May 6, 2009

Problem for Poisson Distribution


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.

No comments:

Post a Comment