# Computing $\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}$ where $X_1,X_2,X_3$ are i.i.d $U(0,1)$

Suppose that $$X_1 , X_2 , X_3$$ are independent $$U (0, 1)$$-distributed random variables and let $$(X_{(1)} , X_{(2)} , X _{(3)} )$$ be the corresponding order statistic. Compute $$\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}$$.

I have the joint of the distribution which is $$3!$$

Also, I have:

$$\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}=1-\mathbb{P}\{X_{(1)}+X_{(2)} \le X_{(3)}\}$$ with $$\mathbb 0

But when I do:

$$1-6\left(\int_0^{1}\int_0^{x_3}\int_0^{x_2} \,dx_1\,dx_2\,dx_3 +\int_0^{1}\int_0^{x_3}\int_0^{x_3-x_2} \,dx_1\,dx_2\,dx_3\right)=-3$$

I do not really know what I'm doing wrong.

## 2 Answers

There could possibly be a simpler argument that does not require doing the integration explicitly.

For the integration, I felt it convenient to use a change of variables.

You are looking for

$$P(X_{(1)}+X_{(2)}

Change variables $$(x,y,z)\to(u,v,w)$$ with $$u=x+y\,,\,v=y\,,\,w=z$$

Then, $$x+y

Therefore,

\begin{align} \iiint\mathbf1_{x+y

• why does v goes from 0 to w/2 ? Mar 8, 2019 at 22:14
• I get -1 as answer Mar 8, 2019 at 23:13
• @StubbornAtom What happened to the 6? It looks like the 6 got dropped and the answer should be 6/12 not 1/12, right? Oct 14, 2021 at 5:32

There’s a symmetry among the intervals that the $$X_i$$ divide the unit interval into. You can see this by imagining that you uniformly randomly choose $$4$$ points on a circle, then uniformly randomly choose one of the $$4$$ points at which to cut the circle into an interval, with the remaining $$3$$ points yielding the $$X_i$$ uniformly randomly independently distributed on the unit interval. Since the $$4$$ resulting intervals were on an equal footing in the circle, they are still so in the interval.

Denoting these intervals by $$a$$, $$b$$, $$c$$ and $$d$$ in order, we obtain $$X_{(1)}+X_{(2)}=a+a+b$$ and $$X_{(3)}=a+b+c$$, so $$X_{(1)}+X_{(2)}\gt X_{(3)}$$ is equivalent to $$a\gt c$$. Due to the symmetry among the intervals, the probability for this is $$\frac12$$.