You are waiting for a bus at a bus station. The buses arrive at the station according to a Poisson process with an average arrival time of 10 minutes ($\lambda$ = 0.1/min). If the buses have been running for a long time and you arrive at the bus station at a random time, what is your expected waiting time? On average, how many minutes ago did the last bus leave?
In the solution to this problem, it states: "Taking advantage of the memoryless property of exponential distribution, we know that the expected waiting time is 10min. If you look back in time, the memoryless property still applies. So on average, the last bus arrived 10 minutes ago as well."
Could someone explain the derivation for memoryless looking "backwards in time"? I don't understand that part. Thanks!
This follows from some elementary results in renewal theory.
Let $\{T_n\}$ be a renewal process, i.e. $T_n=\sum_{i=1}^n X_i$ where $\{X_n\}$ is a sequence of i.i.d. random variables with $\mathbb P(X_1\geqslant 0)=1$ and $\mathbb P(X_1>0)>0$. Let $F(t):=\mathbb P(X_1\leqslant t)$ be the distribution function of $X_1$. Let $N_t := \sum_{n=1}^\infty \mathsf 1_{\{T_n\leqslant t\}}$ be the number of renewals in the interval $[0,t]$ and $M(t)=\mathbb E[N_t]$ the renewal function. For a given time $t\in[0,\infty)$ let $A_t:=t-T_{N_t}$ be the age and $R_t:=T_{N_t+1}-t$ the remaining life.
Arriving at a random time $t$, the waiting time for the next bus is $R_t$ and the time since the last bus left is $A_t$. To find the distribution of $R_t$, let $y\in[0,\infty)$ and condition on the time of the first renewal:
\begin{align} \mathbb P(R_t>y) &= \int_0^\infty \mathbb P(R_t>y\mid T_1=s)\ \mathsf d F(s)\\ &= \int_0^t \mathbb P(R_{t-s}>y)\ \mathsf d F(s) + \int_t^{t+y}0\ \mathsf dF(s) + \int_{t+y}^\infty \mathsf 1\ \mathsf dF(s)\\ &= 1-F(t+y) + \int_0^t (1-F(t+y-s))\ \mathsf dM(s), \end{align} where the last equality follows from having shown that $r_y(t) := \mathbb P(R_t>y)$ satisfies the renewal equation $r_y=1-F(y) + r_y\star F$, which has unique solution $1-F(y)+((1-F(y))\star M$.
From the above and the fact that $\mathbb P(A_t \geqslant y)=\mathbb P(R_{t-y}>y)$ for $y\in[0,t]$ we have $$ \mathbb P(A_t\geqslant y) = 1-F(t) + \int_0^{t-y} (1-F(t-s))\ \mathsf dM(s),\ y\in[0,t]. $$ For a Poisson process we have $F(t) = (1-e^{-\lambda t})\mathsf 1_{[0,\infty)}(t)$ and $M(t) = \lambda t$, so $$ P(R_t>y) = e^{-\lambda(t+y)} + \int_0^t \lambda e^{-\lambda(t+y-s)}\ \mathsf ds = e^{-\lambda y} $$ and $$ \mathbb P(A_t\geqslant y) = e^{-\lambda t}+\int_0^{t-y} \lambda e^{-\lambda(t-s)}\ \mathsf ds = e^{-\lambda y}. $$ From this it follows that the distribution of the remaining life and age at a random time $t$ is exponentially distributed with mean $\frac1\lambda$, as was to be shown.