# 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 $$$\begin{split} &P(\text{an event happen in (t,t+h)}|X(t)=\text{odd})=\lambda_1h+o(h)\\ &P(\text{an event happen in (t,t+h)}|X(t)=\text{even})=\lambda_2h+o(h) \end{split} \tag{25.1}$$$ 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 $$$\begin{split} P(X(t+h)=\text{odd})&=\sum_iP(X(t+h)=\text{odd}|X(t)=i)P(X(t)=i)\\ &=P(X(t+h)=\text{odd}|X(t)=\text{even})P(X(t)=\text{even})\\ &+P(X(t+h)=\text{odd}|X(t)=\text{odd})P(X(t)=\text{odd}) \end{split} \tag{25.2}$$$ using (25.1), we have $$$\begin{split} P(X(t+h)=\text{odd})&=(\lambda_2 h+o(h))P(X(t)=\text{even})\\ &+(1-\lambda_1 h+o(h))P(X(t)=\text{odd}) \end{split} \tag{25.3}$$$ Now denote $$P_1(t)=P(X(t)=\text{odd})$$ and $$P_2(t)=P(X(t)=\text{even})$$. By (25.3) we have $$$\begin{split} P_1(t+h)&=(\lambda_2 h+o(h))P_2(t)+(1-\lambda_1 h+o(h))P_1(t)\\ &=\lambda_2 hP_2(t)+(1-\lambda_1 h)P_1(t)+o(h) \end{split} \tag{25.4}$$$ which implies $$$\frac{P_1(t+h)-P_1(t)}{h}=-\lambda_1 P_1(t)+\lambda_2 P_2(t)+\frac{o(h)}{h} \tag{25.5}$$$
Take the limit as $$h\to 0$$ on both side, we get $$$P_1^{\prime}(t)=-\lambda_1P_1(t)+\lambda_2P_2(t) \tag{25.6}$$$

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

Equation (25.6) and (25.7) can be combined to write in matrix form as $$$\mathbf{P}^{\prime}(x)=\begin{pmatrix} P^{\prime}_1(x)\\P^{\prime}_2(x)\end{pmatrix}= \begin{pmatrix} -\lambda_1 & \lambda_2\\ \lambda_1 & -\lambda_2 \end{pmatrix} \begin{pmatrix} P_1(x)\\P_2(x)\end{pmatrix}=\boldsymbol{\Lambda}\mathbf{P}(x) \tag{25.9}$$$
Suppose $$\mathbf{P}(x)=\boldsymbol{\eta}e^{rt}$$, then $$\mathbf{P}^{\prime}(x)=r\boldsymbol{\eta}e^{rt}$$, substitute into (25.9), we have $$$(\boldsymbol{\Lambda}-r \mathbf{I})\boldsymbol{\eta}e^{rt}=\mathbf{0} \tag{25.10}$$$ since $$e^{rt}$$ are not zero, we can drop it and finally get
$$$(\boldsymbol{\Lambda}-r \mathbf{I})\boldsymbol{\eta}=\mathbf{0} \tag{25.11}$$$
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 $$$\mathbf{P}(x)=c_1\begin{pmatrix} \frac{\lambda_2}{\lambda_1}\\ 1\end{pmatrix}+c_2e^{-(\lambda_1+\lambda_2)t}\begin{pmatrix} -1\\ 1\end{pmatrix}^T \tag{25.12}$$$
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 $$$\begin{pmatrix} P_1(x)\\P_2(x)\end{pmatrix}=\begin{pmatrix} \frac{\lambda_2}{\lambda_1+\lambda_2}(1-e^{-(\lambda_1+\lambda_2)t})\\ \frac{\lambda_2}{\lambda_1+\lambda_2}(\frac{\lambda_1}{\lambda_2}+e^{-(\lambda_1+\lambda_2)t})\end{pmatrix} \tag{25.13}$$$

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, $$$W=\sum_{i=1}^{N(T)}(T-S_i) \tag{25.14}$$$ is a compound Poisson process. Thus, $$$E(W)=E(T-S_1+T-S_2+\cdots+T-S_{N(T)})=E(N(T))E(T-S_1)=\lambda TE(T-S_1) \tag{25.15}$$$

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

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