A birth-death process is a special form of a [[Markov Chain]] where transitions are restricted to moving up, moving down, or staying in the current state. For example, it can model a supermarket queue where the number of people increases by one (birth), decreases by one (death), or remains the same. ![[birth-death-process.png|center|400]] Since a birth-death process is [[Convergence of Markov Chains#Recurrent & Transient States|recurrent]] and [[Convergence of Markov Chains#Periodicity|aperiodic]], it has well-defined steady-state probabilities when $n$ (the number of steps) becomes large. ![[birth-death-process-split.png|center|200]] ## Steady-State Probabilities We can draw a random boundary in the middle of the Markov chain, and analyze the frequencies of crossing that boundary. If we cross e.g. 100 times from left to right, I have to cross 99, 100 or 101 times from right to left. So if $n$ is large we can say that: $ \begin{align} \pi_{i+1}*q_{i+1} &= \pi_i*p_i \\[4pt] \pi_{i+1} &= \pi_i* \frac{p_i}{q_{i+1}} \end{align} $ where: - $p_i$: Transition rate from state $i$ to state $(i+1)$, i.e. "birth". - $q_i$: Transition rate from state $i$ to state $(i-1)$, i.e. "death". Using this relation we can express any $\pi_j$ in terms of $\pi_0$ and the $\{p, q\}$ terms. $ \pi_j=\pi_0* \prod_{i=0}^{j-1} \frac{p_i}{q_{i+1}}$ Since we know that all steady-state probabilities need to sum to $1$, we have an additional constraint. $ \sum_j \pi_j=1$ Substituting $\pi_j$: $ \begin{align} \pi_0 + \overbrace{\pi_0*\frac{p_0}{q_1}}^{\pi_1}+ \overbrace{\pi_0*\frac{p_0*p_1}{q_1*q_2}}^{\pi_2} + \dots &= 1 \\ \pi_0*\left(1+\frac{p_0}{q_1}+\frac{p_0*p_1}{q_1*q_2}+ \dots\right)&=1 \end{align} $ ## Special Cases of Birth-Death Processes **Constant Rates:** Let the "birth-rate" $p$ be constant across all $i$. Also the "death-rate" $q$ is constant across all $i$. We define $\rho$ as the ratio of these two constant rates. $\rho=\frac{p}{q}$ The normalization condition simplifies to a [[Convergence of Series#Geometric Series|Geometric Series]]. $ \begin{align} \pi_0*\left(1+\frac{p}{q}+\frac{p*p}{q*q}+ \dots\right)&=1 \\[4pt] \pi_0*\Big(1+\rho+\rho^2+ \dots\Big)&=1 \end{align} $ **Symmetric Random Walk:** When $p=q$ we call it a symmetric random walk, where $\rho=1$ and all $\pi_i$ are equally likely. $ \begin{align} \pi_0*(1+\rho+\rho^2+ \dots+\rho^m)&=1\\ \pi_0*(1+1+1^2+ \dots+1^m)&=1 \end{align} $ Since $1^n=1$, we are basically summing up ones for $(m+1)$ times. $\pi_0=\frac{1}{1+m}$ By the recursive formula we see that all $\pi_i$ are equal in that scenario: $ \pi_{i+1} = \pi_i* \frac{p}{q}, \quad \text{where}\: \frac{p}{q}=1$ **Geometric Series:** When $p < q$ then it follows that $\rho < \vert 1 \vert$. For $m \to \infty$, we have a infinite geometric series that converges to a finite sum. $ \begin{align} \pi_0*(\overbrace{1+\rho+\rho^2+ \dots+\rho^m}^{\text{geometric}})&=1\\[6pt] \pi_0* \sum_{i=1}^\infty \rho^i&=1 \\[4pt] \pi_0* \frac{1}{1-\rho}&=1\\[6pt] \pi_0&=1-\rho \end{align} $