Chapter 3.6, Problem 93E

1. a. In a Poisson process, what has to happen in both the time interval (0, t) and the interval (t, t + Δt) so that no events occur in the entire interval (0, t + Δt)? Use this and Assumptions 1-3 to write a relationship between P₀(t + Δt) and P₀(t).
2. b. Use the result of part (a) to write an expression for the difference P₀(t + Δt) − P₀(t). Then divide by Δt and let Δt → 0 to obtain an equation involving (d/dt)P₀(t), the derivative of P₀(t) with respect to t.
3. c. Verify that P₀(t) = e^−αt satisfies the equation of part (b).
4. d. It can be shown in a manner similar to parts (a) and (b) that the P_k(t)s must satisfy the system of differential equations

   $\begin{array}{ccc}\frac{d}{dt}{P}_k(t)& =& \alpha P_{k-1}(t)\text{ - }\alpha P_k(t)\\ k& =& 1,\text{ 2, 3,}\mathrm{...}\end{array}$

   Verify that P_k(t) = e^−αt(αt)^k/k! satisfies the system. (This is actually the only solution.)

To determine

Find what will happen in both of the intervals (0, t) and $\left(t,t+\Delta t\right)$ if no events occur in the interval $\left(0,t+\Delta t\right)$

Find the relationship between $P_0\left(t+\Delta t\right)$ and $P_0\left(t\right)$.

Answer to Problem 93E

The relationship between $P_0\left(t+\Delta t\right)$ and $P_0\left(t\right)$ is $P_0\left(t+\Delta t\right)=P_0\left(t\right)\times \left\{1-\alpha \times \Delta t-o\left(\Delta t\right)\right\}$

Explanation of Solution

Given info:

In a Poisson process no events occur in the interval $\left(0,t+\Delta t\right)$.

Calculation:

It is clear that no events occur in the interval $\left(0,t+\Delta t\right)$ if no event occur in the interval (0, t) and no event occur in the interval $\left(t,t+\Delta t\right)$.

The assumptions of Poisson process:

1. 1. There exists a parameter $\alpha >0$ such that for any short time interval of length $\Delta t$ , the probability that exactly one event occurs is $\alpha \times \Delta t+o\left(\Delta t\right)$
2. 2. The probability of more than one event occurring during $\Delta t$ is $o\left(\Delta t\right)$ [the probability of no events during $\left(\Delta t\right)$ is $1-\alpha \times \Delta t-o\left(\Delta t\right)$ ]
3. 3. The number of events occurring during the time interval $\Delta t$ is independent of the number that occur prior to this time interval.

Hence, using the 3^rd assumption,

$\begin{array}{c}{P}_0\left(t+\Delta t\right)=P\left(\text{no events occur in the interval}\left(0,t+\Delta t\right)\right)\\ =P\left(\text{no events occur in the interval}\left(0,t\right)\right)\times P\left(\text{no events occur in the interval}\left(t,t+\Delta t\right)\right)\\ =P_0\left(t\right)\times P\left(\text{no events occur in the interval}\left(t,t+\Delta t\right)\right)\end{array}$

From the second assumption,

$P\left(\text{no events occur in the interval}\left(t,t+\Delta t\right)\right)=1-\alpha \times \Delta t-o\left(\Delta t\right)$

Hence, $P_0\left(t+\Delta t\right)=P_0\left(t\right)\times \left\{1-\alpha \times \Delta t-o\left(\Delta t\right)\right\}$.

To determine

Find the expression of $P_0\left(t+\Delta t\right)-P_0\left(t\right)$ from part a.

Find the equation involving $\left(\frac{d}{dt}\right)P_0\left(t\right)$.

Answer to Problem 93E

The expression is $P_0\left(t+\Delta t\right)-P_0\left(t\right)=-P_0\left(t\right)\left\{\alpha \times \Delta t+o\left(\Delta t\right)\right\}$.

The equation is $\frac{d{P}_0\left(t\right)}{dt}=-\alpha P_0\left(t\right)$.

Explanation of Solution

Calculation:

From part a,

$\begin{array}{c}{P}_0\left(t+\Delta t\right)=P_0\left(t\right)\times \left\{1-\alpha \times \Delta t-o\left(\Delta t\right)\right\}\\ =P_0\left(t\right)-P_0\left(t\right)\left\{\alpha \times \Delta t+o\left(\Delta t\right)\right\}\\ P_0\left(t+\Delta t\right)-P_0\left(t\right)=-P_0\left(t\right)\left\{\alpha \times \Delta t+o\left(\Delta t\right)\right\}\end{array}$

Dividing both side of $P_0\left(t+\Delta t\right)-P_0\left(t\right)=-P_0\left(t\right)\left\{\alpha \times \Delta t+o\left(\Delta t\right)\right\}$ by $\Delta t$, it can be found that,

$\begin{array}{c}\frac{P_0\left(t+\Delta t\right)-P_0\left(t\right)}{\Delta t}=\frac{-P_0\left(t\right)\left\{\alpha \times \Delta t+o\left(\Delta t\right)\right\}}{\Delta t}\\ =\frac{-P_0\left(t\right)\times \alpha \times \Delta t}{\Delta t}-\frac{o\left(\Delta t\right)}{\Delta t}\end{array}$

It is known that $\frac{o\left(\Delta t\right)}{\Delta t}\to 0,\text{ when }\Delta t\to 0$

Hence, $\frac{P_0\left(t+\Delta t\right)-P_0\left(t\right)}{\Delta t}=-P_0\left(t\right)\times \alpha$

It is known that $\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}=\frac{df\left(t\right)}{dt}$

Using this formula,

$\begin{array}{l}\frac{P_0\left(t+\Delta t\right)-P_0\left(t\right)}{\Delta t}=-P_0\left(t\right)\times \alpha \\ \frac{d{P}_0\left(t\right)}{dt}=-\alpha P_0\left(t\right)\end{array}$

Hence, the equation is $\frac{d{P}_0\left(t\right)}{dt}=-\alpha P_0\left(t\right)$.

To determine

Check whether $P_0\left(t\right)=e^{-\alpha t}$ satisfies the equation of the part b.

Answer to Problem 93E

The equation is satisfied.

Explanation of Solution

Calculation:

According to part b the equation involving $\left(\frac{d}{dt}\right)P_0\left(t\right)$ is $\frac{d{P}_0\left(t\right)}{dt}=-\alpha P_0\left(t\right)$.

Putting $P_0\left(t\right)=e^{-\alpha t}$,

$\begin{array}{c}\frac{d{P}_0\left(t\right)}{dt}=\frac{d}{dt}\left(e^{-\alpha t}\right)\\ =-\alpha e^{-\alpha t}\\ =-\alpha P_0\left(t\right)\end{array}$

Hence, the equation is satisfied.

To determine

Verify whether $P_k\left(t\right)=e^{-\alpha t}\frac{{\left(\alpha t\right)}^k}{k!}$ satisfies the system.

Answer to Problem 93E

$P_k\left(t\right)=e^{-\alpha t}\frac{{\left(\alpha t\right)}^k}{k!}$ satisfies the system.

Explanation of Solution

Given info:

$\frac{d}{dt}{P}_k\left(t\right)=\alpha P_{k-1}\left(t\right)-\alpha P_k\left(t\right),k=1,\mathrm{2...3}$

Calculation:

Putting $P_k\left(t\right)=e^{-\alpha t}\frac{{\left(\alpha t\right)}^k}{k!}$,

$\begin{array}{c}\frac{d}{dt}{P}_k\left(t\right)=\frac{d}{dt}\left\{e^{-\alpha t}\frac{{\left(\alpha t\right)}^k}{k!}\right\}\\ =\frac{{\alpha }^k}{k!}\times \frac{d}{dt}\left\{e^{-\alpha t}{\left(t\right)}^k\right\}\\ =\frac{{\alpha }^k}{k!}\times \left\{e^{-\alpha t}k{\left(t\right)}^{k-1}+{\left(t\right)}^k{e}^{-\alpha t}\left(-\alpha \right)\right\}\\ =\frac{e^{-\alpha t}{\alpha }^k{\left(t\right)}^{k-1}}{\left(k-1\right)!}-\alpha \frac{e^{-\alpha t}{\left(\alpha t\right)}^k}{k!}\end{array}$

$\begin{array}{c}=\alpha \frac{e^{-\alpha t}{\left(\alpha t\right)}^{k-1}}{\left(k-1\right)!}-\alpha \frac{e^{-\alpha t}{\left(\alpha t\right)}^k}{k!}\\ =\alpha P_{k-1}\left(t\right)-\alpha P_k\left(t\right)\end{array}$

Hence, $P_k\left(t\right)=e^{-\alpha t}\frac{{\left(\alpha t\right)}^k}{k!}$ satisfies the system.