I'm stuck computing the expected value of the last hitting time before a time $s$ in the waiting time paradoxon. Suppose we come to a bus stop at a time $s \in \mathbb{R}$, where buses are randomly stopping according to a Poisson process with intensity $\gamma$. With $\tau_s$ we denote the time, the last bus before $s$ has left. The distribution is
$\mathbb{P}_{\tau_s} = \mathbb{P}_{s-X}$ where $X\sim Expo(\gamma)$. I want to compute the expected value of $\tau_s$, which should be $s-\frac{1}{\gamma}(1-e^{-\gamma s})$, as in @Did's answer in this post: Expected time of last bus left
Since my attempts don't output this well-known formula, there must be some mistake, which I fail to see:
$\mathbb{E}[\tau_s] = \int_\Omega \tau_s(\omega)dP(\omega) = \int_\Omega s-X(\omega)dP(\omega) = s-\int_\Omega 1_{(X(\omega)\leq s)}X(\omega)dP(\omega) = s- \mathbb{E}[X, X\leq s] = s-\int_0^s x\cdot f_X(x) dx = s-\int_0^s x\gamma e^{-\gamma x} = s-\frac{1}{\gamma}(1-e^{-\gamma s})+se^{-\gamma s}$
Please, where is my mistake?
With some help from the comments, I was able to solve the question:
Since $\tau_s$ cannot be negative-valued, its distribution is really $\tau_s \sim (s-\min\{X, s\})$ with $x\sim Expo(\gamma)$. Thus
$\mathbb{E}[\tau_s] = \mathbb{E}[s-\min\{X, s\}] = \mathbb{E}[s-X, X< s]$ since if $X\geq s$ then $s-\min\{X, s\} = s- s = 0$.
$\mathbb{E}[s-X, X< s] = \mathbb{P}[X<s]\cdot s - \mathbb{E}[X, X< s] = (1-e^{-\gamma s})s - \int_0^s x\gamma e^{-\gamma x}dx$
hence
$\mathbb{E}[s-X, X< s] = (1-e^{-\gamma s})s-\frac{1}{\gamma}(1-e^{-\gamma s})+se^{-\gamma s} = s- \frac{1}{\gamma}(1-e^{-\gamma s})$
as desired.