Noise-Driven Switching in Cellular State Models

Why Switching Is Not the Same as an Equilibrium

A deterministic model can describe the average fraction of cells in each of two phenotypic states. Its equilibrium might assign 80% of cells to state B and 20% to state A. That result does not mean an individual cell is “80% in B,” nor does it say how often cells change state. A stochastic model describes individual transitions and can therefore distinguish two systems with the same long-run occupancy but very different switching speeds.

Here we study a cell that can be in state A or state B, with random transitions in either direction. The model is deliberately minimal: its value is that the relationship among transition rates, occupancy, and dwell times can be derived exactly. The introductory material on cellular continuous-time Markov chains and transition propensities supplies the context for this formulation.

Definition: A two-state continuous-time Markov chain \(X(t)\) has state space \(\{A,B\}\). When the cell is in A, it transitions to B at constant rate \(k_{AB}>0\); when it is in B, it transitions to A at constant rate \(k_{BA}>0\). The Markov assumption means that, conditional on the current state, the distribution of the future does not depend on how long the cell has already occupied that state. Rates have units of inverse time.

The model assumes that the rates are constant over the time interval of interest, transitions occur one at a time, and the state labels capture all information needed to predict the next transition rate. It also assumes that the two states communicate: both rates are positive, so each state can eventually be reached from the other. These assumptions are testable modelling choices, not consequences of observing two labels in data.

Occupancy and the Deterministic Limit

Let \(q_B(t)=P(X(t)=B)\). Over a short interval of length \(\Delta t\), a cell in A has probability \(k_{AB}\Delta t+o(\Delta t)\) of switching to B, while a cell in B has probability \(k_{BA}\Delta t+o(\Delta t)\) of switching to A. The probability of two or more transitions in that interval is \(o(\Delta t)\). Balancing probability flow into and out of B gives

\frac{dq_B}{dt}=k_{AB}(1-q_B)-k_{BA}q_B =k_{AB}-(k_{AB}+k_{BA})q_B.

This equation also describes the expected fraction in B for a large population of independent, identically modelled cells. Its equilibrium is a population-level quantity. Defining \(\lambda=k_{AB}+k_{BA}\), solution of the linear equation gives

q_B(t)=\pi_B+\bigl(q_B(0)-\pi_B\bigr)e^{-\lambda t}, \qquad \pi_B=\frac{k_{AB}}{k_{AB}+k_{BA}}, \qquad \pi_A=\frac{k_{BA}}{k_{AB}+k_{BA}}.

The stationary probabilities \(\pi_A\) and \(\pi_B\) are the long-run fractions of time a single cell spends in the respective states, as well as the limiting probabilities of finding it in those states. They are not the fraction of time that every cell spends in both states simultaneously. At any instant, one cell is in A or B; its observed state keeps changing even when the population-level probability has reached equilibrium.

Result: The occupancy ratio depends on the relative rates, since \(\pi_B/\pi_A=k_{AB}/k_{BA}\). The rate of relaxation toward the stationary occupancy depends on their sum, \(k_{AB}+k_{BA}\). Thus stationary occupancy alone does not determine how quickly switching occurs.

Worked Example: Occupancy and a Transient Probability

Suppose a hypothetical cell switches from A to B at \(k_{AB}=0.08\ \mathrm{h}^{-1}\), and from B to A at \(k_{BA}=0.02\ \mathrm{h}^{-1}\). Find its stationary occupancy and the probability it is in B after 10 hours if it starts in A.

The total rate is \(\lambda=0.08+0.02=0.10\ \mathrm{h}^{-1}\). Therefore

\pi_B=\frac{0.08}{0.10}=0.800, \qquad \pi_A=\frac{0.02}{0.10}=0.200.

Starting in A means \(q_B(0)=0\). Substituting into the transient solution at \(t=10\ \mathrm{h}\),

q_B(10)=0.800+(0-0.800)e^{-0.10(10)} =0.800(1-e^{-1}) \approx 0.5057.

Thus the probability of being in B at 10 hours is about 0.506. This is not the probability of having made at least one A-to-B transition: that probability is \(1-e^{-0.08(10)}\approx0.5507\). A cell that has switched to B may have switched back before the 10-hour observation.

Dwell Times and Switching Rates

A dwell time is the duration of one uninterrupted visit to a state. Under a constant transition rate, remaining in A for at least time \(t\) requires that no A-to-B transition has occurred by \(t\). The constant hazard \(k_{AB}\) therefore gives an exponential dwell-time distribution. The corresponding statement holds in B.

Result: For a cell entering A, its dwell time \(T_A\) satisfies \(P(T_A>t)=e^{-k_{AB}t}\), so \(T_A\sim\operatorname{Exp}(k_{AB})\), with mean \(1/k_{AB}\) and variance \(1/k_{AB}^2\). For a cell entering B, \(T_B\sim\operatorname{Exp}(k_{BA})\), with mean \(1/k_{BA}\) and variance \(1/k_{BA}^2\). The rate is the inverse mean dwell time, not the mean dwell time itself.

The exponential distribution has a constant hazard: conditional on still being in a state, the chance of leaving during the next short interval is approximately the same regardless of the time already spent there. This “memoryless” property is a direct implication of the model’s Markov assumption. It is not a general biological law.

Worked Example: The Same Occupancy, Faster Switching

Compare the rates in the previous example with a second hypothetical model having \(k_{AB}=0.40\ \mathrm{h}^{-1}\) and \(k_{BA}=0.10\ \mathrm{h}^{-1}\). Calculate the occupancy and mean dwell times for both models.

For the second model, \(\pi_B=0.40/(0.40+0.10)=0.800\) and \(\pi_A=0.200\), identical to the first model. But its mean dwell times are

E[T_A]=\frac{1}{0.40}=2.50\ \mathrm{h}, \qquad E[T_B]=\frac{1}{0.10}=10.0\ \mathrm{h}.

In the first model, the corresponding means were \(12.5\ \mathrm{h}\) in A and \(50.0\ \mathrm{h}\) in B. Both models have the same long-run occupancy, but the second switches five times as fast: both of its rates are multiplied by five. Occupancy cannot by itself reveal switching speed.

For example, starting in A, the probability of at least one transition out of A within one hour is \(1-e^{-0.40(1)}\approx0.3297\) in the second model, compared with \(1-e^{-0.08(1)}\approx0.07688\) in the first. These are probabilities of leaving A, not probabilities of being in B at the end of the hour.

Estimating Rates from Dwell-Time Data

If complete dwell times are observed and the exponential model is appropriate, a rate can be estimated from durations in its originating state. Suppose \(m\) independent, complete dwell times in A are \(t_1,\ldots,t_m\). Their likelihood is proportional to \(k_{AB}^m\exp(-k_{AB}\sum_i t_i)\). Maximizing its log gives

\ell(k_{AB})=m\log(k_{AB})-k_{AB}\sum_{i=1}^{m}t_i, \qquad \frac{d\ell}{dk_{AB}}=\frac{m}{k_{AB}}-\sum_{i=1}^{m}t_i=0, \qquad \widehat{k}_{AB}=\frac{m}{\sum_{i=1}^{m}t_i}.

The estimate is the number of completed exits divided by the total observed time at risk in A. Estimating \(k_{BA}\) uses the B dwell times in the same way. This derivation assumes the observed dwell episodes are complete and the rate is constant. If a recording ends while a cell remains in a state, that episode is right-censored: its observed duration is not a completed dwell time and should not simply be treated as one.

Worked Example: Estimating Two Rates from Complete Dwell Times

In a hypothetical experiment, four complete A dwell times are \(9, 11, 14,\) and \(16\ \mathrm{h}\). Four complete B dwell times are \(38, 44, 52,\) and \(66\ \mathrm{h}\). Estimate the transition rates and the stationary occupancy under the two-state model.

The total time in A is \(9+11+14+16=50\ \mathrm{h}\), so

\widehat{k}_{AB}=\frac{4}{50\ \mathrm{h}} =0.0800\ \mathrm{h}^{-1}.

The total time in B is \(38+44+52+66=200\ \mathrm{h}\), giving

\widehat{k}_{BA}=\frac{4}{200\ \mathrm{h}} =0.0200\ \mathrm{h}^{-1}.

The estimated stationary probability of B is \(0.0800/(0.0800+0.0200)=0.800\), and the estimated mean dwell times are \(12.5\ \mathrm{h}\) in A and \(50.0\ \mathrm{h}\) in B. These estimates summarize the hypothetical observations under the assumed model; they do not establish that biological switching is truly memoryless.

What the Deterministic Equilibrium Leaves Out

A deterministic equation for the fraction in B reports an expected or population-level trajectory. For a finite population of \(N\) independent cells at stationarity, the number in B has a binomial distribution with parameters \(N\) and \(\pi_B\); the fraction in B has mean \(\pi_B\) and variance \(\pi_B(1-\pi_B)/N\). A single cell does not follow the smooth fractional trajectory: its path consists of discrete state changes and random dwell times. The deterministic equilibrium is useful for average occupancy, but it does not replace a description of individual switching.

The rates also determine how quickly occupancy loses memory of its initial condition. For the indicator \(I_B(t)\), which is 1 when the cell is in B and 0 otherwise, stationarity gives

\operatorname{Cov}\bigl(I_B(t),I_B(t+s)\bigr) =\pi_A\pi_B e^{-(k_{AB}+k_{BA})s}, \qquad s\geq 0.

The decay time for this correlation is \(1/(k_{AB}+k_{BA})\). It concerns temporal persistence in state, whereas \(\pi_B\) concerns occupancy. Keeping these quantities separate helps prevent a common mistake: inferring that stable population proportions imply stable individual cell states.

Quantity	What it describes	Depends on
Stationary probability \(\pi_B\)	Long-run occupancy of B	Rate ratio \(k_{AB}/k_{BA}\)
Mean dwell in A	Duration of a visit to A	\(1/k_{AB}\)
Mean dwell in B	Duration of a visit to B	\(1/k_{BA}\)
Relaxation time	Decay of occupancy transients and state correlation	\(1/(k_{AB}+k_{BA})\)

Assumptions, Diagnostics, and Common Pitfalls

The model is most useful when the two labels represent biologically meaningful states and transitions are sufficiently well observed to estimate their timing. Several features of data can signal a poor fit:

Non-exponential dwell times. If the hazard of leaving a state changes with time already spent there, a constant-rate Markov model is inadequate. A dwell-time survival curve that is not consistent with exponential decay is one warning.
Hidden substates or cell-to-cell differences. A mixture of cells with different rates, or an unobserved intermediate state, can produce dwell-time patterns that are not exponential even when each underlying component has constant rates.
Rates that change with time or conditions. A treatment, signal, or changing environment may alter transition rates. Pooling observations across such conditions can make a constant-rate estimate misleading.
Observation error or missed transitions. Limited temporal resolution can merge multiple transitions into one apparent dwell, while noisy state classification can create false switches.

A deterministic occupancy equation is an alternative when only mean population fractions are needed and stochastic fluctuations are negligible for the question at hand. A two-state stochastic model is appropriate when dwell times, individual trajectories, or finite-population variation matter. The models answer different questions; agreement in equilibrium occupancy does not make their predictions interchangeable.

Common mistake: “The deterministic equilibrium is 80% B, so each cell spends 80% of every observation period in B.” The equilibrium specifies a long-run probability or population fraction, not an exact schedule for each cell. Individual dwell times remain random, and two models can have the same 80% occupancy but very different dwell-time distributions.

Key takeaway: In a two-state continuous-time Markov model, the rate ratio sets stationary occupancy, while each individual rate sets the dwell time in its originating state. A deterministic equilibrium describes average occupancy; it does not describe the timing or randomness of cellular switches.

Check Your Understanding

Answer each question, then compare your reasoning with the brief solutions.

A cell switches A to B at \(0.12\ \mathrm{h}^{-1}\) and B to A at \(0.03\ \mathrm{h}^{-1}\). What is the stationary probability of B, and what is the mean dwell time in A?
Two models have rates \((0.06,0.02)\ \mathrm{h}^{-1}\) and \((0.30,0.10)\ \mathrm{h}^{-1}\), respectively, where each pair is \((k_{AB},k_{BA})\). Which has greater stationary occupancy in B? Which has faster switching?
Why is \(1-e^{-k_{AB}t}\) not generally the probability of being in B at time \(t\) when starting in A?
Three complete A dwell times total \(30\ \mathrm{h}\). Under the constant-rate model, what is the maximum-likelihood estimate of \(k_{AB}\), and what mean dwell time does it imply?

Solutions. (1) \(\pi_B=0.12/(0.12+0.03)=0.800\); the mean A dwell is \(1/0.12=8.33\ \mathrm{h}\). (2) Both have \(\pi_B=0.06/0.08=0.30/0.40=0.750\); the second switches five times faster because both rates are five times larger. (3) The expression is the probability of at least one exit from A, but after entering B the cell can return to A before time \(t\). (4) \(\widehat{k}_{AB}=3/(30\ \mathrm{h})=0.100\ \mathrm{h}^{-1}\), implying a mean dwell of \(1/0.100=10.0\ \mathrm{h}\).