Dear Lazyweb

[jpmassar]

If a murder occurs in Oakland, CA approximately once every three days, and the interarrival time between murders is assumed to be Poisson distributed, then what is the probability that in some 18 hour interval over the year-to-date (or over the whole year) that there will have been at least 5 murders

Comments

[clutch_c]

The mean and variance of the Poisson are the same. So if the mean for an 18 hour period is 0.25 and the variance is 0.25, the standard deviation is sqrt(0.25)=0.5. Five murders = 0.25 + 4.75, or 0.25 + 9.5 sd. The probability of of a 9.5 sd event on the Poisson distribution is 0.00066%. There are 487 non-overlapping 18 hour periods in a year. So the probability 5 murders will NOT happen is (1-.0000066)^487 = 99.68%. So about a 0.3% chance of seeing it one or more times in a year.

This ignores the fact that there are more overlapping 18 hour periods and that 6+ murders could occur. So 0.3% might be a little low.

(Edit: my original answer assumed 18 hours was half of three days. It's actually 25% of three days, as edited.)

[jpmassar]

Thanks. As an offhand guess, I would have expected it to be higher. In terms of overlapping intervals, can we solve it as follows?

We know there are (approximately) 120 starting points for intervals that could contain five murders (i.e., the initial murder) Figure out the probability that four murders do not occur within an 18 hours interval given the Poisson parameters, exponentiate that by 120, and one-minus the result.

[clutch_c]

I don't think it's that simple because the periods aren't independent if they overlap. For example, if period 1 is midnight to Noon, period 2 is 6 am to 6 pm and period 3 is noon to midnight, and periods 1 and 3 have no events, then it isn't possible for period 2 to have any events. Maybe a professional statistician would know the answer to this. Dr. Bayes?

[jpmassar]

I just realized I could simulate it. Doh!

[markgritter]

I don't see any reason to view murders as any sort of arrival process. Wouldn't it be simpler to take the attitude that there are N murders per year, randomly distributed among the 18-hour intervals? (Or, randomly distributed among hours and then look for an 18-hour window.)

With, say, 120 murders/year we are looking at p=0.0137 that any given hour contains a murder. So the binomial distribution tells us that after any given murder, we should expect to see four or more in the next 18 "trials" with frequency 9.2 * 10^-5, or about 1 per 11,000. So maybe 1% chance per year?

But, of course, murders aren't uniformly distributed since even murderers have to sleep.

[hgfalling]

Isn't the model JP suggested just the limit of the process you describe? If you let the time step get smaller and smaller, then you get a binomial with a smaller and smaller success rate over more and more trials, which converges to the Poisson process.

[hgfalling]

I wasn't able to find a clean closed-form formula after a brief search, although if you want to go looking for stuff on M/D/inf queues you might be able to.

I assume also that you meant that the interarrival times are exponentially distributed, which makes the number of murders in a given period Poisson. (this might matter if you try to simulate it).

Anyway, in the real world murders aren't like this at all, the conditional probability of a second murder happening given one murder is happening is way, way higher than this model would predict.

[tom_bayes]

Just saw this post. I don't really have anything to add that Jerrod didn't already say. The fact that we would have overlapping intervals and that real homicides are going to "cluster" more than the Poisson would predict is problematic and is why this doesn't just reduce to a standard homework problem.

I just had a half-baked thought cross my mind that using some sort of zero-inflated Poisson model might be helpful. I'll post back if I have any further thoughts.