Probability of 1 Phone Call: Solve w/ Exponential Distrib.

[Providence88]

Homework Statement

Suppose that on average, I will receive one phone call every night. Under reasonable assumptions (not given in this question prompt), show that the probability of receiving exactly one phone call, tonight, is e^-1.

Homework Equations

Possibly the gamma distribution function or exponential distribution function, but I'm not sure, as this problem arises out of material far out of the introductory distribution functions.

That is,

f(y) = \frac{1}{\Gamma(\alpha)\beta^{\alpha}} * y^{\alpha-1}e^{\frac{-y}{\beta}

is a gamma distribution for y>0. An exponential distribution, of course, is just a gamma with alpha = 1.

The Attempt at a Solution

Well, I thought that if \beta = 1, then you simply compute f(1), where y is the number of calls in a day. This, however, seems too easy, considering that this class is Mathematical Statistics 2, and we learned that sort of thing in M.S. 1.

Thanks!

 

[Dick]

I think you are supposed to derive the probability without using a given arbitrary distribution. I would guess that the reasonable assumption is that there is a pool of N people that might call you, each one with probability p. So just use the binomial distribution. Now let N go to infinity.

 

[Providence88]

I think you are supposed to derive the probability without using a given arbitrary distribution. I would guess that the reasonable assumption is that there is a pool of N people that might call you, each one with probability p. So just use the binomial distribution. Now let N go to infinity.

I guess you're talking about Poisson approximation? I had to look that one up. It's mentioned in the book, but not under that term. Very clever. I'll have to try that out.

 

[Dick]

Yes, I am. But you shouldn't need to look it up. I think they want you to derive it from the binomial distribution. That would be worthy of a Stat 2 exercise.

 

[Providence88]

Right. I'll try and derive it without looking back at the book. Thanks!