# Poisson Process Question relating to time between arrivals

The number of insurance claims arriving electronically in a small insurance office has been modelled as a Poisson Process with rate $5$ per hour.

(I) What is the distribution of the arrival time between claims?

(II) What is the distribution of the time it takes for $10$ claims to arrive?

(III) Given that the last claim arrived $15$ seconds ago, what is the mean time from now until the next claim arrives?

(IV) What is the distribution of the number of claims which arrive in an eight hour day?

I know the answer to (I) is that the distribution time between claims is I.I.D. exponential with mean 12 because there are 5 arriving per hour, so one every 12 minutes. I also know that the answer to (II) is that the distribution of the time taken for 10 claims to arrive is gamma with parameter $\alpha = 10$ and $\beta = 12$.

I do not, however, know what the answers to part (III) ad (IV) are. My lecturer mentioned something about the "$15$ seconds ago" part being irrelevant because the distribution is memoryless.

Any help is greatly appreciated.

## 1 Answer

(III) The exponential distribution has the property that

$$P(X>x+u\mid X>u)=P(X>x); \ \ x,u \ge 0.$$ This property is called the memoryless nature of the exponential distribution.

Proof

$$P(X>x+u \mid X>u)=\frac{P(X>x+u)}{P(X>u)}=\frac{e^{-\lambda(x+u)}}{e^{-\lambda u}}=e^{-\lambda x}.$$

So, given that $X>15$ sec, the probability that $X>x+15$ sec is still $e^{-\lambda x}$. That is, the waiting time after $15$ seconds is exponential with parameter $\lambda$ as if nothing had happened. The exponential distribution forgets its past. The expectation is then $\lambda=12$ min even after any waiting time.

(IV) The Poisson process remains Poisson with parameter $5\times 8=40$.