From Holding Times to a Pure-Birth Path
A pure-birth process moves through ordered states one at a time: from state \(n\), its only possible jump is to \(n+1\). The rate \(\lambda_n\) controls the waiting time for that jump. Assigning a random holding time to each state gives a direct way to construct a sample path, not merely to specify its rates. The construction also identifies exactly which sums of waiting times determine the state at a given time.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let the state space be \(S=\{i,i+1,i+2,\ldots\}\), and take continuous time \(t\ge0\). For each \(n\in S\), fix a finite rate \(\lambda_n>0\). On the probability space, take independent random variables \(E_n\) with exponential distributions of rates \(\lambda_n\), so that \(\mathbb{P}(E_n>x)=e^{-\lambda_n x}\) for \(x\ge0\). Start at state \(i\), and define the successive birth times by \(T_0=0\) and \(T_k=\sum_{r=i}^{i+k-1}E_r\) for \(k\ge1\).
This indexing makes the event-time calculation transparent: the \(k\)th birth after time zero occurs at \(T_k\), and reaching state \(i+k\) takes \(k\) holding times. Equivalently, the state at time \(t\) is \(i\) plus the number of birth times \(T_k\), \(k\ge1\), that are at most \(t\), provided the path has not accumulated infinitely many births by then. The convention \(T_k\le t<T_{k+1}\) handles event times without ambiguity.
Why the Construction Has the Required Rates
For a pure-birth process with rate \(\lambda_n\) in state \(n\), the holding time in that state should be exponential with rate \(\lambda_n\), followed by a jump to \(n+1\). The construction assigns exactly that holding-time distribution and makes successive holding times independent. In generator notation, its only nonzero off-diagonal rate in row \(n\) is \(q_{n,n+1}=\lambda_n\); the diagonal entry is \(q_{nn}=-\lambda_n\), and all other entries in that row are zero. The row sum is \(\lambda_n-\lambda_n=0\).
Proof. By construction, the interval from \(T_k\) up to but not including \(T_{k+1}\) has length \(E_{i+k}\). This is exponential with rate \(\lambda_{i+k}\), and the next state is \(i+k+1\) with probability \(1\). It remains to check the Markov property at an arbitrary time \(t\). If \(X_t=n\), the process has spent some elapsed time \(a\ge0\) in state \(n\), and has not yet left it. For \(x\ge0\), the exponential survival formula gives \(\mathbb{P}(E_n>a+x\mid E_n>a)=e^{-\lambda_n(a+x)}/e^{-\lambda_n a}=e^{-\lambda_n x}\). Thus the remaining time in state \(n\), conditional on its elapsed time and survival so far, again has the exponential distribution of rate \(\lambda_n\). The holding times in states greater than \(n\) are independent of \(E_n\) and of the holding times already used, so their conditional joint distribution is unchanged by the observed past. Therefore the future, given \(\mathcal{F}_t\), is distributed as a fresh construction started at \(n=X_t\). This is the Markov property. The assumed divergence of the holding-time sum ensures that every finite time belongs to one of the specified holding intervals, so the process is defined for all \(t\ge0\). \(\square\)
The memoryless calculation is essential. Independence of holding times alone would not make the process Markov if the holding-time distributions were not exponential: the elapsed time in the current state could then affect the distribution of the remaining wait. Here the exponential lack-of-memory property removes that dependence.
Worked Example: Locating the State Along a Realized Path
Start at state \(0\), with rates \(\lambda_0=2\), \(\lambda_1=3\), and \(\lambda_n=1\) for \(n\ge2\). The generator row at each state \(n\) has entries \(q_{nn}=-\lambda_n\), \(q_{n,n+1}=\lambda_n\), and zeros elsewhere, so every row sums to zero. Consider a realized path for which \(E_0=\tfrac14\), \(E_1=\tfrac16\), and \(E_2=\tfrac18\). These are illustrative realized values, not positive-probability events of exact equality under continuous distributions.
The first birth time is \(T_1=E_0=\tfrac14\). The second is \(T_2=E_0+E_1=\tfrac14+\tfrac16=\tfrac5{12}\). The third is \(T_3=E_0+E_1+E_2=\tfrac5{12}+\tfrac18=\tfrac{10}{24}+\tfrac3{24}=\tfrac{13}{24}\). Since \(\tfrac5{12}=\tfrac{10}{24}<\tfrac12<\tfrac{13}{24}\), the defining interval is \(T_2\le\tfrac12<T_3\), and therefore \(X_{1/2}=2\).
Check: The first two births have occurred by time \(\tfrac12\), while the third has not. Counting two births from the initial state \(0\) gives state \(2\), agreeing with the interval definition.
Times to Reach Higher States
Because the time to reach state \(i+k\) is the sum \(T_k=E_i+\cdots+E_{i+k-1}\), its expectation and variance follow from the corresponding quantities for independent exponential variables. For a rate-\(\lambda\) exponential variable \(E\), \(\mathbb{E}[E]=1/\lambda\) and \(\operatorname{Var}(E)=1/\lambda^2\). Independence makes the expectation of a sum the sum of expectations and makes its variance the sum of variances. These calculations concern a fixed finite number of births and do not require a formula for the full distribution of \(X_t\).
Proof. By definition, \(T_k=\sum_{r=i}^{i+k-1}E_r\), a finite sum. Each summand has finite mean \(1/\lambda_r\) and finite variance \(1/\lambda_r^2\). Linearity of expectation gives \(\mathbb{E}[T_k]=\sum_{r=i}^{i+k-1}\mathbb{E}[E_r]=\sum_{r=i}^{i+k-1}1/\lambda_r\). For the variance, expanding the square gives the sum of the individual variances plus twice the sum of pairwise covariances. Independence implies \(\operatorname{Cov}(E_r,E_s)=0\) for \(r\ne s\), so \(\operatorname{Var}(T_k)=\sum_{r=i}^{i+k-1}\operatorname{Var}(E_r)=\sum_{r=i}^{i+k-1}1/\lambda_r^2\). \(\square\)
Worked Example: Mean and Variance of a Three-Birth Time
Start at state \(1\), with \(\lambda_1=2\), \(\lambda_2=4\), \(\lambda_3=5\), and \(\lambda_n=1\) for \(n\ge4\). The generator row sums are zero because each row has diagonal entry \(-\lambda_n\) and its only off-diagonal entry is \(\lambda_n\). Find the mean and variance of the time to reach state \(4\).
Three births are needed, so \(T_3=E_1+E_2+E_3\). The mean is \(\mathbb{E}[T_3]=\tfrac12+\tfrac14+\tfrac15=\tfrac{10}{20}+\tfrac5{20}+\tfrac4{20}=\tfrac{19}{20}\). The variance is \(\operatorname{Var}(T_3)=\tfrac1{2^2}+\tfrac1{4^2}+\tfrac1{5^2}=\tfrac14+\tfrac1{16}+\tfrac1{25}=\tfrac{100}{400}+\tfrac{25}{400}+\tfrac{16}{400}=\tfrac{141}{400}\).
Check: The three terms in the mean are the reciprocal rates for states \(1,2,3\); the three variance terms are their squared reciprocals. Substituting the rates into the proposition gives exactly \(\tfrac{19}{20}\) and \(\tfrac{141}{400}\).
State Probabilities from Event Times
The event-time construction also yields transient probabilities by conditioning on when births occur. To be in state \(i\) at time \(t\), the first holding time must exceed \(t\). To be in state \(i+1\), the first birth must occur at some time \(s\le t\), and the next holding time must last at least \(t-s\). In general, probabilities for higher states involve the distributions of sums of holding times. The next calculation uses only the first three states, so it does not require a general population-size formula.
Worked Example: Probabilities in the First Three States
Start at state \(0\), with \(\lambda_0=1\), \(\lambda_1=2\), \(\lambda_2=3\), and \(\lambda_n=3\) for \(n\ge2\). At every state the generator has diagonal entry \(-\lambda_n\) and one off-diagonal entry \(\lambda_n\), so the row sum is zero. Compute the probabilities of states \(0,1,2\) at \(t=\log 2\).
The first-state probability is the chance the first holding time exceeds \(t\): \(\mathbb{P}(X_t=0)=\mathbb{P}(E_0>t)=e^{-t}\). At \(t=\log 2\), this is \(e^{-\log 2}=\tfrac12\).
For state \(1\), the first birth occurs at time \(s\) with density \(e^{-s}\), and the process remains in state \(1\) until \(t\) with probability \(e^{-2(t-s)}\). Hence \(\mathbb{P}(X_t=1)=\int_0^t e^{-s}e^{-2(t-s)}\,ds=e^{-2t}\int_0^t e^s\,ds=e^{-2t}(e^t-1)=e^{-t}-e^{-2t}\). At \(t=\log 2\), this equals \(\tfrac12-\tfrac14=\tfrac14\).
For state \(2\), first find the density of the second birth time \(T_2=E_0+E_1\). Conditioning on the first birth at \(s\), the second occurs at \(u>s\) with density \(2e^{-2(u-s)}\). Thus, for \(u\ge0\), \(f_{T_2}(u)=\int_0^u e^{-s}2e^{-2(u-s)}\,ds=2e^{-2u}\int_0^u e^s\,ds=2(e^{-u}-e^{-2u})\). After the second birth, the process stays in state \(2\) for at least \(t-u\) with probability \(e^{-3(t-u)}\). Therefore \(\mathbb{P}(X_t=2)=\int_0^t2(e^{-u}-e^{-2u})e^{-3(t-u)}\,du=e^{-3t}(e^t-1)^2\). At \(t=\log 2\), this is \(\tfrac18(2-1)^2=\tfrac18\).
Check: The three probabilities are nonnegative, and their sum is \(\tfrac12+\tfrac14+\tfrac18=\tfrac78\). The remaining probability \(\tfrac18\) is for states at least \(3\), so the probabilities of the first three states do not exceed the total probability \(1\).
Common Pitfalls and Scope
Common mistake: treating a rate as a mean waiting time. If the rate in a state is \(2\), the mean holding time is not \(2\); it is \(1/2\). For example, with rates \(2\) and \(4\) in the first two states, the expected time for two births is \(\tfrac12+\tfrac14=\tfrac34\), not \(2+4=6\). The correction is to use the exponential mean \(1/\lambda_n\) for each holding time.
A second source of errors is confusing the number of births with the state label. Starting at \(i\), after \(k\) births the state is \(i+k\), not \(k\); the event time is \(T_k\), the sum of the first \(k\) holding times. Finally, this construction specifies the path up to any possible accumulation of infinitely many births. To use it as a process on every finite time interval, nonexplosion must hold. The construction does not by itself prove that condition for arbitrary rates; a later treatment of explosion gives criteria for checking it.
Check Your Understanding
Use the holding-time construction and check each result against the event-time or rate definition.
- Starting at state \(3\), write the sum of holding times that gives the time of the second birth, and state the process value at that event time.
- If the rates in states \(0,1,2\) are \(1,2,4\), find the mean and variance of the time to reach state \(3\).
- Explain why the remaining holding time in the current state has the same exponential distribution even when the process has already spent a positive amount of time there.
- For rates \(1\) and \(2\) in states \(0\) and \(1\), derive the probability of being in state \(1\) at time \(t\) by conditioning on the first birth time.
- What condition on the infinite sum of holding times is needed for this construction to define an \(S\)-valued process for every \(t\ge0\)?