# Conditional probability Poisson process and dependent random variables

This is for homework, but I'm stuck cause I can't find something like this on my book, or in the internet for that matter.

Say I have a restaurant that's open the 24 hours, and we know that costumers enter the establishment according a Poisson process at a rate of 5 costumers/hour.

1. Given that 6 costumers arrived from 1:00 am and 2:30 am, what's the probability that less than 3 customers arrive from 2:30 am to 4:00 am?

I have modeled this question like this: P(C<3|X=6), being C the amount of customers that could arrive and X the customers that have already arrived. so I use the definition of conditional probability: $$\frac{P(C<3\bigcap X=6)}{P(X=6)}$$

I ended up using this property that seems shady: $$P(A\bigcap B)=P(A) P(B)$$

Make the math and I end up with $$\frac{0.1367 \cdot 0.059}{0.1367}=0.059$$ which seems reasonable, but I don't think is right. So help me here if it's not ok.

2. If we know that 30 customers arrived from 10:00 pm to 4:00 am, what's the probability that 20 customers arrived from 1:30 am to 3:15 am?

This questions has me stumped. I think I should break the times in something like: $$[10:00, 1:30)\bigcup [1:30, 3:15]\bigcup (3:15, 4:00]$$

And then do the probabilities separate, and add them up at the end, but that seems wrong, and I don't know what to do...

-

The textbook on Poisson processes you use surely mentions that the number of points falling in some disjoint time sets are independent. Re 1., this tells you that the conditional probability you are after is the same as the (unconditional) probability that less than 3 customers arrive from 2:30am to 4:00am. This is the probability that a Poisson random variable with parameter 5 per hour $\times$ 1.5 hour $=$ 7.5, is less than 3.

Re 2., the same textbook surely explains that once you know $n$ points fall in an interval [a,b], their locations in [a,b] are $n$ i.i.d. random variables distributed according to the density of the Poisson process. Here the Poisson process is homogeneous hence the number of customers who arrived from 1:30am to 3:15am is binomial $(n,p)$ where $n=30$ and $p$ denotes the proportion of the overall time the time span of interests represents, that is $p=\frac{1.75}6=\frac7{24}$.

-
Thank you very much. –  alvinbaena Mar 11 '12 at 0:34

There is nothing shady about the property you used in part a). In a Poisson process, events occurring in one time frame are independent of events occurring in another, disjoint, time frame.

Note that in your answer to a), the probabilities $P[X=6]$ cancel, as they should, leaving you with $P[C<3\mid X=6]=P[C<3]$. No suprise here, as this is a direct consequence of the independence property. (I did not check that you computed $P[C<3]$ correctly.)

Although Didier Piau's approach to part b) is a better approach to the problem; I think it's worthwhile to compute the probability directly. In fact, this computation may help you see why the property mentioned by Didier is true.

So I offer:

Let $\lambda=5$ and let:

$\ \ \ A$ be the number of events occurring in the 4.25 hour time period from 10-1:30 and 3:15-4.

$\ \ \ B$ be the number of events occurring in the 1.75 hour time period from 1:30-3:15

and

$\ \ \ C$ be the number of events occurring in the 6 hour time period from 10-4.

Note $A$ has Poisson distribution with parameter $4.25\lambda$, $B$ has Poisson distribution with parameter $1.75\lambda$, and $C$ has Poisson distribution with parameter $6\lambda$.

In your problem, you wish to find $P[\,B=20\mid C=30\,]$.
We have: \eqalign{ P[\,B=20\mid C=30\,]&={P\bigl[\,(B=20 )\cap (C=30)\,\bigr]\over P[C=30] } \cr &={P[ (A=10)\cap (B=20)]\over P[C=30] } \cr &={P[ A=10]P[B=20]\over P[C=30] } \cr &={{\textstyle(4.25\lambda)^{10} e^{-4.25\lambda}\over\textstyle 10!} \cdot {\textstyle(1.75\lambda)^{20} e^{-1.75\lambda}\over \textstyle20!} \over{\textstyle(6\lambda)^{30} e^{-6\lambda}\over\textstyle 30!} } \cr &={{(4.25 )^{10} \lambda^{10} \over 10!} \cdot {(1.75 )^{20}\lambda^{20} e^{-6\lambda}\over 20!} \cdot {30!\over 6 ^{10}6^{20}\lambda^{30} e^{-6\lambda} } } \cr &=\Bigl({4.25\over 6}\Bigr)^{10} \Bigl({1.75\over 6}\Bigr)^{20} {30!\over 10! 20!}. }

-