Thread: Expectation/Variation of a Poissan Process

1. Expectation/Variation of a Poissan Process

An ISP provider offers special discounts to every 3rd connecting customer. Its customers connect according to a Poissan process, with lambda = 5 customers per minute.

a) What is the probability that no offer is made during the first 2 minutes?
Since lambda is 5 per minute, i doubled it to 10 (since its for 2 minutes) then just evaluated the P(x=0) to get a very small decimal number...i think this is right

b) What is the expectation and variation of the time of the first offer?
I'm stuck here...I think the expected value of the time of the first offer is .6 minutes, since there are an expected 5 connections per minute, and one offer every 3 connections (so 3/5 = 0.6), is this right? If so, how would I find the variance?

Thank you!

2. a)

I'll give you the $\displaystyle \lambda_{2min} = 10$, but you need to reread the problem.

You do not need only p(0). You require also p(1) and p(2).

b)

Let's try $\displaystyle \lambda_{12sec} = 1$

Where does that leave us?

It may be even more prodent to define $\displaystyle \lambda_{36sec} = 3$

Just a thought.

3. Originally Posted by mistykz
An ISP provider offers special discounts to every 3rd connecting customer. Its customers connect according to a Poissan process, with lambda = 5 customers per minute.

[snip]

b) What is the expectation and variation of the time of the first offer?
I'm stuck here...I think the expected value of the time of the first offer is .6 minutes, since there are an expected 5 connections per minute, and one offer every 3 connections (so 3/5 = 0.6), is this right? If so, how would I find the variance?

Thank you!
The interval between connecting customers has an exponential distribution with a mean of 1/5 min and a variance of (1/5)^2, and the intervals are independent random variables.

You require the sum of 3 such intervals so the mean is 3(1/5), and the variance is 3(1/5)^2.

4. Alright, thanks for the help everyone