Suppose that busses arrive at a bus stop as a Poisson process with rate $\lambda$ starting from time $t=0$ (that is, the interarrival time between busses is exponentially distributed with parameter $\lambda$).
You arrive at the bus stop at a particular (given) time $t$. Let $D_t$ be the amount of time since the last bus has departed, and $A_t$ be the amount of time until the next bus arrives.
What is the distribution of $D_t$ and $A_t$?
Show that $\Bbb E[D_t+A_t]>\dfrac 1\lambda$.
I think that we should have $A_t \sim \exp (\lambda)$, because the Poisson process has the memoryless property. So at any given time $t$, the next arrival disregards whatever happened before $t$ and arrives in $\exp(\lambda)$ time starting from $t$.
However, I feel that this is not quite right, or else the last part would be rather trivial:
$$\Bbb E[D_t + A_t] = \Bbb E[D_t] + \Bbb E[A_t] \geq \Bbb E[A_t] = \frac 1\lambda$$
As for $D_t$, I have no idea.