# Compute $P(X_{(2)} ≤ 3X_{(1)})$ by using the integration technique

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

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 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: $$\mathbf {\int_{x_2\le3x_1}}24[-\frac{x_1^2}{2}+x_1x_2-x_2+\frac{1}{2}]\mathbf {d^2x}$$

• 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)? – Minus One-Twelfth Mar 7 '19 at 22:50
• corrected my typo, thank you for pointing it out – Mahamad A. Kanouté Mar 7 '19 at 22:54
• By the way, there is a formula for the joint distribution of two order statistics of the uniform distribution. See this for example. – Minus One-Twelfth Mar 7 '19 at 22:57
• 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. – 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 $$X_{1}, is simply $$n! {\bf 1}_{\{0.

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 $$2/3$$.

• 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 – Mahamad A. Kanouté Mar 8 '19 at 3:52
• Correct. Hope this helps. – Fnacool Mar 8 '19 at 13:36
• 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 ? – Mahamad A. Kanouté Mar 8 '19 at 15:39