Suppose that $X_1,X_2,X_3.X_4$ are independent $U\in(0,1)$-distributed random variables and let $(X_{(1)}X_{(2)}X_{(3)}X_{(4)})$ be the corresponding order statistic.

Compute $P(X_{(2)} ≤ 3X_{(1)})$

Things I know:

Thought that writing $P(X_2-3X_1\le0)$ And then use transformations to get a joint of $U=X_2-3X_1$ and a dummy $V=X_2$.

  • With $f_{UV}(u,v)=f_{X_{(1)}X{(2)}}(\frac{v-u}{3},v)(-3)$
  • Also, $$f_{X_{(1)}X_{(2)}X_{(3)}X_{(4)}}(x_1,x_2,x_3,x_4)=24$$ with $0<x_1<x_2<x_3<x_4<1$

Things I'm not sure about

I don't really know how to get the bounds of integration in order to get the joint of just $X_{(1)}$ and $X_{(2)}$

I will try $x_2<x_3<x_4 ,\quad x_1<x_4<1$ as the bounds of integration.

Does anyone have some tip on how to take the bounds without making errors?

I don't really know how these bounds work so any explanation would be appreciated.

Update: I managed to get $f_{X_{(1)}X_{(2)}}(x_1,x_2)=24\int_{x_{1}}^1\int_{x_2}^{x_{4}}d_{x_3}d_{x_4}=24[-\frac{x_1^2}{2}+x_1x_2-x_2+\frac{1}{2}]$

Still not getting the answer after: $$\mathbf {\int_{x_2\le3x_1}}24[-\frac{x_1^2}{2}+x_1x_2-x_2+\frac{1}{2}]\mathbf {d^2x}$$

  • 1
    $\begingroup$ Do you want $P\left(X_{\color{red}{2}} \le 3X_{\color{red}{1}}\right)$ or $P\left(X_{\color{blue}{(2)}} \le 3X_{\color{blue}{(1)}}\right)$ (parentheses in the subscripts for order statistics)? $\endgroup$ – Minus One-Twelfth Mar 7 '19 at 22:50
  • $\begingroup$ corrected my typo, thank you for pointing it out $\endgroup$ – Mahamad A. Kanouté Mar 7 '19 at 22:54
  • 2
    $\begingroup$ By the way, there is a formula for the joint distribution of two order statistics of the uniform distribution. See this for example. $\endgroup$ – Minus One-Twelfth Mar 7 '19 at 22:57
  • $\begingroup$ Just peeked at the formulas, but my main concern is to be able to derive the stuff from scratch because I always make errors on the bounds. $\endgroup$ – Mahamad A. Kanouté Mar 7 '19 at 23:08

I'll replace the constant $3$ with a constant $a>1$ and will solve for a general number of RVs.

Some introduction.

Observe that the joint distribution of $(X_1,X_2,\dots,X_n)$ is $1$.

From this, the joint distribution conditioned on $X_{1}<X_2<X_3\dots < X_n$, is simply $n! {\bf 1}_{\{0<x_1<x_2<x_3\dots<x_n<1\}}$.

But by permutation of indices it follows that the above density is also the joint density of $(X_{(1)},\dots,X_{(n)})$.

With this density we conclude that the distribution of $X_{(1)}$ conditioned on $X_{(2)}$ is uniform on $[0,X_{(2)}]$.

As a result, $$ P(X_{(1)} > \frac{X_{(2)}}{a}) = E[ \frac{X_{(2)}-X_{(2)}/a}{X_{(2)}}]=1-1/a.$$

The answer does not depend on $n$. In the specific case the answer is then $2/3$.

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  • $\begingroup$ So you're solving the problem for a general case then? It might look like I can use that formula for other cases whenever I get a uniform dist $\endgroup$ – Mahamad A. Kanouté Mar 8 '19 at 3:52
  • $\begingroup$ Correct. Hope this helps. $\endgroup$ – Fnacool Mar 8 '19 at 13:36
  • $\begingroup$ Do you know I may apply this to a more general case like for example: if I have the second and the fourth random variable instead of the first and second ? $\endgroup$ – Mahamad A. Kanouté Mar 8 '19 at 15:39

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