Assume customers arrive according to a Poisson process with rate $\lambda = 3$.
Given three customers arrive in the first hour, find the probability that the third customer arrives in the second half of the hour.
So that is if $X_3$ is the third arrival time: $P(X_3 \in [0.5,1] \mid N(1) = 3)$
and we have $X_3 \sim$ Gamma$(3,3)$.
To solve this I have cases : $\bigg \{X_1,X_2 \in [0,.5]$ and $X_3 \in [.5,1]\bigg\}$ or
$\bigg \{X_1 \in [0,.5]$ and $X_2,X_3 \in [.5,1]\bigg\}$ or $\bigg \{X_1,X_2, X_3 \in [.5,1] \bigg \}$
However, isn't there a way to solve this using the Uniform distribution?
Best Answer
I think the key to solving this problem is to identify the following relationship: $$\{N(1)=3\}=\{X_3\leq 1,X_4>1\}$$ This way we can say $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)}{P(N(1)=3)}$$ Using the total law of probability, $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)f_{X_3}(t)dt$$ Since inter$-$arrival times are independent and $\sim \exp(3)$ we have $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)=\int_{1-t}^{\infty}3e^{-3t}dt=e^{3t-3}$$ Using the fact that $X_3 \sim \text{Erlang}(3,3)$ and $N(1)\sim \text{Poisson}(3)$ we get $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1e^{3t-3}f_{X_3}(t)dt=\frac{63}{16e^3}$$ $$P(N(1)=3)=\frac{27}{6e^3}$$ Dividing these two gives us our desired probability: $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{7}{8}$$