Poisson Distribution: What's the probability of getting a first week without any events when you are told that 5 events occurred within a month?

Question

I'm told that the probability of getting $n$ murders per month in London can be modelled as a Poisson distribution with rate $\lambda$. I'd like to calculate the probability that, in a month with 5 murders, there were no murders within the first week. I thought this would simply equal the probability that there are no murders in a week multiplied by the probability that there are 5 murders in the (remaining) three weeks. Is this correct? If not, what should I be doing instead?

Answer 1

First, you have to convert the monthly event rate to a weekly event rate (assuming that there are 4 weeks in a month). So if the monthly event rate is $\lambda$, the weekly event rate is $\lambda/4$. Then you want a conditional probability: if the random number of events per week is $$X \sim \operatorname{Poisson}(\lambda/4),$$ then you want $$\Pr[X_1 = 0 \mid X_1 + X_2 + X_3 + X_4 = 5],$$ where $X_1, X_2, X_3, X_4$ are independent and identically distributed Poisson variables with rate $\lambda/4$.

Answer 2

Assume a four week month and divide the month into two periods the first week and the remaining three weeks. The Poisson distribution of the number of murders in a period of time is essentially a consequence of the probability model in which the events occur randomly in time. Thus if we have 5 events in the month the probability that they all occur in the second period is $(3/4)^5$. This is the same result as you will get from the conditional probability calculation using Bayes' rule.