Problem for Poisson Distribution

這個問題是我在教課時，自己發現的。或許是一個顯而易見的問題，但似乎很少人去想過，它花了我不少時間去釐清，也發覺自己的統計能力還是未到家（汗）歡迎各方不吝指教。

Suppose a Poisson process of customers arriving a store has an arrival rate of 60 people in one hour. Then, if random variable X is the number of arrival customers in one hour, then by Poisson distribution, the mean and the variance are both equal to the arrival rate.

E(X) = V(X) = 60.

Now, if the unit is changed from hour to minute, the arrival rate becomes 1 customer per minute. Suppose random variable X' is the number of arrival customer in one minute. Then,

E(X') = V(X') = 1.

But, wait a minute. Since X is the number of arrivals in one hour and X' is the number of arrivals in one minutes, then we may say that X = 60X', which gives

V(X) = V(60X') = 60^2 V(X') = 3600.

Now, this is not consistent with what we obtained previously. What makes it happened? Any thing wrong with this derivation?

Click to show my explanation.

Generally, the first two statements are correct. The expectation and the variance of a Poisson random variable are the same, no matter what unit is used. The main problem comes from the last statement: X = 60X'. The arrival rate in one hour is 60 times the rate in one minute, for sure. However, as random variables, the number of arrivals in one hour is not necessary 60 times the number in one minutes. Fundamentally, if X' has Poisson distribution, X = 60X' does not has Poisson distribution. Thus, the variance of a Poisson random variable cannot apply to X = 60X' and that's why the variance of X is not equal to the mean of X in the last equation.