4.15.2 Problem 9.5

Let the interr arival time between each car be \(T_{i}\) where \(i\) is the interval as indicated by this diagram

Figure 4.5: Inter-arrival times are random variable which is exponential distributed

For the number of cars that pass through the intersection to be \(n\) it must imply that the interval between the first \(n\) cars was less than \(\tau \) and that the \(\left ( n+1\right ) ^{th}\) car arrived after than \(n^{th}\) car after more than \(\tau \) units of time. Therefore

\begin{equation} \Pr \left ( X=n\right ) =\Pr \left ( T_{1}<\tau ,T_{2}<\tau ,T_{3}<\tau ,\cdots ,T_{n}<\tau ,T_{n+1}\geq \tau \right ) \tag {1}\end{equation}

But since \(X\) is a Poisson random number with parameter \(\lambda \) , then the time between increment \(T\) is an exponential random number with parameter \(\lambda \) (and they are independent from each others). Hence

\[ \Pr \left ( T_{i}\geq \tau \right ) =e^{-\lambda \tau }\]

and

\[ \Pr \left ( T_{i}<\tau \right ) =1-e^{-\lambda \tau }\]

Hence (1) becomes

\[ \Pr \left ( X=n\right ) =\left ( 1-e^{-\lambda \tau }\right ) ^{n}e^{-\lambda \tau } \]

This is a small program which plots the probability above as function of \(n\) for some fixed \(\lambda ,\tau \). It shows as expected the probability of \(n\) becomes smaller the larger \(n\) gets.

Now

\begin{align*} E\left ( X\right ) & ={\displaystyle \sum \limits _{n=0}^{\infty }} n\Pr \left ( X=n\right ) \\ & ={\displaystyle \sum \limits _{n=0}^{\infty }} n\left ( 1-e^{-\lambda \tau }\right ) ^{n}e^{-\lambda \tau }\\ & =e^{-\lambda \tau }{\displaystyle \sum \limits _{n=1}^{\infty }} n\left ( 1-e^{-\lambda \tau }\right ) ^{n}\end{align*}

Let \(1-e^{-\lambda \tau }=z\) then the above becomes

\[ E\left ( X\right ) =e^{-\lambda \tau }{\displaystyle \sum \limits _{n=1}^{\infty }} nz^{n}\]

The above sum converges since by ratio test the \(\left ( k\right ) ^{th}\) term over the \(\left ( k+1\right ) ^{th}\) term is less than one. (I can find a closed form expression for this sum?)

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