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Chapter 3: Q93E (page 136)

a. In a Poisson process, what has to happen in both the time interval \({\rm{(0,t)}}\) and the interval \({\rm{(t,t + \Delta t)}}\) so that no events occur in the entire interval \({\rm{(0,t + \Delta t)}}\)? Use this and Assumptions \({\rm{1 - 3}}\) to write a relationship between \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t)}}\) and \({{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)

b. Use the result of part (a) to write an expression for the difference \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\) Then divide by \({\rm{\Delta t}}\) and let to obtain an equation involving \({\rm{(d/dt)}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\), the derivative of \({{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)with respect to \({\rm{t}}\).

c. Verify that \({{\rm{P}}_{\rm{0}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}\)satisfies the equation of part (b).

d. It can be shown in a manner similar to parts (a) and (b) that the \({{\rm{P}}_{\rm{k}}}{\rm{(t)}}\)must satisfy the system of differential equations

\(\begin{array}{*{20}{c}}{\frac{{\rm{d}}}{{{\rm{dt}}}}{{\rm{P}}_{\rm{k}}}{\rm{(t) = \alpha }}{{\rm{P}}_{{\rm{k - 1}}}}{\rm{(t) - \alpha }}{{\rm{P}}_{\rm{k}}}{\rm{(t)}}}\\{{\rm{k = 1,2,3, \ldots }}}\end{array}\)

Verify that \({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{{\rm{(\alpha t)}}^{\rm{k}}}{\rm{/k}}\) satisfies the system. (This is actually the only solution.)

Short Answer

Expert verified

(a) \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) = }}{{\rm{P}}_{\rm{0}}}{\rm{(t) \times (1 - \alpha \Delta t - o(\Delta t))}}{\rm{.}}\)

(b) \(\frac{{{\rm{d}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{dt}}}}{\rm{ = - \alpha }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}{\rm{.}}\)

(c) The equation is satisfied.

look inside for proof

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the both the time interval \({\rm{(0,t)}}\) and the interval \({\rm{(t,t + \Delta t)}}\) so that no events occur in the entire interval \({\rm{(0,t + \Delta t)}}\)

Proposition: A Poisson Random Variable with the parameter \({\rm{\mu = \alpha t}}\)can be used to describe the number of events that occur over a time interval of length \({\rm{t}}\). This implies that

\({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times }}\frac{{{{{\rm{(\alpha t)}}}^{\rm{k}}}}}{{{\rm{k!}}}}\)

Poisson process is another name for it (not formally defined).

Assume no events occur at regular intervals.

\({\rm{(0,t + \Delta t)}}\)

This is only possible if no single event occurs at regular intervals.

\({\rm{(0,t)}}\)and\({\rm{(t,t + \Delta t)}}\)

We know that the events described are independent because of assumption \({\rm{3}}\) on page \({\rm{134}}\), and we know that

\(\begin{array}{*{20}{c}}{{\rm{P(\{ \;no events in interval\;(0,t + \Delta t)\} )}}}\\{{\rm{ = P(\{ \;no events in interval\;(0,t)\} {C}\{ \;no events in interval\;(t,t + \Delta t)\} )}}}\\{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(\{ \;no events in interval\;(0,t)\} ) \ast P(\{ \;no events in interval\;(t,t + \Delta t)\} )}}}\end{array}\)

(1) The occurrences are unconnected (assumption \({\rm{3}}\)).

The following statement is correct:

\({\rm{P(\{ no events in interval(t,t + \Delta t)\} ) = 1 - \alpha \Delta t - o(\Delta t)}}\)

where assumption \({\rm{1}}\) was used in conjunction with assumption \({\rm{2}}\) (page \({\rm{134}}\)).

Finally, the following relationship holds when we substitute everything in the equation above.

\({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) = }}{{\rm{P}}_{\rm{0}}}{\rm{(t) \ast (1 - \alpha \Delta t - o(\Delta t))}}{\rm{.}}\)

Where,

\({\rm{P(\{ no events in interval(0,t + \Delta t)\} ) = }}{{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t)}}{\rm{.}}\)

03

Determining the expression for the difference \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)

We have the following information:

\({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) = }}{{\rm{P}}_{\rm{0}}}{\rm{(t) \times (1 - \alpha \Delta t - o(\Delta t))}}\)

or, alternatively,

\({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) = }}{{\rm{P}}_{\rm{0}}}{\rm{(t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)(\alpha \Delta t - o(\Delta t))}}{\rm{.}}\)

As a result, the difference is expressed as

\({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t) = - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)(\alpha \Delta t - o(\Delta t))}}{\rm{.}}\)

We get \({\rm{\Delta t}}\)when we divide the expression.

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{\Delta t}}}}{\rm{ = - }}\frac{{{{\rm{P}}_{\rm{0}}}{\rm{(t)(\alpha \Delta t - o(\Delta t))}}}}{{{\rm{\Delta t}}}}}\\{\frac{{{{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{\Delta t}}}}{\rm{ = - \alpha }}{{\rm{P}}_{\rm{0}}}{\rm{(t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t) \times }}\frac{{{\rm{o(\Delta t)}}}}{{{\rm{\Delta t}}}}}\end{array}\)

We have \({\rm{\Delta tn0}}\) on the right side.

\(\frac{{{\rm{o(\Delta t)}}}}{{{\rm{\Delta t}}}}{\rm{n0}}\)

When \({\rm{\Delta tn0}}\), the left side expression converges to the derivative.

\(\frac{{{{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{\Delta t}}}}{\rm{n}}\frac{{{\rm{d}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{dt}}}}{\rm{.}}\)

When we combine those two outcomes, we obtain the following:

\(\frac{{{\rm{d}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{dt}}}}{\rm{ = - \alpha }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}{\rm{.}}\)

04

Determining that \({{\rm{P}}_{\rm{0}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}\)satisfies the equation of part (b).

with the assumption

\({{\rm{P}}_{\rm{0}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{,}}\)

We have it.

\(\frac{{{\rm{d}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}}}{{{\rm{dt}}}}{\rm{ = }}\frac{{\rm{d}}}{{{\rm{dt}}}}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ = - \alpha }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ = - \alpha }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)

As a result, the equation is satisfied. This is the likelihood that \({\rm{(0,t)}}\)contains no events.

05

Determining Verify that \({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{{\rm{(\alpha t)}}^{\rm{k}}}{\rm{/k}}\) satisfies the system

We have it.

\(\begin{array}{*{20}{c}}{\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{ - \alpha }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}{\rm{ + }}\frac{{{\rm{k\alpha }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{{\rm{k - 1}}}}}}{{{\rm{k!}}}}}\\{{\rm{ = - \alpha }}\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}{\rm{ + \alpha }}\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{{\rm{k - 1}}}}}}{{{\rm{(k - 1)!}}}}}\\{{\rm{ = - \alpha }}{{\rm{P}}_{\rm{k}}}{\rm{(t) + \alpha }}{{\rm{P}}_{{\rm{k - 1}}}}{\rm{(t)}}}\end{array}\begin{array}{*{20}{c}}{\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{ - \alpha }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}{\rm{ + }}\frac{{{\rm{k\alpha }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{{\rm{k - 1}}}}}}{{{\rm{k!}}}}}\\{{\rm{ = - \alpha }}\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{\rm{k}}}}}{{{\rm{k!}}}}{\rm{ + \alpha }}\frac{{{{\rm{e}}^{{\rm{ - \alpha t}}}}{\rm{ \times (\alpha t}}{{\rm{)}}^{{\rm{k - 1}}}}}}{{{\rm{(k - 1)!}}}}}\\{{\rm{ = - \alpha }}{{\rm{P}}_{\rm{k}}}{\rm{(t) + \alpha }}{{\rm{P}}_{{\rm{k - 1}}}}{\rm{(t)}}}\end{array}\)

\({\rm{k = 1,2,3, \ldots }}\)as we needed to show. (1): originates from the derivatives product rule.

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