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.

**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.

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