# Probability Theory: Order statistics and triple integrals

## Homework Statement

Let $U_1, U_2, U_3$ be independent uniform on $[0,1]$.

a) Find the joint density function of $U_{(1)}, U_{(2)}, U_{(3)}$.

b) The locations of three gas stations are independently and randomly placed along a mile of highway. What is the probability that no two gas stations are less than 1/3 mile apart?

## The Attempt at a Solution

Let $U_{(1)}, U_{(2)}, U_{(3)}$ take on values $x,y,z$ respectively

a) The answer was 6, with $0<x<y<z<1$

b) I defined the limits as such. $0<x<1/3$, since this is the maximum the first station can be placed before the other 2 get too close to each other. Then $x+1/3 < y <z - 1/3$, and finally $2/3<z<1$, so $y$'s possible positions are constrained by the positions of its neighbours.

However, integrating the joint density function over these limits returned a 0, which was clearly wrong. I know the right limits to integrate over were $0<x<y-1/3$ then $1/3 < y < z - 1/3$, and $2/3<z<1$. What was wrong with my reasoning?

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StoneTemplePython
Gold Member
2019 Award

## Homework Statement

Let $U_1, U_2, U_3$ be independent uniform on $[0,1]$.

a) Find the joint density function of $U_{(1)}, U_{(2)}, U_{(3)}$....
Let $U_{(1)}, U_{(2)}, U_{(3)}$ take on values $x,y,z$ respectively
....
a) The answer was 6, with $0<x<y<z<1$
To be honest I have no idea what this means. Based on the wording of the problem I would have thought you wanted something related to the Beta distribution here.

b) The locations of three gas stations are independently and randomly placed along a mile of highway. What is the probability that no two gas stations are less than 1/3 mile apart?....

b) I defined the limits as such. $0<x<1/3$, since this is the maximum the first station can be placed before the other 2 get too close to each other. Then $x+1/3 < y <z - 1/3$, and finally $2/3<z<1$, so $y$'s possible positions are constrained by the positions of its neighbours.

However, integrating the joint density function over these limits returned a 0, which was clearly wrong. I know the right limits to integrate over were $0<x<y-1/3$ then $1/3 < y < z - 1/3$, and $2/3<z<1$. What was wrong with my reasoning?
This too feels like there's no actual work being shown which makes things somewhat difficult.

Forgetting order statistics for the moment: at a bare minimum you should be able to consider the case of some arbitrary ordering, say the case of $U_1$ then $U_2$ then $U_3$ arriving and set up an integral like

$\int_0^{\frac{1}{3}}\int_{u_1+\frac{1}{3}}^{\frac{2}{3}} \int_{u_2+ \frac{1}{3}}^{1} 1\cdot du_3\cdot du_2\cdot du_1$

and then consider that the ordering is arbitrary, so a natural adjustment is needed...

@StoneTemplePython

Sincerest apologies, allow me to write out everything.

a) Let $U_{(1)}, U_{(2)}, U_{(3)}$ take on values $x,y,z$ respectively with $0<x<y<z<1$. By a differential argument,

$$f_{U_{(1)}, U_{(2)}, U_{(3)}} dxdydz = Pr(x<U_{(1)} < x + dx, \ y<U_{(2)} < y + dy, \ z < U_{(3)} < z + dz )$$
which equals to
$$3! \ P(x<U < x + dx, \ y<U< y + dy, \ z < U< z + dz ) = 3! f(x)f(y)f(z) \ dxdydz$$
where f(x) is the uniform density on $[0,1]$. Finally,
$$f_{U_{(1)}, U_{(2)}, U_{(3)}} = 3! f(x)f(y)f(z) = 6$$

b) To find the probability that no two stations are less than 1/3 miles from each other, I will have to integrate the density over the relevant limits for $x,y,z$. The minimum of $x$ has to be zero, and the maximum I felt, was 1/3 since any further and we would not be able to fit the other stations in. Thus $0<x<1/3$. For $y$, it had to be at least 1/3 miles from both $x$ and $z$, so I defined the limits to be $x+1/3 < y < z - 1/3$. Finally, the minimum of $z$ had to be 2/3, any shorter and the first two stations would not fit. Thus $2/3 < z < 1$. The integral was then

$$\int_{2/3}^{1} \int_{0}^{1/3} \int_{x+1/3}^{z - 1/3} f_{U_{(1)}, U_{(2)}, U_{(3)}} dydxdz = \int_{2/3}^{1} \int_{0}^{1/3} \int_{x+1/3}^{z - 1/3} 6 \ dydxdz$$

This returned 0 which was wrong. The limits I should have integrated across were $0<x<y−1/3$, then $1/3<y<z−1/3$ and finally $2/3<z<1$. May I know why the limits I initially integrated over were incorrect?

Many thanks.

• StoneTemplePython
StoneTemplePython
Gold Member
2019 Award
ok, let's focus on (b) for the moment, and then perhaps in words discuss (a) if you want to -- I understand your result for (a) now as I was originally thinking of marginals not joint distributions last night.

for (b)
I'd start by peeling back the integrals one step at a time, and in particular consider some point-wise results (i.e. consider some specific sample paths and how they relate to the integral vs the problem statement).

Suppose for instance
$U_{(1)}(\omega) = 0.2$
and
$U_{(3)}(\omega) = 0.7$

To make this crisp, let's ignore zero probability events of ties, so we can say

$U_{(1)}\lt U_{(2)} \lt U_{(3)}$

Now you and I can reason through this and see that it is impossible for this to be a 'success' for part b, because a 'success' requires at a minimum a gap of $\frac{2}{3}$ between $U_{(3)}(\omega)$ and $U_{(1)}(\omega)$. Yet the gap is clearly $0.7 - 0.2 = \frac{1}{2}\lt \frac{2}{3}$.

So re-visiting your integral, let's pull out the 6 and re-state it as

$$6\cdot \int_{2/3}^{1} \int_{0}^{1/3} \int_{x+1/3}^{z - 1/3} 1 \ dydxdz = 6\cdot \int_{2/3}^{1} \int_{0}^{1/3} \big(\int_{x+1/3}^{z - 1/3} 1 \ dy\big)dxdz$$

it should be clear that the 'responsibility' of assigning zero probability density of success comes from inner most integral of $\big(\int_{x+1/3}^{z - 1/3} 1 \ dy\big)$

yet

$$\big(\int_{x+1/3}^{z - 1/3} 1 \ dy\big) = (z -x) -\frac{2}{3} = (0.7 - 0.2)- \frac{2}{3} \lt 0$$
which immediately tells you that something is wrong with your limits of integration since any probability or probability density should be real non-negative everywhere in the integral. (Of course if you had something more exotic akin to inclusion-exclusion, this statement would perhaps need to be tweaked, but thankfully this is plain vanilla integration over a probability density.)

- - - - - -
I'm not sure I have a cure-all for limits of integration issues. It is easy to mess them up. Note that the way you set this up has two integrals on the outside

$$\int_{2/3}^{1} \int_{0}^{1/3}$$

and each of them have limits of integration that are standalone. Then you have one inner integral whose limits of integration depend on both variables-- and the interaction between those limits is where the culprit was buried. Tackling two dependencies at a time is a lot trickier than tackling one.

With this in mind, my general approach is to chain on the integrals with one dependency at a time so that I can verify the result (via something like an informal induction) one integral, and one dependency, at a time.

e.g. what I had written in my first response is
$$P_{\text{success}}(u_1, u_2, u_3) = \int_0^{\frac{1}{3}}\int_{u_1+\frac{1}{3}}^{\frac{2}{3}} \int_{u_2+ \frac{1}{3}}^{1} 1\cdot du_3\cdot du_2\cdot du_1$$

and $P_{\text{success}}$ is a symmetric function of iid $u_i$'s, hence there are $3! = 6$ possible orderings -- i.e. the natural adjustment I alluded to. Each ordering of success constitutes an event and the probability of the union of those events is what (b) is looking for. Since the events are mutually exclusive their probabilities add (equality case of Boole's Inequality) and so

$$6\cdot \int_0^{\frac{1}{3}}\int_{u_1+\frac{1}{3}}^{\frac{2}{3}} \int_{u_2+ \frac{1}{3}}^{1} 1\cdot du_3\cdot du_2\cdot du_1$$

does give you the answer. My suggested approach here does deprecate order statistics somewhat. But as is common in my posts, it does emphasize the role of events-- which can really help focus your thinking.

Last edited:
• WWCY
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Let $U_1, U_2, U_3$ be independent uniform on $[0,1]$.

a) Find the joint density function of $U_{(1)}, U_{(2)}, U_{(3)}$.

b) The locations of three gas stations are independently and randomly placed along a mile of highway. What is the probability that no two gas stations are less than 1/3 mile apart?

## The Attempt at a Solution

Let $U_{(1)}, U_{(2)}, U_{(3)}$ take on values $x,y,z$ respectively

a) The answer was 6, with $0<x<y<z<1$

b) I defined the limits as such. $0<x<1/3$, since this is the maximum the first station can be placed before the other 2 get too close to each other. Then $x+1/3 < y <z - 1/3$, and finally $2/3<z<1$, so $y$'s possible positions are constrained by the positions of its neighbours.

However, integrating the joint density function over these limits returned a 0, which was clearly wrong. I know the right limits to integrate over were $0<x<y-1/3$ then $1/3 < y < z - 1/3$, and $2/3<z<1$. What was wrong with my reasoning?

The central station's position $u_2$ must reserve a symmetric band of width 2/3 (1/3 on each side) within which no other station can fall. Thus, we need $1/3 \leq u_2 \leq 2/3,$ and for any such $u_2$ the first station's position $u_1$ must lie in $[0,u_2-1/3]$, while the third station's position must lie in $[u_2+1/3,1].$
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