The UVA astronomy club is watching a meteor shower. Meteors appear at an average rate of per hour.
(a) Write a short explanation to justify the use of a Poisson distribution to model the appearance of meteors. Why should appearances be Poisson distributed?
(b) What is the probability that the club sees more than 2 meteors in a single hour?
(c) Suppose we find out that over a four hour evening, 13 meteors were spotted. What is the probability that none of them happened in the first hour, conditioned on that information?
Poisson processes model events that occur randomly when you know the mean number of events within any given interval and all disjoint intervals (of any size) are independent.
Since meteors arrive independently of each other, a Poisson process is reasonable.
(b)
Compute probability:
We have .
(c)
Let Window A be the first hour, Window B be the second through fourth hours, and Window C be the complete four hours.
The background rate is , so we have , , and . Define three Poisson variables accordingly: , , .
Now the desired probability is:
The last line follows because . By independence of Poisson process windows:
Suppose that vehicle lifetimes follow an exponential distribution with an expected lifetime of 10 years.
Suppose you have one car that is 5 years old, and one that is 15 years old, at the present moment.
What is the probability that the first car outlives the second? (I.e. that the second breaks at an earlier time than the first breaks, both starting now.)
By the memoryless property of exponential distributions, elapsed time has no effect on future events. The fact that one car is older than the other has no effect on the remaining lifetimes.
Since both cars have the same remaining lifetime distribution, the probability that either car outlives the other is 0.5.
Consider the Poisson process of phone calls coming to a call center at an average rate of 1 call every 6 minutes.
Let us model the wait time for 5 calls to come in. You may use Desmos or similar to perform the integration numerically.
(a) Method One: An arrival of ‘1-call’ comes in at an average rate of calls per hour. So a Bundle of ‘5-calls’ comes in at an average rate of Bundles per hour. Use an exponential variable with to determine the probability that the wait time for a Bundle (of 5 calls) is at most .
(b) Method Two: Use calls per hour with an Erlang distribution at to determine the probability that the wait time for 5 calls is at most .
(c) Compare the results of (a) and (b). Can you explain why they agree or disagree? Which is correct??
Compute probability the wait time for a Bundle is at most 1 hr:
(b)
Erlang distribution:
Desired probability:
This integral requires iterated IBP.
(c)
The results disagree. Method 2 is the correct approach assuming individual calls arrive according to a Poisson process.
It turns out that if you have some Poisson process, bundles of arrivals do not themselves follow a Poisson process.
Think of a Poisson process as some dots scattered on a timeline. Place a heavy dot at the location of each dot in order, and erase all other dots. These heavy dots do not follow a Poisson process.