You are waiting for the bus. You model the arrival process with a Poisson process N. of intensity λ with arrival times $(T_k)_{k≥0}$, $T_0$ = 0 and inter-arrival times $(X_k)_{k≥1}$. More precisely assume that we started our clock when the first bus of the day arrived (time zero.) Then the Poisson counting process counts the buses arriving within the time intervals (0, s], s ≥ 0. For instance $N_s = 0$ means that the previous bus was the initial one arriving at time 0.
You arrive at the bus stop at a fixed time t > 0. What is the expected value of the time passed since the last bus came? (if no buses besides the initial came, then t time has passed; if the bus arrives immediately after you, then 0 time has passed)
Best Answer
Let $A_t = t - S_{N_t}$ be the age at time $t$, $F(t) = 1-e^{-\lambda t}$ the interarrival distribution, and $m(t) = \mathbb E[N_t]$ the renewal function. We know that $m(t) = \lambda t$ (see here for a proof), and for $0<x\leqslant t$ we have \begin{align} \mathbb P(A_t\leqslant x) &= F(t) - \int_0^{t-x} (1-F(t-s))\ \mathsf dm(s)\\ &=1-e^{-\lambda t} - \int_0^{t-x}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &=1-e^{-\lambda t} - (e^{-\lambda x}-e^{-\lambda t})\\ &=1 - e^{-\lambda x}. \end{align} Now, $\mathbb P(A_t\leqslant x)=1$ for $x>t$, so we have $$ \mathbb E[A_t] = \int_0^t e^{-\lambda x}\ \mathsf dx = \frac1\lambda(1-e^{-\lambda t}). $$