Birth-death process

A birth-death process moves only between neighbouring nonnegative integer states, with state-dependent birth and death rates.

A pure birth process on ${\mathbb{N}}_0$ jumps only from $n$ to $n+1$, at a strictly positive rate ${\lambda }_n$. Its successive holding times are independent exponential variables of rates ${\lambda }_n$.

For a pure birth process with fixed initial state and finite statewise rates, the jump-count compensator gives $dM=(e^{\theta }-1)M_{-}[dN-\lambda (X_{-})dt]$, so $M$ is a local martingale up to explosion. If $\lambda \le C$, a rate-$C$ Poisson candidate clock with accepted jumps of probability $\lambda (X_{-})/C$ prevents explosion and dominates the jump count. Its finite exponential moments dominate every stopped $M$ on each finite horizon, upgrading the local martingale to a true martingale. A random initial state requires $\mathbb{E}{e}^{\theta X_0}<\infty$; uniformly bounded rates do not supply this missing initial integrability.

For a carrying capacity $N$, each of $k$ cells splits into two at rate $1-k/N$, so the population increases by one at the displayed total rate and is absorbed at $N$. Starting from population density $p\in (0,1)$, its normalized density converges on finite horizons to $p{e}^t/(1-p+p{e}^t)$ by the Kurtz fluid limit theorem.

// Target: analysis.bigb

A pure birth process is nonexplosive exactly whenIts explosion time is the sum of independent exponential holding times. If the displayed series converges, that sum has finite expectation. If it diverges, the Laplace transform of the sum isforcing the sum to be infinite almost surely.

The pure birth process with rates ${\lambda }_n=\alpha n+\beta$, where $\alpha ,\beta >0$, combines linear reproduction with immigration. Starting from zero, its mean satisfiesand hence

Add independent rate-$\mu$ resets to zero to a linear birth process with immigration. The resulting irreducible continuous-time Markov chain is nonexplosive by domination of births and finiteness of Poisson reset counts. When $\mu =\lambda$, an excursion from one reaches $n$ with probability $2/(n+1)$, so its jump chain is recurrent with infinite expected jump count. The elapsed return-cycle time instead has meanThus the continuous-time chain is positive recurrent while its jump chain is null recurrent. Its stationary probabilities are ${\pi }_0=1/2$ and ${\pi }_n=1/[(n+1)(n+2)]$ for $n\ge 1$.

The Yule process is the pure birth process with rate $n\lambda$ in population state $n$. Starting from one individual, it is nonexplosive because the independent holding times have divergent total almost surely.

For the Yule process starting from one individual, the generator applied to $f(n)=n$ givesand hence $\mathbb{E}{X}_t=e^{\lambda t}$.

Given $X_t=n+1$, the $n$ birth times in a Yule process have the law of the order statistics of independent variables with densityThis follows by dividing the joint holding-time density by
$\mathbb{P}(X_t=n+1)=e^{-\lambda t}(1-e^{-\lambda t}{)}^n$.

For individual birth and death rates $\lambda ,\mu$ and one initial individual, the finite-time extinction probability satisfiesWhen $\lambda \ne \mu$,

For birth rates $b_i$ and death rates $d_i$, a reversible measure satisfiesThus ${\pi }_{i+1}/{\pi }_i=b_i/d_{i+1}$, and it is an invariant distribution when the resulting measure is summable and the chain is nonexplosive.

For birth rates $b_n$, death rates $d_n$, and state probabilities $P_n$,

Since births and deaths change the state by $+1$ and $-1$,

For per-capita birth and death rates $B$ and $D$, the rates are $b_n=Bn$ and $d_n=Dn$. If $\mu =\mathbb{E}{N}_t$ and ${\sigma }^2=\mathrm{var}{N}_t$, thenThe expected value sees only the net growth rate $B-D$, whereas the variance also sees the total event rate $B+D$.

The $r$th falling factorial moment is $\mathbb{E}[(N{)}_r]$, where $(N{)}_r=N(N-1)\cdots (N-r+1)$.

For a nonnegative integer-valued random variable, the displayed inversion is valid when $\mathbb{E}(X{)}_s{s}^k/s!\to 0$. Indeed the gap between consecutive Jordan-Bonferroni exact-occurrence inequalities is $\mathbb{E}(X{)}_{k+m+1}/[k!(m+1)!]$, which tends to zero under this hypothesis. No exchange of an infinite signed sum with an expectation is needed.

If $N$ is Poisson with mean $\lambda$, then $\mathbb{E}[(N{)}_r]={\lambda }^r$; in particular, $\mathbb{E}N=\lambda$ and $\mathbb{E}[N(N-1)]={\lambda }^2$.

If nonnegative integer-valued random variables $X_n$ have all fixed factorial moments converging to ${\lambda }^k$, then their convergence in distribution is to the Poisson distribution with mean $\lambda$. The first-moment bound gives uniform tightness, and bounds on higher moments give uniform integrability of each fixed lower moment. Every subsequential limit therefore has the indicated factorial moments. For such a limit $X$ and $t\ge 0$, the nonnegative binomial expansion and the monotone convergence theorem give $\mathbb{E}(1+t{)}^X={\sum }_{k\ge 0}{t}^k{\lambda }^k/k!=e^{\lambda t}$. This identifies its moment-generating function as that of the Poisson distribution, uniquely determining the limit.

Approximating the population law by a Poisson law replaces its first two factorial moments by $\lambda$ and ${\lambda }^2$, closing a quadratic first-moment equation as a Riccati equation for $\lambda$.

For unit jumps with birth and death rates $b(x),d(x)$, the Fokker--Planck drift and diffusion in the conventionare $u=b-d$ and $D=(b+d)/2$.

Near a stable deterministic equilibrium, replace the drift by its linearization and the diffusion coefficient by its equilibrium value. The resulting Ornstein-Uhlenbeck approximation has a stationary normal distribution.

A reaction's stoichiometric vector outer product multiplied by its reaction propensity function is its contribution to the infinitesimal state-increment covariance. Summing over channels gives $B$. Its normalized form $D$ divides by the two mean copy numbers. The Fokker-Planck equation writes the associated second-derivative term with $B/2$; the covariance evolution and normalized stationary Lyapunov equation instead use $B$ and $D$ themselves. A simultaneous unit loss of two species contributes positive cross noise even when their long-time stationary covariance is negative.

Let $A$ be the drift Jacobian at a stable mean-field equilibrium, $S=\mathrm{diag}({\mu }_i)$, and $B={\sum }_r{a}_r{\nu }_r{\nu }_r^{\mathsf{T}}$ the reaction noise matrix. For normalized fluctuations, put $M=-S^{-1}AS$, $D=S^{-1}B{S}^{-1}$, and ${\eta }_{ij}=\mathrm{Cov}(X_i,X_j)/({\mu }_i{\mu }_j)$. Stationarity of the linear-noise covariance gives the displayed Continuous Lyapunov equation. It applies to driven reaction networks without requiring thermal equilibrium. A mode with zero restoring eigenvalue and nonzero noise has no stationary covariance.

// Target: mathematical-biology.bigb

For RNA birth/death rates ${\lambda }_1,{\beta }_1{x}_1$ and protein rates ${\lambda }_2{x}_{1}f(x_2),{\beta }_2{x}_2$, linearize at stationary means $m_1,m_2$. Put $H=1-\partial \log f/\partial \log x_2=1+h$. The relative-fluctuation drift is $\left(\begin{array}{cc}-{\beta }_1& 0\\ {\beta }_2& -{\beta }_{2}H\end{array}\right)$ and the diffusion matrix is $\mathrm{diag}(2{\beta }_1/m_1,2{\beta }_2/m_2)$. The stationary covariance equation gives the displayed normalized protein variance. Its extrinsic part factors as $H^{-2}[{\beta }_{2}H/({\beta }_1+{\beta }_{2}H)]m_1^{-1}$: static susceptibility squared times a time-averaging factor times environmental noise. Faster response alone weakens averaging; feedback reduces susceptibility and thereby lowers noise. The formula is exact for affine rates when $H=1$, and is a linear noise approximation for nonlinear feedback.

For a unit-jump population chain with one-step probabilities $\alpha k/n$ and $\beta k^2/n^2$, its equilibrium density is $u_0=\alpha /\beta$. Speed time by $n$ and center and scale the population as $(k-n{u}_0)/\sqrt{n}$. The accelerated conditional drift is $-\alpha x-\beta x^2/\sqrt{n}$, and the second moment is $2{\alpha }^2/\beta +3\alpha x/\sqrt{n}+\beta x^2/n$. Their compact-set limits give the displayed Ornstein-Uhlenbeck process. If $2{\alpha }^2<\beta$, the invalid-probability threshold stays a fixed positive density distance above equilibrium. Before that threshold, the generator of $x^2$ is at most one because its drift contribution is nonpositive. The stopped Lyapunov function bound proves compact containment and makes the probability of hitting that threshold on a fixed accelerated horizon tend to zero. This supplies the global hypotheses of the diffusion approximation theorem.

If boundary fluxes decay sufficiently fast, two integrations by parts givewhile the diffusion term contributes zero to the first moment.

For constant birth rate $\lambda$ and death rate $\delta$ away from zero, the chain is transient when $\lambda >\delta$, null recurrent when $\lambda =\delta$, and positive recurrent when $\lambda <\delta$.

The queue length in an $M/M/1$ queue is a birth-death process with constant arrival rate $\lambda$ and service rate $\mu$ while nonempty.

With packet arrival rate lambda, independent exponential payloads of mean one and byte-service rate C, let $\rho =\lambda /C<1$. The number N of unfinished packets has law $(1-\rho ){\rho }^n$. Conditional on N=n, the remaining payloads are n independent unit-mean exponentials, including the in-service residual by memorylessness. Summing their Laplace transforms gives $\mathbb{E}{e}^{-sW}=(1-\rho )(1+s)/(1+s-\rho )$. Thus W has mass $1-\rho$ at zero and, conditional on positivity, is exponential with rate $1-\rho$. The probability of at least r headers is instead ${\rho }^r$ for integer r.

If potential arrivals form a Poisson process of rate $\lambda$ and an arrival seeing $n$ customers joins with probability $p(n)$, the queue length is a birth-death process with birth rate $b_n=\lambda p(n)$ and death rate $d_n=\mu$ for $n\ge 1$. Its stationary ratios, when normalizable, are

An $M/M/1$ queue is positive recurrent exactly when its traffic intensity $\rho =\lambda /\mu$ is below one.

When $\rho <1$, the equilibrium queue length has distribution ${\pi }_n=(1-\rho ){\rho }^n$ and mean $\rho /(1-\rho )$.

At equilibrium, the departure process of a stable $M/M/1$ queue is Poisson with the same rate as the arrival process.

If a served customer returns with probability $p$, thinning service completions gives effective external departure rate $\mu (1-p)$ for the queue-length process.

In equilibrium, a Bernoulli-feedback $M/M/1$ queue has the same queue-length process as an ordinary queue with service rate $\mu (1-p)$, so its external departures form a Poisson process of rate $\lambda$.

Starting with one customer, the mean busy period of an $M/M/1$ queue iswhen $\lambda <\mu$. It is infinite when $\lambda \ge \mu$.

An $M/M/\infty$ queue has Poisson arrivals of rate $\lambda$, independent exponential service times of rate $\mu$, and infinitely many servers. Its occupancy is a birth-death process with birth rate $\lambda$ and death rate $n\mu$ in state $n$.

For every $\lambda ,\mu >0$, an $M/M/\infty$ queue is positive recurrent and has the Poisson distribution with mean $\rho =\lambda /\mu$ as its unique stationary distribution:

In stationarity, the departure process of an $M/M/\infty$ queue is a Poisson process of rate $\lambda$. By time reversibility, the departures before any fixed time are independent of the occupancy at that time.

If $X_0=n$ and $p_t=e^{-\mu t}$, thenwith independent summands. The binomial term counts surviving initial customers; Poisson thinning gives the surviving later arrivals.

Writing $\rho =\lambda /\mu$, the stationary empty probability is $e^{-\rho }$. Alternating idle periods of mean $1/\lambda$ with busy periods and applying the renewal-reward theorem gives

An $M/M/s/K$ queue has arrivals forming a Poisson process of rate $\lambda$, $s$ servers with independent service times having an exponential distribution of rate $\mu$, and total system capacity $K\ge s$. Its occupancy is the finite birth-death process on $\{0,\dots ,K\}$ with

Put $\rho =\lambda /\mu$. The unique stationary distribution of an $M/M/s/K$ queue iswhere $Z$ is the sum of the displayed weights. Adjacent terms satisfy detailed balance for a birth-death process, so the queue is reversible in equilibrium.