4.15.1 Problem 9.3

Start by showing that the processes \(N_{1}\left ( t\right ) \) and \(N_{2}\left ( t\right ) \) are each a Poisson process. Next show that they are independent by showing that the product of these 2 distributions is equal to the joined distribution.

Given: \(N\left ( t\right ) =N_{1}\left ( t\right ) +N_{2}\left ( t\right ) \), Where are told that \(N\left ( t\right ) \) is a Poisson process. Need to find \(\Pr \left ( N_{1}\left ( t\right ) =n\right ) \) and \(\Pr \left ( N_{2}\left ( t\right ) =m\right ) \).

By law of total probabilities

\begin{align*} \Pr \left ( N_{1}\left ( t\right ) =n\right ) & =\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =0\right ) \text { }\\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =1\right ) \\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =2\right ) \\ & \vdots \\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =\infty \right ) \end{align*}

Hence

\begin{equation} \Pr \left ( N_{1}\left ( t\right ) =n\right ) ={\displaystyle \sum \limits _{m=0}^{\infty }} \text { }\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) \tag {A1}\end{equation}

Similarly,

\begin{equation} \Pr \left ( N_{2}\left ( t\right ) =m\right ) ={\displaystyle \sum \limits _{n=0}^{\infty }} \text { }\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) \tag {A2}\end{equation}

Now find expression for the joined distribution \(\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) \) to complete the above evaluation. Condition on \(N\left ( t\right ) \) hence we obtain

\begin{align*} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) & =\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =0\right ) \Pr \left ( N\left ( t\right ) =0\right ) \\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =0\right ) \Pr \left ( N\left ( t\right ) =1\right ) \\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =0\right ) \Pr \left ( N\left ( t\right ) =2\right ) \\ & \vdots \\ & or\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =0\right ) \Pr \left ( N\left ( t\right ) =\infty \right ) \end{align*}

or

\[ \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) ={\displaystyle \sum \limits _{k=0}^{\infty }} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =k\right ) \Pr \left ( N\left ( t\right ) =k\right ) \]

But since \(N\left ( t\right ) =N_{1}\left ( t\right ) +N_{2}\left ( t\right ) \), then the above reduces to one case which is

\begin{equation} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) =\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =n+m\right ) \Pr \left ( N\left ( t\right ) =n+m\right ) \tag {1}\end{equation}

And all the other probabilities must be zero.

Now in (1), we are given that \(\Pr \left ( N\left ( t\right ) =n+m\right ) \) is a Poisson process with some rate \(\lambda \) Hence the rate adjusted for duration \(t\) must be \(\lambda t\,\), hence from definition of Poisson process with rate \(\lambda t\,\) we write

\begin{equation} \Pr \left ( N\left ( t\right ) =n+m\right ) =\frac {\left ( \lambda t\right ) ^{n+m}e^{-\left ( \lambda t\right ) }}{\left ( n+m\right ) !} \tag {2}\end{equation}

Now we need to evaluate the term \(\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =n+m\right ) \) in (1). This terms asks for the probability of getting the sum \(\left ( n+m\right ) \). If we think of \(n\) as number of successes and \(m\) as number of failures, then this is asking for probability of getting \(n\) success out of \(n+m\) trials. But this is given by Binomial distribution

\[ \Pr \left ( X=n\right ) =\begin {pmatrix} n+m\\ n \end {pmatrix} p^{n}q^{m}\]

Where \(p\) is the probability of event type \(I\), and \(q\) is the probability of not getting this event, which is the probability of event \(II\) \(\ \) which is given by \(q=\left ( 1-p\right ) \) hence the above becomes

\begin{equation} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\ |\ N\left ( t\right ) =n+m\right ) =\begin {pmatrix} n+m\\ n \end {pmatrix} p^{n}q^{m} \tag {3}\end{equation}

Substitute (2) and (3) into (1) we obtain

\begin{align} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) & =\begin {pmatrix} n+m\\ n \end {pmatrix} p^{n}q^{m}\frac {\left ( \lambda t\right ) ^{n+m}e^{-\left ( \lambda t\right ) }}{\left ( n+m\right ) !}\nonumber \\ & =\frac {\left ( n+m\right ) !}{m!n!}p^{n}q^{m}\frac {\left ( \lambda t\right ) ^{n+m}e^{-\left ( \lambda t\right ) }}{\left ( n+m\right ) !}\nonumber \\ & =\frac {\left ( p\lambda t\right ) ^{n}}{n!}\frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda t\right ) } \tag {4}\end{align}

But \(p+q=1\), hence \(e^{-\left ( \lambda t\right ) }=e^{-\left ( \lambda t\left ( p+q\right ) \right ) }=e^{-\left ( \left ( \lambda tp\right ) +\left ( \lambda tq\right ) \right ) }=e^{-\left ( \lambda tp\right ) }e^{-\left ( \lambda tq\right ) }\) hence (4) becomes

\begin{equation} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) =\left ( \frac {\left ( p\lambda t\right ) ^{n}}{n!}e^{-\left ( \lambda tp\right ) }\right ) \left ( \frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda tq\right ) }\right ) \tag {5}\end{equation}

The above is the joined probability of \(N_{1}\left ( t\right ) \) and \(N_{2}\left ( t\right ) .\) We know can determine the probability distribution of \(N_{1}\left ( t\right ) \) and \(N_{2}\left ( t\right ) \) from substituting (5) into (A1) and (A2)

\begin{align*} \Pr \left ( N_{2}\left ( t\right ) =m\right ) & ={\displaystyle \sum \limits _{n=0}^{\infty }} \Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) \\ & ={\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac {\left ( p\lambda t\right ) ^{n}}{n!}e^{-\left ( \lambda tp\right ) }\right ) \left ( \frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda tq\right ) }\right ) \end{align*}

We remove terms outside sum which do not depend on \(n\) and obtain

\[ \Pr \left ( N_{2}\left ( t\right ) =m\right ) =\left ( \frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda tq\right ) }\right ) e^{-\left ( \lambda tp\right ) }{\displaystyle \sum \limits _{n=0}^{\infty }} \frac {\left ( p\lambda t\right ) ^{n}}{n!}\]

But \({\displaystyle \sum \limits _{n=0}^{\infty }} \frac {\left ( p\lambda t\right ) ^{n}}{n!}=e^{p\lambda t}\) by definition, hence the above becomes

\[ \Pr \left ( N_{2}\left ( t\right ) =m\right ) =\frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda tq\right ) }\]

Therefore, we see that \(\Pr \left ( N_{2}\left ( t\right ) =m\right ) \) satisfies the Poisson formula. To show it is a Poisson distribution, we must also show that it satisfies the following

1. \(N_{2}\left ( 0\right ) =0\). We see that at \(t=0\), the above becomes \(\Pr \left ( N_{2}\left ( 0\right ) =0\right ) =\frac {\left ( q\lambda \times 0\right ) ^{0}}{0!}e^{-\left ( \lambda q\times 0\right ) }=0^{0}\times 1,\)But\(\footnote {$0^{0}$ depends on the context.\ I\ checked a reference that in this context, it is ok to define $0^{0}=1$ otherwise, $0^{0}$ is taken as undefined.}\) \(0^0=1\), hence \(\Pr \left ( N_{2}\left ( 0\right ) =0\right ) =1\), Therefore \(N_2\left ( 0\right ) =0.\)
2. Increments are independents of each others.  Since the original process \(N\left ( t\right ) \) is already given to be Poisson process, then the increments of \(N\left ( t\right ) \) are independent of each others. But \(N_{2}\left ( t\right ) \) increments are a subset of those increments. Therefore, \(N_{2}\left ( t\right ) \) increments must by necessity be independent of each others.

Similar arguments show that

\[ \Pr \left ( N_{1}\left ( t\right ) =n\right ) =\frac {\left ( q\lambda t\right ) ^{n}}{n!}e^{-\left ( \lambda tq\right ) }\]

and that it satisfies the Poisson definition\(.\)We now need to show independence. We see that

\[ \Pr \left ( N_{2}\left ( t\right ) =m\right ) \Pr \left ( N_{1}\left ( t\right ) =n\right ) =\frac {\left ( q\lambda t\right ) ^{m}}{m!}e^{-\left ( \lambda tq\right ) }\frac {\left ( q\lambda t\right ) ^{n}}{n!}e^{-\left ( \lambda tq\right ) }\]

But from (5) above, we see this is the same as \(\Pr \left ( N_{1}\left ( t\right ) =n,N_{2}\left ( t\right ) =m\right ) \), therefore

\[ \Pr \left ( N_{2}\left ( t\right ) =m\right ) \Pr \left ( N_{1}\left ( t\right ) =n\right ) =\Pr \left ( N_{1}=n,N_{2}=m\right ) \]

Hence \(N_{1}\left ( t\right ) \) and \(N_{2}\left ( t\right ) \) are 2 independent Poisson processes.