数学代写|运筹学作业代写operational research代考|Birth and Death Process

2023年4月7日

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数学代写|运筹学作业代写operational research代考|Birth and Death Process

An important special case of a continuous-time Markov chain is the birth and death process. Suppose we have a system whose state at any time is the number of people present at that time. When there are $i$ people present, births (new arrivals) occur at an exponential rate $\lambda_i$, while deaths (departures) occur at an exponential rate $\mu_i$. That is, in state $i$, the time $B_i$ until the next birth is exponentially distributed with mean $1 / \lambda_i$, and the time $D_i$ until the next death is exponentially distributed with mean $1 / \mu_i$, and these times are independent of each other. Such a process is called a birth and death process, in which $\lambda_i, i=0,1, \ldots$, are called the birth rates and $\mu_i, i=1,2, \ldots$, the death rates.

A birth and death process is a continuous-time Markov chain with state space $I={0,1, \ldots}$ in which the only transitions possible from state $i$ are to states $i-1$ and $i+1$ for $i>0$ and to state 1 for $i=0$. The Markov jump chain description of the birth and death process is:
\begin{aligned} \nu_0 & =\lambda_0 \ \nu_i & =\lambda_i+\mu_i \quad \text { for } i>0 \ p_{0,1} & =1 \ p_{i, i+1} & =\mathbb{P}\left(B_i0, \ p_{i, i-1} & =\mathbb{P}\left(B_i>D_i\right)=\frac{\mu_i}{\lambda_i+\mu_i} \quad \text { for } i>0 . \end{aligned}
The time until either a birth or a death occurs, which is the time until state $i$ is left, is $\min \left(B_i, D_i\right)$ and is exponentially distributed with rate $\nu_i=\lambda_i+\mu_i$ for $i>0$. Naturally, nobody can depart if there are no people present, so $\nu_0=\lambda_0$ and correspondingly $p_{0,1}=1$. Further, the transition from state $i$ is to state $i+1$ if a birth occurs before a death, and the probability that an exponential random variable with rate $\lambda_i$ occurs earlier than an (independent) exponential random variable with rate $\mu_i$ is $\lambda_i /\left(\lambda_i+\mu_i\right)$. The last transition probability is explained analogously.
A birth and death process for which $\mu_i=0$ for all $i$ is called a pure birth process. A special case is the Poisson process: the pure birth process with constant birth rates $\lambda_i=\lambda$, see Appendix $\mathrm{C}$.

数学代写|运筹学作业代写operational research代考|Equilibrium Distribution

The infinitesimal transition rates of the birth and death process are
$$q_{i j}= \begin{cases}\lambda_i & \text { if } j=i+1, i \geq 0 \text { (birth rates) } \ \mu_i & \text { if } j=i+1, i \geq 1 \quad \text { (death rates) } \ -\nu_i & \text { if } j=i, i \geq 0\end{cases}$$
The balance equations are
\begin{aligned} & \lambda_0 \pi_0=\mu_1 \pi_1 \ & \lambda_i \pi_i+\mu_i \pi_i=\lambda_{i-1} \pi_{i-1}+\mu_{i+1} \pi_{i+1} \quad \text { for } i=1,2, \ldots \end{aligned}
Combining the equation $\lambda_0 \pi_0=\mu_1 \pi_1$ for $i=0$ with the equation for $i=1$, we observe that it must be that $\lambda_1 \pi_1=\mu_2 \pi_2$. We can use induction to find that
$$\lambda_i \pi_i=\mu_{i+1} \pi_{i+1} \quad \text { for } i=0,1, \ldots$$
These equations are called detailed balance equations as they relate the average number of transitions between neighboring states.

We may now solve the detailed balance equations. To this end, note that (8.11) implies that
$$\pi_{i+1}=\frac{\lambda_i}{\mu_{i+1}} \pi_i \quad \text { for } i=0,1, \ldots$$
Iterating this equation gives that
$$\pi_i=\pi_0 \prod_{j=0}^{i-1} \frac{\lambda_j}{\mu_{j+1}} \quad \text { for } i=1,2, \ldots$$
The birth and death process is ergodic if
$$G:=1+\sum_{i=1}^{\infty} \prod_{j=0}^{i-1} \frac{\lambda_j}{\mu_{j+1}}<\infty .$$
In that case, $\pi_0=G^{-1}$ and $\pi_i$ in (8.12) is the equilibrium distribution.

运筹学代考

数学代写|运筹学作业代写operational research代考|Birth and Death Process

$\backslash$ begin ${$ aligned $} \backslash$ nu_0 \& $=\backslash$ lambda_0 $\backslash \backslash n u_{-} i$ \& $=\backslash$ lambda

数学代写|运筹学作业代写operational research代考|Equilibrium Distribution

$$q_{i j}=\left{\lambda_i \quad \text { if } j=i+1, i \geq 0 \text { (birth rates) } \mu_i\right.$$

$$\lambda_0 \pi_0=\mu_1 \pi_1 \quad \lambda_i \pi_i+\mu_i \pi_i=\lambda_{i-1} \pi_{i-1}+\mu_{i+1}$$

$$\lambda_i \pi_i=\mu_{i+1} \pi_{i+1} \quad \text { for } i=0,1, \ldots$$

$$\pi_{i+1}=\frac{\lambda_i}{\mu_{i+1}} \pi_i \quad \text { for } i=0,1, \ldots$$

$$\pi_i=\pi_0 \prod_{j=0}^{i-1} \frac{\lambda_j}{\mu_{j+1}} \quad \text { for } i=1,2, \ldots$$

$$G:=1+\sum_{i=1}^{\infty} \prod_{j=0}^{i-1} \frac{\lambda_j}{\mu_{j+1}}<\infty .$$

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