Course 3 Poisson Process

[aniasat]

Hi Guys...... I have some problem with the following sum.....

Events occur according to a Poisson Process with 2 expected occurences per hour.
If you are given that two events occured in the first hour calculate the probability that the second event occured in the last half hour.
(Ans : 0.75)

My reasoning goes as follows.

The waiting time till the second event is a Gamma with alpha = 2 and theta = 0.5
So the required prob. = [ F(1) - F(0.5) ] / [Pr of two events in the first hour]

I am not getting the correct answer.

Any sugestions and solutions ???

Thanks in advance for all the help.

Cheers,
Anita.

[mopete]

You're very close! I think your problem is that you assume the the first arrival occured in the first 30 minutes, and you calculated the probability of the second arriving in the last 30 minutes given two in total. There are actually two cases:

1. Two arrivals in the last 30 minutes to yield two arrivals for the hour.
2. One arrival in the first 30 minutes and one arrival in the second 30 min interval to yield two in total for the hour. (you had this one I think)

Use a conditional probability argument as follows:

Pr{2 arrivals in last 30 min | no arrivals in first 30 min}*Pr(no arrivals in first 30 min) + Pr{1 arrival in last 30 min| one arrival in first 30 min}*Pr(1 arrival in first 30 min} / Pr{2 arriv. in 1h}

i.e. Let N(t1, t2) denote the number of arrivals in interval of length t2-t1

=( Pr{N(30,60)=2}*Pr{N(0,30)=0} + Pr{N(30,60)=1}*Pr{N(0,30)=1})/Pr{N(0,60) = 2}

=(0.5*e^-1*e^-1 + e^-1*e^-1) / e^-2*4/2

=3/4

Sorry, I'm no good with formatting. Hopefully you get my solution though.

[Higher Authority]

An easier approach is consider each of the two event times to be uniformly distributed on the 0-1 hr interval. The probability of not having any claims occur in the 2nd 30min is .5 *.5 = .25.
Hence the probability of having the 2nd claim occur in the 2nd 30 min. is 1-.25 = .75

[aniasat]

Thank you so much!!

ANita