Chapter 25 Quiz 2: Poisson Process: Problems and Tentative Solutions

Exercise 25.1 (Birth process, probability of even and odd number of occurence) Let $X(t)$ be a pure birth continuous time Markov chain. Assume that $$\begin{equation} \begin{split} &P(\text{an event happen in (t,t+h)}|X(t)=\text{odd})=\lambda_1h+o(h)\\ &P(\text{an event happen in (t,t+h)}|X(t)=\text{even})=\lambda_2h+o(h) \end{split} \tag{25.1} \end{equation}$$ where $\frac{o(h)}{h}\to 0$ as $h\to 0$. Take $X(0)=0$. Find the following probabilities: $P_1(t)=P(X(t)=\text{odd})$ and $P_2(t)=P(X(t)=\text{even})$.

Proof. We start with consider $P(X(t+h)=\text{odd})$, we have $$\begin{equation} \begin{split} P(X(t+h)=\text{odd})&=\sum_iP(X(t+h)=\text{odd}|X(t)=i)P(X(t)=i)\\ &=P(X(t+h)=\text{odd}|X(t)=\text{even})P(X(t)=\text{even})\\ &+P(X(t+h)=\text{odd}|X(t)=\text{odd})P(X(t)=\text{odd}) \end{split} \tag{25.2} \end{equation}$$ using (25.1), we have $$\begin{equation} \begin{split} P(X(t+h)=\text{odd})&=(\lambda_2 h+o(h))P(X(t)=\text{even})\\ &+(1-\lambda_1 h+o(h))P(X(t)=\text{odd}) \end{split} \tag{25.3} \end{equation}$$ Now denote $P_1(t)=P(X(t)=\text{odd})$ and $P_2(t)=P(X(t)=\text{even})$. By (25.3) we have $$\begin{equation} \begin{split} P_1(t+h)&=(\lambda_2 h+o(h))P_2(t)+(1-\lambda_1 h+o(h))P_1(t)\\ &=\lambda_2 hP_2(t)+(1-\lambda_1 h)P_1(t)+o(h) \end{split} \tag{25.4} \end{equation}$$ which implies $$\begin{equation} \frac{P_1(t+h)-P_1(t)}{h}=-\lambda_1 P_1(t)+\lambda_2 P_2(t)+\frac{o(h)}{h} \tag{25.5} \end{equation}$$
Take the limit as $h\to 0$ on both side, we get $$\begin{equation} P_1^{\prime}(t)=-\lambda_1P_1(t)+\lambda_2P_2(t) \tag{25.6} \end{equation}$$

Very similarly, we can get the other differential equation $$\begin{equation} P_2^{\prime}(t)=\lambda_1P_1(t)-\lambda_2P_2(t) \tag{25.7} \end{equation}$$ The boundary conditions are $$\begin{equation} \left\{\begin{aligned} & P_1(0)=0\\ &P_2(0)=1 \end{aligned}\right. \tag{25.8} \end{equation}$$

Equation (25.6) and (25.7) can be combined to write in matrix form as $$\begin{equation} \mathbf{P}^{\prime}(x)=\begin{pmatrix} P^{\prime}_1(x)\\P^{\prime}_2(x)\end{pmatrix}= \begin{pmatrix} -\lambda_1 & \lambda_2\\ \lambda_1 & -\lambda_2 \end{pmatrix} \begin{pmatrix} P_1(x)\\P_2(x)\end{pmatrix}=\boldsymbol{\Lambda}\mathbf{P}(x) \tag{25.9} \end{equation}$$
Suppose $\mathbf{P}(x)=\boldsymbol{\eta}e^{rt}$, then $\mathbf{P}^{\prime}(x)=r\boldsymbol{\eta}e^{rt}$, substitute into (25.9), we have $$\begin{equation} (\boldsymbol{\Lambda}-r \mathbf{I})\boldsymbol{\eta}e^{rt}=\mathbf{0} \tag{25.10} \end{equation}$$ since $e^{rt}$ are not zero, we can drop it and finally get
$$\begin{equation} (\boldsymbol{\Lambda}-r \mathbf{I})\boldsymbol{\eta}=\mathbf{0} \tag{25.11} \end{equation}$$
The matrix $\boldsymbol{\Lambda}$ has two distinct eigenvalues $r_1=0$ and $r_2=-\lambda_1-\lambda_2$, with corresponding eigenvector $\begin{pmatrix} \frac{\lambda_2}{\lambda_1}& 1\end{pmatrix}^T$ and $\begin{pmatrix} -1& 1\end{pmatrix}^T$, therefore, the solution to (25.9) is $$\begin{equation} \mathbf{P}(x)=c_1\begin{pmatrix} \frac{\lambda_2}{\lambda_1}\\ 1\end{pmatrix}+c_2e^{-(\lambda_1+\lambda_2)t}\begin{pmatrix} -1\\ 1\end{pmatrix}^T \tag{25.12} \end{equation}$$
Since the boundary condition is $\mathbf{P}(0)=\begin{pmatrix} 0& 1\end{pmatrix}^T$, we can find $c_1=\frac{\lambda_1}{\lambda_1+\lambda_2}$ and $c_2=\frac{\lambda_2}{\lambda_1+\lambda_2}$. Thus, we finally get $$\begin{equation} \begin{pmatrix} P_1(x)\\P_2(x)\end{pmatrix}=\begin{pmatrix} \frac{\lambda_2}{\lambda_1+\lambda_2}(1-e^{-(\lambda_1+\lambda_2)t})\\ \frac{\lambda_2}{\lambda_1+\lambda_2}(\frac{\lambda_1}{\lambda_2}+e^{-(\lambda_1+\lambda_2)t})\end{pmatrix} \tag{25.13} \end{equation}$$

Exercise 25.2 (Expectation of Compound Poisson Process) Assume that passengers arrive at a bus station as a Poisson process with rate $\lambda$. The only bus departs after a deterministic time $T$. Let $W$ be the combined waiting time for all passengers. Compute $E(W)$.

Hint: Make use of the fact that given the number of events between $[0,T]$ for a Poisson process, $N(T)=k$, the arrival times $S_1,\cdots,S_k$ are distributed as order statistic of a $Unif(0,T)$ distribution.

Proof. Suppose passenger $i$ arrives at the station at time $S_i$, his waiting time is then $W_i=T-S_i$. Therefore, $$\begin{equation} W=\sum_{i=1}^{N(T)}(T-S_i) \tag{25.14} \end{equation}$$ is a compound Poisson process. Thus, $$\begin{equation} E(W)=E(T-S_1+T-S_2+\cdots+T-S_{N(T)})=E(N(T))E(T-S_1)=\lambda TE(T-S_1) \tag{25.15} \end{equation}$$

Using the hint, we have $$\begin{equation} E(T-S_1)=\int_0^T(T-s_1)\cdot\frac{1}{T}ds_1=T-\frac{1}{T}\int_0^Ts_1ds_1=\frac{T}{2} \tag{25.16} \end{equation}$$ Therefore, $E(W)=\frac{\lambda T^2}{2}$.

Stein, Michael. 1999. Interpolation of Spatial Data: Some Theory of Kriging. New York, NY: Springer.