probability terminology for parameter in a Markov process Suppose $$P(\text{feature present at time} \ t \ \text{and} \ t+\Delta t) = \beta^{2}+\beta(1-\beta) \exp(\Delta t/\tau)$$
where $\tau = 1/(\pi_{01}+\pi_{10})$. What is $\tau$?
 A: I don't recognize the formula exactly but $1/\tau$ is proportional to a rate at which something happens and $\tau$ a time scale (as Didier suggests in the comment).
Imagine a light or other two-state system (e.g., feature on / feature off) that is being activated, de-activated, or reversed, at random times according to a random process with some characteristic rate of occurrence of each action per unit of time.  The events ON (1), OFF (0), ON-reversal (10), and OFF-reversal (01) could have rates (or probabilities) $r_0, r_1, \pi_{10}, \pi_{01}$, and the events could happen according to a Poisson process or suitable Markov chain.
If, on average, the light is on for a fraction $\beta$ of the time, then for large $\Delta t$ what happens at the two times is independent and the probability is close to $\beta^2$.   For small $\Delta t$ the probability of a change of state is also small, and the probability of light being on at both times is almost the same as that of the light being on at time $t$, which is $\beta$.
$\exp(-\Delta t/\tau) = \exp (-\Delta t (\pi_{01} + \pi_{10}))$ looks like a probability that no reversals occur.  For example, it is that probability for an interval of length $\Delta t$ if, on average, one person per unit time randomly walks by the light switch and a fraction $\pi_{10}$ of people always will turn the light off (if on) and $\pi_{01}$ will always turn it on (if off).
A: As suggested earlier, this formula can be realized by a continuous-time Markov process with two states.
Start with a discrete-time Markov chain on states 0 and 1, with the matrix of transition probabilities $M = \pi_{ij}$.  Construct the continuous-time Markov chain that makes a transition according to $M$ at Poisson distributed times, with an average rate of one transition per unit of time.  The continuous process has infinitesimal generator $M - I$, ie. the probability distribution evolves by $e^{t(M-I)}$. 
$M$ has eigenvalues $1$ and $(1-(\pi_{01}+\pi_{10}))$ and its spectral gap (also called eigenvalue gap) is the difference between these, $\pi_{01}+\pi_{10} $.  The reciprocal of this gap, written as $\tau$ in the formula, is known as the relaxation time in the Markov chain literature (at least for reversible chains, and sometimes for non-reversible chains such as our $M$).
If the $\pi_{ij}$ are chosen so that $\beta = \pi_{01} / (\pi_{01} + \pi_{10})$ then


*

*the stationary distribution of the chain assigns probability $\beta$ to state $1$;

*given that the chain is in state $1$ at time $t$, the probability of being in state $1$ at time $t + |\Delta t|$ is $\beta + (1-\beta)\exp(-|\Delta t|(\pi_{01}+\pi_{10}))$.   
This is equivalent to the formula in the question after correcting the signs in the exponent.  The Markov process should be started in its equilibrium distribution, which means either starting the chain at time $-\infty$, or starting at finite time $t_0$ after initializing the chain to state 1 or 0 with probabilities $\beta$ and $1 - \beta$ respectively. 
