Intuition for minimising expected time in Poisson Process

Question

Trains arrive at a station according to some Poisson process with rate $\lambda$.
If we take the train from that station then it takes a time $r$ to get home, measured from the time at which you enter the train to arrive home. If you walk from the train station to home then it takes a time $w$ to arrive home. Suppose we have a policy that when arriving at the train station, we wait a time $s$ and if the train has not arrived, then we walk home.
What is the intuition where we need only consider the cases $s=0$ and $s=\infty$ when minimising the expected wait time? I know that we can take cases $w>\frac{1}{\lambda} - r$ and $w<\frac{1}{\lambda} - r$
I computed the expected wait time $\mathbb{E}(T)$ where $T$ is the total time to be $$ \mathbb{E}(T) = r + \frac{1}{\lambda} + e^{-\lambda s}(w-\frac{1}{\lambda}-r). $$

Answer 1

Let $S_1$ be the arrival time of the first train, then $S_1\sim\mathsf{Exp}(\lambda)$. Assuming $r<w$, we have $$ T = (S_1+r)\cdot\mathsf1_{\{S_1<s\}} +(s+w)\cdot\mathsf 1_{\{S_1\geqslant s\}}. $$ For $0<t<s+r$ we have $$ \mathbb P(T\leqslant t) = \mathbb P(S_1+r\leqslant t) = \mathbb P(S_1\leqslant t-r) = 1-e^{-\lambda(t-r)}, $$ and $$ \mathbb P(T=s+w) = \mathbb P(S_1>s) = e^{-\lambda s}. $$ It follows that the distribution of $T$ is $$ F_T(t) = \left(1 - e^{-\lambda(t-r)}\right)\cdot\mathsf 1_{[0,s+r)}(t) + \left(1- e^{-\lambda s}\right)\cdot \mathsf 1_{[s+r),s+w)}(t) + \mathsf 1_{[s+w,\infty)}(t), $$ and hence \begin{align} E[T] &= \int_0^\infty (1-F_T(t))\ \mathsf dt\\ &= \int_0^{s+r} e^{-\lambda(t-r)}\ \mathsf dt + \int_{s+r}^{s+w}e^{-\lambda s}\ \mathsf dt + \\ &= \frac1\lambda(e^{r\lambda} - e^{-s\lambda}) + (w-r) e^{-s\lambda}. \end{align}