What is the probability that the largest of 3 independent uniformly distributed random variables is greater than the sum of the other two?

Question

I wanted to solve this using order statistics, therefore, I need to compute: $$\mathbb{P}\{X_{(1)}+X_{(2)} < X_{(3)}\}$$ The probability density function of $n$ order statistics is: $$n!f(x_1)\cdots f(x_n)$$ Therefore I need to solve the integral $$\iiint_{x_1+x_2 < x_3}3!\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3$$ Since $X_1,X_2,X_3$ are independent and uniformly distributed on $(0,1)$ To calculate the bounds, I do the following: $$\{0<x_1<1\ ,\ 0<x_2<1\ ,\ 0<x_3<1\ ,\ x_1+x_2<x_3\}$$ $$\{0<x_1<\min(1,x_3-x_2) \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$ $$\{0<x_1<1, 1<x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\} \cup \{0<x_1<x_3-x_2, 1>x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$ $$\{0<x_1<1 \ ,\ 0<x_2<\min(1,x_3-1)\ ,\ 0<x_3<1\} \cup \{0<x_1<x_3-x_2 \ ,\ \max(0, x_3-1)<x_2<1\ ,\ 0<x_3<1\}$$ Which reduces to $$\{0<x_1<x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$ This implies that $$3!\int_0^1\int_0^1\int_0^{x_3-x_2}\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3=0$$ What am I doing wrong? Can I even apply order statistics to this problem or not?

Answer 1

The joint pdf for all three sufficient statistics $(X_{(1)},X_{(2)},X_{(3)})$ is $$ f(x_1,x_2,x_3) = 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]}. $$ This is one part you're missing, I think youre integrating incorrectly over regions where your pdf should give zero.

Then the desired probability is $$ \int 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]} \cdot \mathbb{1}_{[ x_1 + x_2 \le x_3]} \; dx_3 \, dx_2 \, dx_1 $$ and it may be helpful to rewrite this as $$ \int 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]} \cdot \mathbb{1}_{[0 \le x_1 + x_2 \le x_3 \le 1]} \; dx_3 \, dx_2 \, dx_1 $$ to get the integral bounds. Can you take it from here? ( I can post more if needed)

Answer 2

Which reduces to $\{0<x_1<x_3−x_2 , 0<x_2<1 , 0<x_3<1\}$

No, it reduces to $\{0<x_1<\min\{x_2,x_3-x_2\}, 0<x_2<\bbox[2pt,border:red 1pt dotted]{x_3}, 0<x_3<1\}$

PS: recalling that $\{x_1<x_2<x_3\}$

Answer 3

The joint density function of the order statistics $X_{(i)}$ and $X_{(j)}$ when $i<j$ is $$ f_{X_{(i)}, X_{(j)}}\left(x_{i}, x_{j}\right)=\frac{n !}{(n-j) !(j-i-1) !(i-1) !}\left[F\left(x_{i}\right)\right]^{i-1}\left[F\left(x_{j}\right)-F\left(x_{i}\right)\right]^{j-i-1}\left[1-F\left(x_{j}\right)\right]^{n-j} f\left(x_{i}\right) f\left(x_{j}\right) $$ for all $ 0<x_{(i)}<x_{(j)}<1 $. So, we need to calculate integration: $$ \mathbb{P}(X_1+X_2\leq X_3)=\int_0^{1/2}\int_{x_1}^{1-x_1}\int_{x_1+x_2}^1 (1-x_1-x_2)(1-x_2) dx_3dx_2dx_1 = 1/2 $$