HW Math 504. Spring 2008. CSUF

Problems 9.3 and 9.5

by Nasser Abbasi

Problem 9.3


prob_9_3.png

Start by showing that the processes MATH and MATH 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: MATH, Where are told that $N\left( t\right) $ is a Poisson process. Need to find MATH and MATH.

By law of total probabilities

MATH

Hence

MATH

Similarly,

MATH

Now find expression for the joined distribution MATH to complete the above evaluation. Condition on $N\left( t\right) $ hence we obtain

MATH

or

MATH

But since MATH, then the above reduces to one case which is

MATH

And all the other probabilities must be zero.

Now in (1), we are given that MATH 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 MATH

Now we need to evaluate the term MATH 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

MATH

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

MATH
hence the above becomes

MATH

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

MATH

But $p+q=1$, hence MATH hence (4) becomes

MATH

The above is the joined probability of MATH and MATH We know can determine the probability distribution of MATH and MATH from substituting (5) into (A1) and (A2)

MATH

We remove terms outside sum which do not depend on $n$ and obtain

MATH

But MATH by definition, hence the above becomes

MATH

Therefore, we see that MATH satisfies the Poisson formula. To show it is a Poisson distribution, we must also show that it satisfies the following

  1. MATH
    We see that at $t=0$, the above becomes MATHButMATH $0^0=1$, hence

    MATH
    , Therefore MATH

  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 MATH increments are a subset of those increments. Therefore, MATH increments must by necessity be independent of each others.

Similar arguments show that MATH and that it satisfies the Poisson definition$.$We now need to show independence. We see that

MATH

But from (5) above, we see this is the same as MATH, therefore

MATH

Hence MATH and MATH are 2 independent Poisson processes.


Problem 9.5


prob_9_5.png

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


d.png
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 MATH car arrived after than $n^{th}$ car after more than $\tau $ units of time. Therefore

MATH

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 MATHand MATH

Hence (1) becomes

MATH

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.


prob_9_3_mma.png

NowMATH

Let MATH then the above becomes

MATH

The above sum converges since by ratio test the MATH term over the MATH term is less than one. (I can find a closed form expression for this sum?)