 Solution

For a stop-phase arrival at time $t$, the delay is $s=r-t$. An interval of delays of width $ds$ therefore corresponds to an interval of arrival phases of the same width, because $|dt/ds|=1$. Its unconditional probability is $ds/R$. Hence the positive-delay part has densityThis ordinary density integrates to $r/R$, not to one: the remaining probability mass is at $s=0$. Thus $S$ has a uniform distribution with an atom at zero, with mixing weight $q=r/R$. Conditional on a positive delay, the density would instead be $1/r$ on $(0,r)$.