# Deriving exponential distribution from arrival time probability.

Consider the following question from "Lectures on Probability Theory and Mathematical Statistics":

The probability that a new customer enters a shop during a given minute is approximately 1%, irrespective of how many customers have entered the shop during the previous minutes. Assume that the total time we need to wait before a new customer enters the shop (denote it by X) has an exponential distribution. What is the probability that no customer enters the shop during the next hour?

The book's solution proceeds by claiming that the rate parameter $$\lambda$$ of the exponential distribution is equal to 1%.

I do not understand how the book arrives at this conclusion. If I try to derive $$\lambda$$ on my own, I get a different solution:

Since the wait time between customer arrivals has an exponential distribution, the customer arrival frequency must have a Poisson distribution. Denote the customer arrival frequency by the random variable A. Then, from the task statement, we have $$P(A\geq1) = 0.01$$. Thus, $$P(A=0) = 0.99$$. This means that $$\frac{e^{-\lambda} \lambda^0}{0!} = 0.99$$. Rearranging, and simplifying, we get $$\lambda = -ln(0.99) \approx 0.01005 \neq 0.01$$. This conflicts with the book's solution, since the parameter $$\lambda$$ of a Poisson distribution is equal to the parameter $$\lambda$$ of its corresponding exponential distribution.

Which solution is correct? Why is it correct? What is the source of the discrepancy?

• You are acute to calculate the rate parameter from properties of exponential/Poisson, rather than simply take the 1% rate and set it equal $\lambda$. I guess the purpose of the problem is the latter, because the idea of a Poisson process is that the average rate over an interval is almost equal to $\lambda$ when the interval is small, and therefore you can plug it in directly. Your answer is exactly correct, while the book's is approximate. Consider calculating $\lambda$ when the problem is framed as (1/60)% chance in the next second, and observe how the discrepancy is even smaller. – Tom Chen Apr 30 at 12:20