Customers arrive at a service facility according to a Poisson process of rate $\lambda$ customers/hour. Let $X(t)$ be the number of customers that have arrived up to time $t$. Let $W_1,W_2,...$ be the successive arrival times of the customers. Determine the conditional mean $E[W_5|X(t)=3]$
Solution:
First we compute the distribution function of $W_5|X(t)=3$, using the cumulative distribution function.
Let $t>u$. Then, we want to compute $P(W_5\ge u|X(t)=3)$ which is the probability of having 5 customers from time $t$ until time $u$.
But this is equivalent to this $P(X(u)\ge 2|X(t)=3)$ probability $\color{fuchsia}{...(1)}$
So,$P(W_5\ge u|X(t)=3)=P(X(u)\ge 2|X(t)=3)=P[(X(u)-X(t))\ge 2]=1- P[(X(u)-X(t))< 2]=1- P[(X(u)-X(t))= 0]-P[(X(u)-X(t))=1]=1-e^{-\lambda (u-t)}-\lambda e^{\lambda (u-t)}$.
So the distribution function is $(u-t)\lambda ^2 e^{-(u-t)\lambda} =f_w(u)$.
Therefore $E[W_5|X(t)=3]=\int_t^{\infty}uf_w(u)du$.
I don't understand the equivalence $\color{fuchsia}{(1)}$, why $P(W_5\ge u|X(t)=3)=P(X(u)\ge 2|X(t)=3)$ ??
Can somebody explain to me this step?
Thanks.