Exponential Holding Times And Jump Probabilities

One Holding Time, Two Possible Jumps

In a continuous-time birth-death process, the process waits in its current state and then moves either one state up or one state down. The rates \(\lambda_n\) and \(\mu_n\) describe those two possible departures from state \(n\). Together they determine how quickly the process leaves \(n\); their relative sizes determine which direction the next jump takes. Keeping these two roles separate is essential when calculating probabilities over a specified time interval.

Let \((X_t)_{t\ge0}\) be a time-homogeneous, nonexplosive continuous-time Markov chain on a countable state space \(S\subseteq\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Assume its only possible transitions from state \(n\) are to \(n+1\) and \(n-1\), with finite rates \(\lambda_n\ge0\) and \(\mu_n\ge0\). At a state with no lower neighbor, set \(\mu_n=0\); a transition to a state outside \(S\) has rate zero. The generator entries are \(q_{n,n+1}=\lambda_n\), \(q_{n,n-1}=\mu_n\), and \(q_{nn}=-(\lambda_n+\mu_n)\), with absent-neighbor entries omitted. Thus each generator row sums to zero.

When the chain starts at \(n\), let \(T_n=\inf\{t>0:X_t\ne n\}\) be its holding time in \(n\). If a jump occurs, let \(J_n=+\) when the jump is to \(n+1\) and \(J_n=-\) when it is to \(n-1\). Write \(r_n=\lambda_n+\mu_n\) for the total departure rate. The holding-time and jump-type statements below concern the chain started at \(n\), under \(\mathbb{P}_n\). If \(r_n=0\), neither transition is possible and the chain remains in \(n\); conditional probabilities given a jump are then undefined.

The theorem separates two questions. The total rate \(r_n\) governs the waiting time, while the ratios of the individual rates to \(r_n\) govern the jump direction. The jump probabilities are conditional on a jump eventually occurring, not probabilities that a jump occurs by a particular finite time.

Why the Holding Time Is Exponential

Define the survival probability \(S_n(t)=\mathbb{P}_n(T_n>t)\). The Markov property says that, if the process has stayed in \(n\) until time \(t\), its future holding time has the same law as a new holding time starting in \(n\). Consequently, for \(s,t\ge0\),

By the definition of the departure rate, the probability of a departure during a short interval of length \(h\), starting in \(n\), is \(r_nh+o(h)\) as \(h\) decreases to zero. Birth and death are disjoint ways to depart, with respective first-departure probabilities \(\lambda_nh+o(h)\) and \(\mu_nh+o(h)\). Therefore \(S_n(h)=1-r_nh+o(h)\). Combining this short-time relation with the multiplicative identity gives the differential equation for survival.

Proof. Assume \(r_n>0\). For \(h>0\), the Markov property and the short-time departure rate give \(S_n(t+h)=S_n(t)S_n(h)=S_n(t)(1-r_nh+o(h))\). Subtract \(S_n(t)\), divide by \(h\), and let \(h\) decrease to zero. This yields the right derivative \(S_n'(t)=-r_nS_n(t)\), with \(S_n(0)=1\). Multiplying by \(e^{r_nt}\) shows that \(e^{r_nt}S_n(t)\) has derivative zero, so it is constant; its value at zero is \(1\). Thus \(S_n(t)=e^{-r_nt}\). The density is \(r_ne^{-r_nt}\), and its mean is \(\int_0^\infty t r_ne^{-r_nt}\,dt=1/r_n\), by integration by parts. If \(r_n=0\), the departure probability is zero and \(S_n(t)=1\) for every \(t\); this is the absorbing-in-place case, not an exponential distribution with positive rate. \(\square\)

Why the Jump Probabilities Are Rate Ratios

For a small interval of length \(h\), the probability that the first departure is a birth is \(\lambda_nh+o(h)\). If the process has already waited \(t\) units without departing, the Markov property gives the same birth rate for the next short interval. Thus the probability of surviving to time \(t\) and then making a birth as its first jump during the next interval is approximately \(S_n(t)\lambda_nh\). This identifies the joint distribution of holding time and jump type.

Proof. Assume \(r_n>0\), and define \(F_+(t)=\mathbb{P}_n(T_n\le t,J_n=+)\). Conditional on \(T_n>t\), the probability that the first departure during the next interval of length \(h\) is a birth is \(\lambda_nh+o(h)\). Hence \(F_+(t+h)-F_+(t)=S_n(t)(\lambda_nh+o(h))\). Using the survival formula just proved gives \(F_+'(t)=\lambda_ne^{-r_nt}\), with \(F_+(0)=0\). Integrating from \(0\) to \(t\) gives \(F_+(t)=\frac{\lambda_n}{r_n}(1-e^{-r_nt})\). For a death, the short-interval probability is \(\mu_nh+o(h)\), so the same argument gives \(F_-(t)=\frac{\mu_n}{r_n}(1-e^{-r_nt})\). As \(t\) tends to infinity, these probabilities tend to \(\lambda_n/r_n\) and \(\mu_n/r_n\). They sum to \(1\), since \(\lambda_n+\mu_n=r_n\). Also, the joint density of a holding time \(t\) and a birth is \(\lambda_ne^{-r_nt}\); this factors as \((\lambda_n/r_n)r_ne^{-r_nt}\). The corresponding density for a death factors as \((\mu_n/r_n)r_ne^{-r_nt}\). Thus the holding time and the type of the next jump are independent when \(r_n>0\). \(\square\)

Worked Calculations

Zero Rates and a Common Mistake

If \(\lambda_n=0\) and \(\mu_n>0\), the next jump is a death with probability \(1\), but the holding time is still exponential, now with rate \(\mu_n\). If \(\mu_n=0\) and \(\lambda_n>0\), the next jump is a birth with probability \(1\), and the holding time has rate \(\lambda_n\). If both rates are zero, then \(r_n=0\), there is no next jump, and neither conditional jump probability is defined. These cases follow directly from the assumptions; the ratios \(\lambda_n/r_n\) and \(\mu_n/r_n\) must not be used when the denominator is zero.

Common mistake. In the second example, one might calculate the probability of a birth by time \(1/2\) as \(\lambda_1t=2(1/2)=1\). This incorrectly treats the birth rate as if it were a probability, and it ignores both the possibility of a death and the possibility of still waiting at time \(1/2\). The correct calculation includes survival up to each possible departure time: \(\int_0^{1/2}2e^{-3s}\,ds=\frac23(1-e^{-3/2})\). The rate \(\lambda_1=2\) is an instantaneous rate, not a probability for a fixed interval. By contrast, the conditional probability that the eventual next jump is a birth is \(2/3\).

The calculations here apply at a fixed state with finite rates in a time-homogeneous birth-death chain. If the total rate is positive, the holding time and next-jump type are governed by the total rate and rate ratios, respectively. If it is zero, the process does not depart from that state. Checking this distinction before using a rate formula prevents both invalid conditional probabilities and errors in finite-time calculations.