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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 $\mathbb 0<x_1<\min\{x_2,x_3-x_2\}, \quad 0<x_2<x_3, \quad0<x_3<1$

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.

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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

$$P(X_{(1)}+X_{(2)}<X_{(3)})=6\iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

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

Then, $$x+y<z\,,\,0<x<y<z<1\implies u<w\,,\,0<u-v<v<w<1$$

Therefore,

\begin{align} \iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z&=\int_0^1\int_0^w\int_v^{\min(2v,w)}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w \\&=\int_0^1\int_0^{w/2}\int_v^{2v}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w+\int_0^1\int_{w/2}^w\int_v^w\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w \\&=\frac{1}{12} \end{align}

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  • $\begingroup$ why does v goes from 0 to w/2 ? $\endgroup$ – Mahamad A. Kanouté Mar 8 at 22:14
  • $\begingroup$ I get -1 as answer $\endgroup$ – Mahamad A. Kanouté Mar 8 at 23:13
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    $\begingroup$ @MahamadA.Kanouté See the breaking of the integral. $\endgroup$ – StubbornAtom Mar 9 at 6:32

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