If three people are randomly placed along a 1 mile road, the probability that no two of them are less than $m$ miles apart for $m \leq \frac{1}{2}$ could be solved by using the density for the order statistics: $f_{X_{(1)},X_{(2)},X_{(3)}}=3!$ for $0 < x_1 < x_2 < x_3 < 1$, and reasoning that:
$0 < x_1 < 1-2m$,
$x_1+m < x_2 < 1-m$
and $x_2+m < x_3 < 1$, and writing outL
$3!\int_0^{1-2m} \int_{x_1+d}^{1-m} \int_{x_2+m}^1 dx_3dx_2dx_1=(1-2m)^3$
However, I sometimes struggle with these problems so I'm trying a more general approach with indicator functions, but I'm having trouble:
First, I set my density to all of $\mathbb{R}^3$ $f_{X_{(1)},X_{(2)},X_{(3)}}=3!I_{0 < x_1 < 1}I_{0 < x_2 < 1}I_{0 < x_3 < 1}$
and then my desired probability is:
${\int \int \int}_{x_2>x_1+m,x_3>x_2+m} 3!I_{0 < x_1 < 1}I_{0 < x_2 < 1}I_{0 < x_3 < 1} dx_1dx_2dx_3$
But I'm not sure how to convert the set being integrated over into Indicator functions in a way that will let me proceed. For instance, if I try:
${\int \int \int} I_{x_2>x_1+m}I_{x_3>x_2+m}3!I_{0 < x_1 < 1}I_{0 < x_2 < 1}I_{0 < x_3 < 1} dx_1dx_2dx_3$, I wouldn't know where to go from here. Is what I'm trying to do possible...that is get the right integration limits for each integral, without having to "reason" what they should be?
Starting with
$$\displaystyle {\int \int \int} I_{x_2>x_1+m}I_{x_3>x_2+m}3!I_{0 < x_1 < 1}I_{0 < x_2 < 1}I_{0 < x_3 < 1} dx_1dx_2dx_3$$
it s a simple reordering to get
$$\displaystyle 3! \int_{x_1} I_{0 < x_1 < 1} \left(\int_{x_2} I_{0 < x_2 < 1} I_{x_2>x_1+m}\left(\int_{x_3} I_{x_3>x_2+m}I_{0 < x_3 < 1} dx_3\right) dx_2\right) dx_1$$
with the inner bracket becoming $$\int_{x_3} I_{x_3>x_2+m}I_{0 < x_3 < 1} dx_3 = (1-(x_2+m))I_{x_2+m\le 1}$$ which is similar to the $\displaystyle \int_{x_2+m}^1 dx_3$ you calculated earlier.
Just continue in this way with the other two integrals and you will be repeating what you did previously.