# Poisson Process Conditonal Probabilities

1. May 8, 2010

### andrew21nz

Hey

I'm really struggling with this:

What is the expected value of a poisson process (rate λ, time t) given that at least one even has occured? I was told the best way was to find the conditional distribution first.

So this is: P(Xt=z | Xt≥1)

= P(Xt=z, Xt≥1) / (PXt≥1)

= P(Xt=z) / (PXt≥1) for z=1, 2....

= (e-λt * (λt)z/z!)
-----------------
........ 1 - e-λt

That's what I keep coming back to. I think it should come out as something poisson? That way the expected value is just the parameter. Clearly I'm doing something wrong and it's really frustrating me

Sorry if it's messy, I couldn't get the Latex thing working

2. May 8, 2010

### mathman

The conditional probability of getting at least one is correct. The easiest way to do it is subtract the probability that none occurred from 1, which gives the same result you got. To complete the problem you need to get expected value under the condition. This will be the unconditional expected value divided by (1 - e-λt).

3. May 8, 2010

### andrew21nz

Oh true, because the expected value is an integral depending on z so the denominator just comes out the front.