mathbf {u}(g_1, {x} _1) \\ \mathbf {u}(g_1, {x} _2) \\ \vdots \\ \mathbf {u}(g_{|\mathcal {G}|}, {x} _{|\mathcal {A}|}) \end{pmatrix} \\&= \begin{pmatrix} \mu _{g_1} p(g_1) \mathbf {v}(g_1, {x} _1,0) \\ \vdots \\ \mu _{g_{|\mathcal {G}|}} p(g_{|\mathcal {G}|}) \mathbf {v}(g_{|\mathcal {G}|}, {x} _{|\mathcal {A}|},0) \end{pmatrix} \end{aligned}$$

and the matrix (written in block form) as:

$$\begin{aligned} \mathbf {C} := \begin{pmatrix} C_{(g_1, {x} _1)\rightarrow (g_1, {x} _1)} &{} C_{(g_1, {x} _2)\rightarrow (g_1, {x} _1)} &{} \ldots \\ C_{(g_1, {x} _1)\rightarrow (g_1, {x} _2)} &{} C_{(g_1, {x} _2)\rightarrow (g_1, {x} _2)} &{} \ldots \\ \vdots &{} \vdots &{} \ddots , \end{pmatrix} \end{aligned}$$

with:

$$\begin{aligned} C_{(g', {x} ')\rightarrow (g, {x} )} = \delta _{g,\epsilon ^+(g', {x} ')} p( {x} '|g') \int _0^{\infty } \phi _{g'}(t) e^{M( {x} ') t} \hbox {d}t, \end{aligned}$$

we then have:

$$\begin{aligned} \mathbf {U} = \text {eig}_1(\mathbf {C}). \end{aligned}$$

(B9)

Finally, we must normalize $\mathbf {u}( {x} )$ appropriately. We do this by recalling that $1^{\top }\mathbf {v}(g, {x} ,0)=1$, since $\mathbf {v}(g, {x} ,0)$ is a vector of probabilities. Then we have:

$$\begin{aligned} \mathbf {u}(g, {x} ) \rightarrow \frac{\mathbf {u}(g, {x} )}{1^{\top }\mathbf {u}(g, {x} )} \mu _{g} p(g). \end{aligned}$$

for each $g, {x}$.

To calculate prediction metrics—i.e., $I_{\hbox {mem}}$ and $I_{\hbox {fut}}$—we need $p( {x} , y )$ and $p( y , \sigma ^-)$. The former is a marginalization of $p( \sigma ^+, y )$ that we just calculated. The second can be calculated via:

$$\begin{aligned} p( \sigma ^-, y ) = \sum _{ \sigma ^+} p( \sigma ^-| \sigma ^+) p( y , \sigma ^+), \end{aligned}$$

where:

$$\begin{aligned} p( \sigma ^-| \sigma ^+)&= p((g_-, {x} _-,\tau _-)|(g_+, {x} _+,\tau _+)) \\&= \delta _{ {x} _+, {x} _-} p(g_-|g_+, {x} _+) \mu _{g_+} \phi _{g_+}(\tau _+ +\tau _-). \end{aligned}$$

Hence, we turn our attention to calculating dissipation metrics, for which we only need:

$$\begin{aligned} \frac{\delta p}{\delta t} = \lim _{\Delta t\rightarrow 0} \frac{\Pr ( {X} _{t+\Delta t}= {x} , Y _t= y ) - \Pr ( {X} _t= {x} , Y _t= y )}{\Delta t}. \end{aligned}$$

Moreover, we can use the Markov chain $Y _t\rightarrow \mathcal {S} ^+_t\rightarrow {X} _{t+\Delta t}$ to compute it:

$$\begin{aligned} \Pr ( {X} _{t+\Delta t}= {x} , Y _t= y ) = \sum _{ \sigma ^+} \Pr ( {X} _{t+\Delta t}= {x} | \mathcal {S} ^+_t= \sigma ^+) \Pr ( Y _t= y , \mathcal {S} ^+_t= \sigma ^+). \end{aligned}$$

We have:

$$\begin{aligned} \Pr ( {X} _{t+\Delta t}= {x} | \mathcal {S} ^+_t= \sigma ^+)&= \Pr ( {X} _{t+\Delta t}= {x} | \mathcal {S} ^+_t=(g', {x} ',\tau ')) \\&= {\left\{ \begin{array}{ll} \frac{\Phi _{g'}(\tau '+\Delta t)}{\Phi _{g'}(\tau ')} &{} {x} = {x} ' \\ \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} p( {x} |\epsilon ^+(g', {x} ')) \Delta t &{} {x} \ne {x} '. \end{array}\right. } \end{aligned}$$

This, combined with $p( \sigma ^+, y )$, gives:

$$\begin{aligned} \Pr ( {X} _{t+\Delta t}= {x} , Y _t= y )&= \sum _{g', {x} '\ne {x} } \int \hbox {d}\tau '~\rho ((g', {x} ',\tau '), y ) \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} \Delta t~p( {x} |\epsilon ^+(g', {x} ')) \\&\quad + \sum _{g'} \int \hbox {d}\tau ' \frac{\Phi _{g'}(\tau '+\Delta t)}{\Phi _{g'}(\tau ')} \rho ((g', {x} ',\tau '), y ) \\&= \Pr ( {X} _{t}= {x} , Y _t= y )\nonumber \\&\quad + \Delta t \Big ( \sum _{g', {x} '\ne {x} } \int d\tau ' p( {x} |\epsilon ^+(g', {x} ')) \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} \rho ((g', {x} ',\tau '), y ) \nonumber \\&\quad -\sum _{g'} \int \hbox {d}\tau ' \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} \rho ((g', {x} ,\tau '), y )\Big ), \end{aligned}$$

correct to $O(\Delta t)$. Recalling that:

$$\begin{aligned} \rho ((g', {x} ',\tau '), y )&= \rho (g', {x} ',\tau ')p( y |g', {x} ',\tau ') \\&= p( {x} '|g') \Phi _{g'}(\tau ') \left( e^{M( {x} ')\tau '}\mathbf {u}(g', {x} ')\right) _{ y }, \end{aligned}$$

gives:

$$\begin{aligned} \frac{\delta p}{\delta t}= & {} \lim _{\Delta t\rightarrow 0} \frac{\Pr ( {X} _{t+\Delta t}= {x} , Y _t= y ) -\Pr ( {X} _{t}= {x} , Y _t= y )}{\Delta t} \nonumber \\= & {} \sum _{g', {x} '\ne {x} } \int \hbox {d}\tau ' p( {x} |\epsilon ^+(g', {x} ')) \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} \rho ((g', {x} ',\tau '), y ) -\sum _{g'} \int \hbox {d}\tau ' \frac{\phi _{g'}(\tau ')}{\Phi _{g'}(\tau ')} \rho ((g', {x} ,\tau '), y ) \nonumber \\\end{aligned}$$

(B10)

$$\begin{aligned}= & {} \sum _{g', {x} '\ne {x} } \int \hbox {d}\tau ' ~p( {x} |\epsilon ^+(g', {x} ')) p( {x} '|g') \phi _{g'}(\tau ') \left( e^{M( {x} ')\tau '}\mathbf {u}(g', {x} ')\right) _{ y } \nonumber \\&\qquad - \sum _{g'} \int \hbox {d}\tau '~p( {x} |g') \phi _{g'}(\tau ') \left( e^{M( {x} )\tau '}\mathbf {u}(g', {x} )\right) _{ y }. \end{aligned}$$

(B11)

From this, Eqs. (7) and (10) can be used to calculate $\dot{{\text {I}}}_\text {np}$ and $\beta P$.

Specialization to Semi-Markov Input

Up to this point, we wrote expressions for the general case of unifilar hidden semi-Markov environment inputs to the sensor. We now specialize to the semi-Markov input case: the environment’s states are directly observed, not hidden. Not surprisingly, a great simplification ensues: hidden states g are the current emitted symbols ${x}$. Recall that, in an abuse of notation, $q( {x} | {x} ')$ is now the probability of observing symbol ${x}$ after seeing symbol ${x} '$.

Hence, forward-time causal states are given by the pair $( {x} ,\tau )$. The analog of Eq. (B5) is:

$$\begin{aligned} \mathbf {p}( y | {x} ,\tau ) = e^{M( {x} )\tau } \mathbf {p}( y | {x} ,0), \end{aligned}$$

and we define vectors:

$$\begin{aligned} \mathbf {u}( {x} ) := \mu _{ {x} } p( {x} ) \mathbf {p}( y | {x} ,0). \end{aligned}$$

The large vector:

$$\begin{aligned} \mathbf {U} := \begin{pmatrix} \mathbf {u}( {x} _1) \\ \vdots \\ \mathbf {u}( {x} _{|\mathcal {A}}) \end{pmatrix} \end{aligned}$$

is the eigenvector $\text {eig}_1(\mathbf{C} )$ of eigenvalue 1 of the matrix:

$$\begin{aligned} \mathbf{C} = \begin{pmatrix} 0 &{} q( {x} _1| {x} _2) \int _0^{\infty } \phi _{ {x} _2}(\tau ) e^{M( {x} _2)\tau }\hbox {d}\tau &{} \ldots \\ q( {x} _2| {x} _1) \int _0^{\infty } \phi _{ {x} _1}(\tau ) e^{M( {x} _1)\tau }\hbox {d}\tau &{} 0 &{} \ldots \\ \vdots &{} \vdots &{} \ddots \end{pmatrix} , \end{aligned}$$

where normalization requires $1^{\top } \mathbf {u}( {x} ) = \mu _{ {x} } p( {x} )$.

We continue by finding $p( y )$, since from this we obtain ${\text {H}}[ Y ]$. We do this via straightforward marginalization:

$$\begin{aligned} p( y )&= \sum _{ \sigma ^+} \rho ( \sigma ^+, y ) = \sum _{ \sigma ^+} p( y | \sigma ^+) \rho ( \sigma ^+) \\&= \sum _{ {x} } \int _0^{\infty }~p( y | {x} ,\tau ) \rho ( {x} ,\tau )~\hbox {d}\tau \\&= \sum _{ {x} } \int _0^{\infty } \left( e^{M( {x} )\tau } \mathbf {v}( {x} ,0)\right) _{ y } \mu _{ {x} } p( {x} ) \Phi _{ {x} }(\tau ) \hbox {d}\tau \\&= \sum _{ {x} } \left( \left( \int _0^{\infty }e^{M( {x} )\tau } \Phi _{ {x} }(\tau ) \hbox {d}\tau \right) \mathbf {u}( {x} )\right) _{ y } . \end{aligned}$$

This implies that:

$$\begin{aligned} \mathbf {p}( y ) = \sum _{ {x} } \left( \int _0^{\infty } e^{M( {x} )\tau } \Phi _{ {x} }(\tau ) \hbox {d}\tau \right) \mathbf {u}( {x} ) . \end{aligned}$$

From earlier, recall that $\mathbf {u}( {x} ) := \mu _{ {x} } p( {x} ) \mathbf {p}( y | {x} ,0)$.

Next, we aim to find $p( {x} , y )$, again via marginalization:

$$\begin{aligned} p( {x} , y )&= \int _0^{\infty } \rho (( {x} ,\tau ), y ) d\tau \nonumber \\&= \int _0^{\infty } \mu _{ {x} } p( {x} ) \Phi _{ {x} }(\tau ) p( y | {x} ,\tau ) d\tau \nonumber \\&= \int _0^{\infty } \mu _{ {x} } p( {x} ) \Phi _{ {x} }(\tau ) \left( e^{M( {x} )\tau } \mathbf {v}( {x} ,0)\right) _{ y } \hbox {d}\tau \nonumber \\&= \left( \left( \int _0^{\infty } e^{M( {x} )\tau }\Phi _{ {x} }(\tau )\hbox {d}\tau \right) \mathbf {u}( {x} ) \right) _{ y } . \end{aligned}$$

(C1)

From the joint distribution $p( {x} , y )$, we easily numerically obtain ${\text {I}}[ {X} ; Y ]$, since $|\mathcal {A}| < \infty$ and $|\mathcal {Y}| <\infty$.

For notational ease, we introduced $\mathcal {T}_t$ in this section as the random variable for the time since last symbol, whose realization is $\tau$. Finally, we require $p( y | \sigma ^-)$ to calculate ${\text {H}}[ Y | \mathcal {S} ^-]$, which we can then combine with ${\text {H}}[ Y ]$ to get an estimate for ${\text {I}}_\text {fut}$. We utilize the Markov chain $Y \rightarrow \mathcal {S} ^+\rightarrow \mathcal {S} ^-$, as stated earlier, and so have:

$$\begin{aligned} p( y | \sigma ^-)&= \sum _{ \sigma ^+} \rho ( y , \sigma ^+| \sigma ^-) \\&= \sum _{ \sigma ^+} p( y | \sigma ^+, \sigma ^-) \rho ( \sigma ^+| \sigma ^-) \\&= \sum _{ \sigma ^+} p( y | \sigma ^+) \rho ( \sigma ^+| \sigma ^-) . \end{aligned}$$

Eq. (B5) gives us $p( y | \sigma ^+)$ as:

$$\begin{aligned} p( y | \sigma ^+)= & {} p( y | {x} _+,\tau _+) \\= & {} \left( e^{M( {x} _+)\tau _+} \mathbf {v}( {x} _+,0)\right) _{ y } \end{aligned}$$

and Eq. (2) gives us $\rho ( \sigma ^+| \sigma ^-)$ after some manipulation:

$$\begin{aligned} \rho ( \sigma ^+| \sigma ^-)&= \rho (( {x} _+,\tau _+)|( {x} _-,\tau _-)) \\&= \delta _{ {x} _+, {x} _-} \frac{\phi _{ {x} _-}(\tau _+ + \tau _-)}{\Phi _{ {x} _-}(\tau _-)} . \end{aligned}$$

Combining the two equations gives:

$$\begin{aligned} p( y | {x} _-,\tau _-)&= \sum _{ {x} _+} \int _0^{\infty }~\delta _{ {x} _+, {x} _-} \frac{\phi _{ {x} _-}(\tau _+ + \tau _-)}{\Phi _{ {x} _-}(\tau _-)} \left( e^{M( {x} _+)\tau _+} \mathbf {v}( {x} _+,0)\right) _{ y }~\hbox {d}\tau _+ \\&= \frac{1}{\Phi _{ {x} _-}(\tau _-)} \left( \left( \int _0^{\infty } \phi _{ {x} _-}(\tau _+ + \tau _-)e^{M( {x} _-)\tau _+} \hbox {d}\tau _+ \right) \mathbf {v}( {x} _-,0)\right) _{ y } . \end{aligned}$$

From this conditional distribution, we compute ${\text {H}}[ Y | \mathcal {S} ^-= \sigma ^-]$, and so ${\text {H}}[ Y | \mathcal {S} ^-]=\langle {\text {H}}[ Y | \mathcal {S} ^-= \sigma ^-]\rangle _{\rho ( \sigma ^-)}$. In more detail, define:

$$\begin{aligned} D_{ {x} }(\tau ) := \int _0^{\infty } \phi _{ {x} }(\tau +s) e^{M( {x} )s} \hbox {d}s , \end{aligned}$$

and we have:

$$\begin{aligned} \mathbf {p}( y | {x} _-,\tau _-) = D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-)/\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-) . \end{aligned}$$

This conditional distribution gives:

$$\begin{aligned} {\text {H}}[ Y | {X} _-= {x} _-,\mathcal {T}_-=\tau _-]&= -\sum _{ y } p( y | {x} _-,\tau _-)\log p( y | {x} _-,\tau _-) \\&= - 1^{\top } \left( \frac{D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-)}{\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-)} \log \left( \frac{D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-)}{\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-)} \right) \right) \\&= -\frac{1}{\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-)} \Big (1^{\top } \left( (D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-))\log (D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-))\right) \\&\qquad -1^{\top } (D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-)) \log (\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-))\Big ) . \end{aligned}$$

We recognize the factor $\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-)$ as $\rho ( {x} _-,\tau _-)$ and so we find that:

$$\begin{aligned} {\text {H}}[ Y | {X} _-,\mathcal {T}_-]&= \sum _{ {x} _-} \int _0^{\infty }~\rho ( {x} _-,\tau _-) {\text {H}}[ Y | {X} _-= {x} _-,\mathcal {T}_-=\tau _-] d\tau _- \\&= -\int _0^{\infty } \left( \sum _{ {x} _-} 1^{\top } \left( (D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-))\log (D_{ {x} _-}(\tau _-)\mathbf {u}( {x} _-)) \right) \right) \hbox {d}\tau _- \\&\qquad + \int _0^{\infty } \left( \sum _{ {x} _-} 1^{\top } D_{ {x} _-}(\tau _-) \mathbf {u}( {x} _-)\log (\mu _{ {x} _-}p( {x} _-)\Phi _{ {x} _-}(\tau _-))\right) \hbox {d}\tau _- . \end{aligned}$$

This, combined with earlier formula for ${\text {H}}[ Y ]$, gives ${\text {I}}_\text {fut}$.

Finally, we wish to find an expression for the nonpredictive information rate $\dot{{\text {I}}}_\text {np}$. We review the somewhat compact derivation of $\delta p / \delta t$ in the more general case, specialized for semi-Markov input. This requires finding an expression for $\Pr ( Y _t= y , {X} _{t+\Delta t}= {x} )$ as an expansion in $\Delta t$. We start as usual:

$$\begin{aligned} \Pr ( Y _t= y , {X} _{t+\Delta t}= {x} ) = \sum _{ {x} '} \int _0^{\infty } \Pr ( Y _t= y , {X} _{t+\Delta t}= {x} , {X} _t= {x} ',\mathcal {T}_t=\tau ) \hbox {d}\tau \end{aligned}$$

and utilize the Markov chain $Y _t\rightarrow \mathcal {S} ^+_t\rightarrow {X} _{t+\Delta t}$, giving:

$$\begin{aligned}&\Pr ( Y _t= y , {X} _{t+\Delta t}= {x} ) = \sum _{ {x} '} \int _0^{\infty }\nonumber \\&\quad \Pr ( Y _t= y | {X} _t= {x} ',\mathcal {T}_t=\tau ) \Pr ( {X} _{t+\Delta t}= {x} | {X} _t= {x} ',\mathcal {T}_t=\tau ) \rho ( {x} ',\tau ) \hbox {d}\tau . \end{aligned}$$

(C2)

We have $\Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau )$ from Eq. (B5). So, we turn our attention to finding $\Pr ( {X} _{t+\Delta t}= {x} | {X} _t= {x} ',\mathcal {T}_t=\tau )$. Some thought reveals that:

$$\begin{aligned} \Pr ( {X} _{t+\Delta t}= {x} | {X} _t= {x} ',\mathcal {T}_t=\tau ) = {\left\{ \begin{array}{ll} \Delta t q( {x} | {x} ') \phi _{ {x} '}(\tau ) / \Phi _{ {x} '}(\tau ) &{} {x} \ne {x} ' \\ \Phi _{ {x} '}(\tau +\Delta t) / \Phi _{ {x} '}(\tau ) &{} {x} = {x} ' \end{array}\right. } , \end{aligned}$$

(C3)

plus corrections of $O(\Delta t^2)$. We substitute Eq. (C3) into Eq. (C2) to get:

$$\begin{aligned} \Pr ( Y _t= y , {X} _{t+\Delta t}= {x} )&= \left( \sum _{ {x} '\ne {x} } \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ',\mathcal {T}_t=\tau ) q( {x} | {x} ') \frac{\phi _{ {x} '}(\tau )}{\Phi _{ {x} '}(\tau )} \rho ( {x} ',\tau )\hbox {d}\tau \right) \Delta t \\&\qquad + \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau ) \frac{\Phi _{ {x} }(\tau +\Delta t)}{\Phi _{ {x} }(\tau )} \rho ( {x} ,\tau )\hbox {d}\tau , \end{aligned}$$

plus corrections of $O(\Delta t^2)$. Recalling:

$$\begin{aligned} \frac{\Phi _{ {x} }(\tau +\Delta t)}{\Phi _{ {x} }(\tau )} = 1-\frac{\phi _{ {x} }(\tau )}{\Phi _{ {x} }(\tau )} \Delta t , \end{aligned}$$

plus corrections of $O(\Delta t^2)$, we simplify further:

$$\begin{aligned} \Pr ( Y _t= y , {X} _{t+\Delta t}= {x} )&= \left( \sum _{ {x} '\ne {x} } \int _0^{\infty }\Pr ( Y _t= y | {X} _t= {x} ',\mathcal {T}_t=\tau ) q( {x} | {x} ')\frac{\phi _{ {x} '}(\tau )}{\Phi _{ {x} '}(\tau )}\rho ( {x} ',\tau )d\tau \right) \Delta t \\&\qquad + \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau ) \rho ( {x} ,\tau )d\tau \\&\qquad -\left( \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau )\frac{\phi _{ {x} }(\tau )}{\Phi _{ {x} }(\tau )}\rho ( {x} ,\tau )d\tau \right) \Delta t , \end{aligned}$$

plus $O(\Delta t^2)$ corrections. We notice that:

$$\begin{aligned} \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau ) \rho ( {x} ,\tau )\hbox {d}\tau = \Pr ( Y _t= y , {X} _t= {x} ) , \end{aligned}$$

so that:

$$\begin{aligned}&\lim _{\Delta t\rightarrow 0} \frac{\Pr ( Y _t= y , {X} _{t+\Delta t}= {x} )-\Pr ( Y _t= y , {X} _t= {x} )}{\Delta t}\\&= \sum _{ {x} '\ne {x} } \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ',\mathcal {T}_t=\tau ) q( {x} | {x} ') \frac{\phi _{ {x} '}(\tau )}{\Phi _{ {x} '}(\tau )} \rho ( {x} ',\tau )\hbox {d}\tau \\&\qquad -\int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau ) \frac{\phi _{ {x} }(\tau )}{\Phi _{ {x} }(\tau )} \rho ( {x} ,\tau )\hbox {d}\tau . \end{aligned}$$

Substituting Eqs. (B5) and (1) into the above expressions yields:

$$\begin{aligned}&\sum _{ {x} '\ne {x} } \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ',\mathcal {T}_t=\tau ) q( {x} | {x} ') \frac{\phi _{ {x} '}(\tau )}{\Phi _{ {x} '}(\tau )} \rho ( {x} ',\tau )\hbox {d}\tau \\&\quad = \sum _{ {x} '} q( {x} | {x} ') \left( \left( \int _0^{\infty } \phi _{ {x} '}(\tau ) e^{M( {x} ')\tau } \hbox {d}\tau \right) \mathbf {u}( {x} ')\right) _{ y } \end{aligned}$$

and:

$$\begin{aligned} \int _0^{\infty } \Pr ( Y _t= y | {X} _t= {x} ,\mathcal {T}_t=\tau ) \frac{\phi _{ {x} }(\tau )}{\Phi _{ {x} }(\tau )} \rho ( {x} ,\tau )\hbox {d}\tau&= \left( \left( \int _0^{\infty } \phi _{ {x} }(\tau ) e^{M( {x} )\tau } \hbox {d}\tau \right) \mathbf {u}( {x} )\right) _{ y } , \end{aligned}$$

so that we have:

$$\begin{aligned} \lim _{\Delta t\rightarrow 0}&\frac{\Pr ( Y _t= y , {X} _{t+\Delta t}= {x} ) -\Pr ( Y _t= y , {X} _t= {x} )}{\Delta t} \\&\qquad \qquad \qquad = \Big ( \sum _{ {x} '} q( {x} | {x} ') \left( \int _0^{\infty } \phi _{ {x} '}(\tau ) e^{M( {x} ')\tau } \hbox {d}\tau \right) \mathbf {u}( {x} ') - \left( \int _0^{\infty } \phi _{ {x} }(\tau ) e^{M( {x} )\tau } \hbox {d}\tau \right) \mathbf {u}( {x} )\Big )_{ y }. \end{aligned}$$

For notational ease, denote the left-hand side as $\delta p( {x} , y ) / \delta t$. The nonpredictive information rate is given by:

$$\begin{aligned} \dot{{\text {I}}}_\text {np}&= \lim _{\Delta t\rightarrow 0} \frac{{\text {I}}[ {X} _t; Y _t]-{\text {I}}[ {X} _{t+\Delta t}; Y _t]}{\Delta t} \\&= \lim _{\Delta t\rightarrow 0} \frac{\left( {\text {H}}[ {X} _t]+{\text {H}}[ Y _t] - {\text {H}}[ {X} _t, Y _t]\right) - \left( {\text {H}}[ {X} _{t+\Delta t}] + {\text {H}}[ Y _t] - {\text {H}}[ {X} _{t+\Delta t}, Y _t]\right) }{\Delta t} \\&= \lim _{\Delta t\rightarrow 0} \frac{{\text {H}}[ {X} _{t+\Delta t}, Y _t] - {\text {H}}[ {X} _t, Y _t]}{\Delta t} , \end{aligned}$$

where we utilize stationarity to assert ${\text {H}}[ {X} _t]={\text {H}}[ {X} _{t+\Delta t}]$. Then, correct to $O(\Delta t)$, we have:

$$\begin{aligned} {\text {H}}[ {X} _{t+\Delta t}, Y _t]&= - \sum _{ {x} , y } \left( p( {x} , y ) + \frac{\delta p( {x} , y )}{\delta t}\Delta t\right) \log \left( p( {x} , y ) + \frac{\delta p( {x} , y )}{\delta t}\Delta t\right) \\&= -\sum _{ {x} , y } p( {x} , y ) \log p( {x} , y ) - \sum _{ {x} , y }p( {x} , y ) \frac{\delta p( {x} , y )/\delta t}{p( {x} , y )}~\Delta t\\&\quad - \sum _{ {x} , y } \frac{\delta p( {x} , y )}{\delta t} \log p( {x} , y ) \Delta t \\&= {\text {H}}[ {X} _t; Y _t] - \sum _{ {x} , y } \frac{\delta p( {x} , y )}{\delta t} \log p( {x} , y ) \Delta t , \end{aligned}$$

which implies:

$$\begin{aligned} \dot{{\text {I}}}_\text {np}&= \sum _{ {x} , y } \frac{\delta p( {x} , y )}{\delta t} \log p( {x} , y ) , \end{aligned}$$

with:

$$\begin{aligned} \frac{\delta p( {x} , y )}{\delta t}&= \left( \sum _{ {x} '} q( {x} | {x} ')\left( \int _0^{\infty } \Phi _{ {x} '}(\tau ) e^{M( {x} ')\tau } \hbox {d}\tau \right) \right. \mathbf {u}( {x} ') - \left( \int _0^{\infty } \phi _{ {x} }(\tau ) e^{M( {x} )\tau } \hbox {d}\tau \right) \left. \mathbf {u}( {x} )\right) _{ y } \end{aligned}$$

(C4)

and $p( {x} , y )$ given in Eq. (C1).