# How to compute $\mathbb P( 3 X_{(1)} \geq X_{(2)}+X_{(3)})$ for order statistics of a uniform distribution?

I am trying to solve a problem for my thesis and I don't see how to do it. I have 4 observations randomly taken from a uniform $(0,1)$ distribution. I want to compute the probability that $3 X_{(1)}\ge X_{(2)}+X_{(3)}$. $X_{(i)}$ is the ith order statistic (I take the order statistic so that my observations are ranked from smallest to biggest). I have solved it for a simpler case but here I am lost to how to do it.

All help would be welcome.

Write the order statistics as $(x_1,x_2,x_3,x_4)$, $0 \le x_1 \le x_2 \le x_3 \le x_4 \le 1$. Begin by noting that $x_1 \le x_2$ implies

$$\Pr[3x_1 \ge x_2+x_3] = 1 - \Pr[3x_1 \lt x_2+x_3] = 1 - \Pr[x_1 \le \min(x_2, \frac{x_2+x_3}{3})].$$

This last event breaks into two disjoint events depending on which of $x_2$ and $(x_2+x_3)/2$ is the larger:

\eqalign{ \Pr[x_1 \le \min(x_2, \frac{x_2+x_3}{3})] &= \Pr[x_2 \le \frac{x_3}{2},\quad x_1 \le x_2] \\ &+ \Pr[\frac{x_3}{2} \le x_2 \le x_3,\quad x_1 \le \frac{x_2+x_3}{3}]. }

Because the joint distribution is uniform on the set $0 \le x_1 \le x_2 \le x_3 \le x_4 \le 1$, with density $4! dx_4 dx_3 dx_2 dx_1$,

$$\Pr[x_2 \le \frac{x_3}{2},\quad x_1 \le x_2] = 4! \int_0^1 dx_4 \int_0^{x_4} dx_3 \int_0^{x_3/2} dx_2 \int_0^{x_2} dx_1 = \frac{1}{4}$$

and

$$\Pr[\frac{x_3}{2} \le x_2 \le x_3,\quad x_1 \le \frac{x_2+x_3}{3}]=4! \int_0^1 dx_4 \int_0^{x_4} dx_3 \int_{x_3/2}^{x_3} dx_2 \int_0^{(x_2+x_3)/2} dx_1 = \frac{7}{12}.$$

(Each integral is straightforward to perform as an iterated integral; only polynomial integrations are involved.)

The desired probability therefore equals $1 - (1/4 + 7/12)$ = $1/6$.

### Edit

A cleverer solution (which simplifies the work) derives from the recognition that when $y_j$ have iid Exponential distributions, $1\le j\le n+1$, then (writing $y_1+y_2+\cdots+y_{n+1} = Y\$), the scaled partial sums

$$x_i = \sum_{j=1}^{i}y_j/Y,$$

$1\le i\le n$, are distributed like the uniform order statistics. Because $Y$ is almost surely positive, it follows easily that for any $n\ge 3$,

\eqalign{ \Pr[3x_1\ge x_2+x_3] &= \Pr[\frac{3y_1}{Y}\ge\frac{y_1+y_2}{Y}+\frac{y_1+y_2+y_3}{Y}]\\ &=\Pr[3y_1\ge(y_1+y_2)+(y_1+y_2+y_3)]\\ &= \Pr[y_1\ge 2y_2+y_3]\\ &= \int_0^\infty \exp(-y_3)\int_0^\infty \exp(-y_2) \int_{2y_2+y_3}^\infty \exp(-y_1) dy_1 dy_2 dy_3\\ &= \int_0^\infty \exp(-y_3)\int_0^\infty \exp(-y_2)\left[\exp(-2y_2-y_3)\right]dy_2 dy_3 \\ &= \int_0^\infty \exp(-2y_3)dy_3 \int_0^\infty \exp(-3y_2)dy_2 \\ &= \frac{1}{2} \frac{1}{3} = \frac{1}{6}. }

• Thanks so much for your help!I was block in my research because of this problem, so again thank you! – sev Oct 30 '11 at 18:29
• +1 The viewpoint added in the recent edit is especially appreciated – Dilip Sarwate Oct 31 '11 at 15:49