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: $$\mathbf {\int_{x_2\le3x_1}}24[-\frac{x_1^2}{2}+x_1x_2-x_2+\frac{1}{2}]\mathbf {d^2x}$$
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 $2/3$.