Poisson Process + Erlang PDF Question

A claims process has one claim every ten hours. This is modeled as a Poisson process with $\displaystyle \lambda = 0.1$

What is the probability that the 3rd claim will be processed in the 5th hour?

How can I get the correct solution using the Erlang PDF function? I tried:

$\displaystyle \int_4^5{\frac{0.1^3 \cdot y^{2} \cdot e^{-0.1y}}{2!} \, dy} \approx 0.0065$

My simpler solution using the Poisson probability mass function is this: The probability of the third claim being processed during the fifth hour is the probability of exactly two claims in four hours combined with the independent probability of at least one in the next hour:

$\displaystyle p_X(2; 0.1 \cdot 4) \cdot (1 - p_X(0; 0.1 \cdot 1))$

$\displaystyle = \frac{0.4^2 \cdot e^{-0.4}}{2!} \cdot \left(1 - \frac{0.1^0 \cdot e^{-0.1}}{0!}\right) \approx 0.0051$

Are both of these approaches correct? Why do these two approaches not match up?

Re: Poisson Process + Erlang PDF Question

the approaches are not equivalent because it is not certain that claims 1&2 will both occur in the first four hours. you could have no claims in the first 4 hours and then 3 claims in hour 5.

The poisson approach is therefore incomplete and gives a lower answer than your erlang method.

Re: Poisson Process + Erlang PDF Question

Perfect! Thanks!

I was just considering the scenario with 2 in the first 4 hours and 1+ in hour 5. When I also consider 0/3+ and 1/2+ cases and sum all the probabilities, the final probability value matches the Erlang integration approach.