Why Linear Birth-Rate Growth Matters
A continuous-time chain can have finite transition rates at every state and still make infinitely many jumps in a finite amount of time. This phenomenon is called explosion. For a birth-death chain on the nonnegative integers, a useful sufficient condition for avoiding explosion is that the birth rate at state \(n\) grows no faster than a constant times \(n+1\). The reason is that the chain cannot climb faster than a suitable pure-birth process, and that comparison process takes infinite time to make infinitely many births.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_t)_{t\ge0}\) be the minimal continuous-time birth-death chain on \(S=\{0,1,2,\ldots\}\), started at a fixed state \(x_0\). At state \(n\), its birth rate is \(\lambda_n\), and its death rate is \(\mu_n\), where \(\lambda_n<\infty\), \(\mu_n<\infty\), and \(\mu_0=0\). The chain jumps from \(n\) to \(n+1\) at rate \(\lambda_n\), and, when \(n\ge1\), from \(n\) to \(n-1\) at rate \(\mu_n\). Its generator has entries \(q_{n,n+1}=\lambda_n\), \(q_{n,n-1}=\mu_n\) for \(n\ge1\), \(q_{n,n}=-(\lambda_n+\mu_n)\), and all other entries zero; at state \(0\), \(q_{0,1}=\lambda_0\) and \(q_{0,0}=-\lambda_0\). Each row sums to zero.
“Finite rates” means that the rate at each individual state is finite; it does not mean that the rates are bounded uniformly over all states. The condition below controls births as the state grows, but it places no uniform upper bound on death rates. Deaths move the chain down, so the comparison will only need to dominate births.
The Comparison Theorem
We first establish the needed fact about the comparison process. Let \((Y_t)_{t\ge0}\) be the pure-birth process on \(S\) with birth rate \(a_n=C(n+1)\) at state \(n\). It has no downward jumps. Its generator row at \(n\) has \(q_{n,n+1}=C(n+1)\) and \(q_{n,n}=-C(n+1)\), so its row sum is zero.
Proof. Start \(Y\) at \(y\), and for an integer \(m>y\) let \(\tau_m=\inf\{t\ge0:Y_t=m\}\). Stop the process when it reaches \(m\). Before stopping, it has only finitely many possible states, so its rates are bounded on the states it can visit. The expected increase in \(Y+1\) per unit time before \(\tau_m\), when the current state is \(n\), is \(C(n+1)\). Consequently, for \(f_m(t)=\mathbb{E}_y[Y_{t\wedge\tau_m}+1]\), the finite-state jump calculation gives \(f_m(t)=y+1+C\int_0^t\mathbb{E}_y[\mathbf{1}_{\{s<\tau_m\}}(Y_s+1)]\,ds\). Because \(Y_{s\wedge\tau_m}+1\) is nonnegative and agrees with \(Y_s+1\) on \(\{s<\tau_m\}\), this implies \(f_m(t)\le y+1+C\int_0^t f_m(s)\,ds\). Iterating this integral inequality gives \(f_m(t)\le(y+1)e^{Ct}\): after \(r\) iterations the terms are bounded by \((y+1)\sum_{j=0}^{r-1}(Ct)^j/j!\) plus a remainder that tends to zero, since \(f_m\) is bounded on each finite time interval for fixed \(m\). On \(\{\tau_m\le t\}\), the stopped value \(Y_{t\wedge\tau_m}+1\) equals \(m+1\). Therefore \[ \mathbb{P}_y(\tau_m\le t)\le \frac{(y+1)e^{Ct}}{m+1}. \] If \(Y\) made infinitely many births by time \(t\), it would reach every \(m>y\) by time \(t\). The displayed bound tends to zero as \(m\to\infty\), so the probability of infinitely many births by time \(t\) is zero. Taking the union over positive integer \(t\) shows that \(Y\) is nonexplosive. \(\square\)
Proof of the theorem. Start the comparison process \(Y\) and the birth-death chain \(X\) at the same state \(x_0\). Construct them jointly while maintaining \(X_t\le Y_t\). Whenever the pair is at \((x,y)\) with \(x\le y\), assign joint births as follows: both coordinates increase by one at rate \(\lambda_x\), and only \(Y\) increases by one at rate \(C(y+1)-\lambda_x\). This second rate is nonnegative because \(\lambda_x\le C(x+1)\le C(y+1)\). Give \(X\) a death from \(x\) to \(x-1\) at rate \(\mu_x\) when \(x\ge1\). Each of these transitions preserves \(X\le Y\): a shared birth preserves the difference, an extra birth of \(Y\) increases it, and a death of \(X\) decreases \(X\). The marginal birth rate of \(X\) is \(\lambda_x\), its death rate is \(\mu_x\), and the marginal birth rate of \(Y\) is \(\lambda_x+C(y+1)-\lambda_x=C(y+1)\), as required.
For precision, this joint construction can first be made up to the time \(Y\) reaches \(m\). Before that time, \(0\le X\le Y\le m\), so the pair has only finitely many possible states and all its transition rates are bounded. A chain with bounded rates on a finite state space cannot make infinitely many jumps in finite time: its successive holding times are bounded below in distribution by exponential holding times with one common finite rate, whose infinite sum is almost surely infinite. Thus the coupled chain is well-defined up to \(\tau_m\). These constructions agree up to their respective stopping times and can be continued as \(m\) increases. By the lemma, \(Y\) does not reach arbitrarily high levels in any bounded time interval. For every fixed \(t<\infty\), it follows that \(Y\) stays in a finite set throughout \([0,t]\), almost surely. On that event \(X\le Y\) also stays in a finite set, where all of its rates are bounded; it therefore cannot make infinitely many jumps by time \(t\). Since this holds for every finite \(t\), \(X\) is nonexplosive. The argument applies to every finite \(x_0\). \(\square\)
The proof explains why no uniform bound on the death rates is needed. A death can add a jump, but it also lowers \(X\). While the chain is confined to finitely many states, all its rates, including its death rates, have a finite maximum. The comparison prevents it from escaping to infinitely high states in finite time, which is the only way infinitely many jumps could accumulate in this birth-death setting.
Worked Comparisons
Worked Example: Linear Births with Linear Deaths
Take \(\lambda_n=n+1\) and \(\mu_n=n\), with \(\mu_0=0\). From every state \(n\ge0\), the birth and death rates are finite. The generator row at \(0\) has entries \(q_{0,1}=1\) and \(q_{0,0}=-1\), whose sum is \(0\). For \(n\ge1\), the row entries are \(q_{n,n+1}=n+1\), \(q_{n,n-1}=n\), and \(q_{n,n}=-(2n+1)\), and their sum is \((n+1)+n-(2n+1)=0\).
The birth-rate condition holds with \(C=1\), since \(\lambda_n=n+1=1\cdot(n+1)\) for every \(n\). Therefore the chain is nonexplosive from every initial state. In the coupling, its births are shared with a pure-birth process of rate \(y+1\) at state \(y\), while the pure-birth process also makes extra births whenever \(y>n\). The death rate \(n\) does not enter the birth-rate comparison.
Check: Starting from \(x_0=2\), the comparison process starts at \(y=2\), so the lemma gives \(\mathbb{P}_2(\tau_m\le t)\le 3e^t/(m+1)\) for \(m>2\). The factor \(3\) is \(y+1=2+1\). This bound tends to zero as \(m\) increases, confirming that the comparison process cannot reach arbitrarily high states by a fixed time.
Worked Example: Large Death Rates Do Not Break the Argument
Take \(\lambda_n=2n+1\) and \(\mu_n=n^2\), with \(\mu_0=0\). At state \(0\), the generator entries are \(q_{0,1}=1\) and \(q_{0,0}=-1\), with sum \(0\). At \(n\ge1\), the entries are \(q_{n,n+1}=2n+1\), \(q_{n,n-1}=n^2\), and \(q_{n,n}=-(2n+1+n^2)\), whose sum is zero. Every rate at each state is finite.
For all \(n\ge0\), \(2n+1\le2n+2=2(n+1)\), so the birth condition holds with \(C=2\). The death rates are not bounded over all states: for example, \(\mu_1=1\), \(\mu_2=4\), and \(\mu_3=9\). This does not affect the theorem, which requires only that each death rate be finite. The chain is nonexplosive from every state.
Check: From \(x_0=0\), the comparison process has \(y=0\) initially, so \(\mathbb{P}_0(\tau_m\le t)\le e^{2t}/(m+1)\). Here \(y+1=1\) and \(C=2\). At \(t=0\), the bound is \(1/(m+1)\), while the probability of reaching \(m>0\) at time zero is \(0\), consistent with the inequality.
Worked Example: A Superlinear Pure-Birth Process Can Explode
Consider a pure-birth process with \(\lambda_n=(n+1)^2\) and \(\mu_n=0\). At every state \(n\ge0\), the generator row has \(q_{n,n+1}=(n+1)^2\), \(q_{n,n}=-(n+1)^2\), and all other entries zero, so each row sums to zero and every rate is finite.
Starting from \(0\), the successive holding times are independent exponentials with respective rates \(1^2,2^2,3^2,\ldots\). If \(H_k\) is the holding time at state \(k-1\), then the explosion time is \(\zeta=\sum_{k=1}^{\infty}H_k\), and \(\mathbb{E}[H_k]=1/k^2\). By monotone convergence, \(\mathbb{E}[\zeta]=\sum_{k=1}^{\infty}1/k^2\). This sum is finite: its first term is \(1\), and for \(k\ge2\), \(1/k^2\le\int_{k-1}^{k}x^{-2}\,dx\), so \(\sum_{k=2}^{\infty}1/k^2\le\int_1^\infty x^{-2}\,dx=1\). Thus \(\mathbb{E}[\zeta]\le2\). A nonnegative random variable with finite expectation is finite almost surely, so this process explodes almost surely.
Check: The birth rates do not satisfy any bound \((n+1)^2\le C(n+1)\) for a fixed finite \(C\), since that inequality would require \(n+1\le C\) for every \(n\). This example shows why the linear-growth hypothesis cannot simply be omitted from the comparison theorem; it does not assert that every process outside the hypothesis explodes.
A Common Comparison Error
A frequent mistake is to compare the total jump rate \(\lambda_n+\mu_n\) with \(C(n+1)\), then conclude that the theorem fails if the death rates make the total rate larger. For the first example, at \(n=1\), the total rate is \(\lambda_1+\mu_1=2+1=3\), while the birth bound with \(C=1\) is \(1(1+1)=2\); the proposed inequality \(3\le2\) is false. That calculation is correct, but it tests the wrong quantity. The coupling compares the upward rate alone: \(\lambda_1=2\le2(1+1)=4\) if \(C=2\), or, for the first example, \(\lambda_1=2=1(1+1)\). The deaths of \(X\) are handled separately and preserve the ordering by moving \(X\) downward. Always check which transitions the comparison is intended to dominate.
The condition is sufficient, not a characterization of all nonexplosive birth-death chains. Its strength is that it gives a direct pathwise comparison requiring no global bound on death rates and no stationary distribution. It is useful before applying calculations that presume the chain exists for all finite times: nonexplosion verifies that no probability mass disappears through infinitely many jumps in finite time.
Check Your Understanding
For each question, identify the state space and compare birth rates only; when checking a generator row, include its diagonal entry.
- For \(\lambda_n=3n+2\) and finite death rates \(\mu_n\), find a constant \(C\) that satisfies the linear-rate condition for every \(n\ge0\).
- Why does the coupling use a shared birth rate \(\lambda_x\) and an extra birth rate \(C(y+1)-\lambda_x\) when \(x\le y\)?
- If \(\lambda_n=4\) for all \(n\ge0\) and \(\mu_n=n^3\), does the theorem imply nonexplosion? State a valid value of \(C\).
- For a pure-birth process with rates \((n+1)^2\), explain why finite rates at every state do not prevent explosion.
- In the first worked example, compute the total rate at \(n=2\) and separately check the birth-rate bound with \(C=1\).