Poisson process with different durations

Question

I am working on a homework problem:

For a given patient seen by a neurologist, the expected number of migraine headaches is two per week. Assume the migraine headaches occur randomly but at a constant rate and independently in time in such a way that the probability of a headache occurring in a given interval is (approximately) proportional to the length of the interval (i.e. the headaches follow a “Poisson process”). Compute the following probabilities by hand:

1. What is the probability for this patient that no headaches will occur in a specified one-week period?
2. What is the probability for this patient that more than three migraines will occur in a specified two-week period?
3. What is the probability, for this patient that exactly one migraine will occur on a specified day?

I can do (1).

$$P(X=0;\lambda=2)=\frac{2^{0}e^{-2}}{0!}\approx.1535$$

I am having trouble with (2) and (3) because the observation period is different than the observation period given the problem. However, since the expected value is a linear operator, it seems that the expected value should scale with the number of weeks. However, I am not sure whether the scaling is direct or inverse.

Answer 1

If $X$ is the random number of headaches experienced in a week, then $X$ is Poisson distributed with rate $\lambda = 2$ per week. This also means that the number of headaches experienced in two weeks is also Poisson, but with rate $2\lambda = 4$ per two-week period. This is one of the characteristic properties of a (homogeneous) Poisson process. In general, for a time increment $t$ where $$X(1) \sim \operatorname{Poisson}(\lambda)$$ is the random number of events such that the mean intensity of events occurring in a unit $(t = 1)$ time interval is $\lambda$, we have $$X(t) \sim \operatorname{Poisson}(\lambda t).$$

So, the second part of the question is asking for $$\Pr[X(2) > 3] = 1 - \Pr[X(2) \le 3],$$ where $X(2) \sim \operatorname{Poisson}(2\lambda = 4)$. And for the third part, since a week is $7$ days, we want to calculate $$\Pr[X(1/7) = 1],$$ where $X(1/7) \sim \operatorname{Poisson}(2/7)$.