Showing independence of ratio of ordered statistics for uniform distribution

Question

From An Intermediate Course in Probability by Allan Gut

Suppose that $X\in U(0,1)$. Let $X_{(1)},X_{(2)},\ldots,X_{(n)}$ be the order variables corresponding to a sample of $n$ independent observations of X, and set \begin{equation*} \begin{split} V_{i}=\frac{X_{(i)}}{X_{(i+1)}},\quad i=1,2,\ldots , n-1, \quad \text{and} \quad V_{n}=X_{(n)}. \end{split} \end{equation*} Show that

(a) $V_{1}, V_{2},\ldots,V_{n}$ are independent.

(b) $V_{i}^{i}\in U(0,1)$ for $i=1,2,\ldots,n$.

I am trying to solve (a) by showing that $U$ and $V$ are independent, i.e. that $f_{U}(u)\cdot f_{V}(v)=f_{U,V}(u,v)$.

$U=\frac{X_{k}}{X_{k+1}},\qquad V=\frac{X_{m}}{X_{m+1}}$

\begin{equation*} \begin{split} f_{X_{k},X_{k+1},X_{m},X_{m+1}}(y_{k},y_{k+1},y_{m},y_{m+1})=\int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-1}}\cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m+1}}^{\infty}\int_{y_{m+2}}^{\infty}\cdots \int_{y_{n-1}}^{\infty}n!\prod_{i=1}^{n}f(y_{i})\\ \times \, dy_{n}\cdots dy_{m+2}\cdots \, dy_{k+2}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ =\int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-1}}\cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m+1}}^{\infty}\int_{y_{m+2}}^{\infty}\cdots \int_{y_{n-1}}^{\infty}n!\prod_{i=1}^{n-2}f(y_{i})\Big[F(y_{n})\Big]_{y_{n-1}}^{\infty}\\ \times \, dy_{n-2}\cdots dy_{m+2}\cdots \, dy_{k+2}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ = \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m+1}}^{\infty}\int_{y_{m+2}}^{\infty}\cdots \int_{y_{n-2}}^{\infty}n!\prod_{i=1}^{n-2}f(y_{i})(1-F(y_{n-1}))\\ \times \, dy_{n-2}\cdots dy_{m+2}\cdots \, dy_{k+2}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ \end{split} \end{equation*} \begin{equation*} \boxed{ \begin{split} s=1-F(y_{n-1})\qquad ds=-f(y_{n-1})\, dy_{n-1}\\ \int_{y_{n-2}}^{\infty}n!\prod_{i=1}^{n-2}f(y_{i})(1-F(y_{n-1}))\times \, dy_{n-2}\\ =\int_{1-F(y_{n+2})}^{0}n!\prod_{i=1}^{n-3}f(y_{i})(-(s))\times ds=n!\prod_{i=1}^{n-3}f(y_{i})\Big[-\frac{s^{2}}{2}\Big]_{1-F(y_{n-2})}^{0}\\ =n!\prod_{i=1}^{n-3}f(y_{i})\frac{(1-F(y_{n-2}))^{2}}{2} \end{split} } \end{equation*} \begin{equation*} \begin{split} =\frac{n!}{2!} \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m+1}}^{\infty}\int_{y_{m+2}}^{\infty}\\ \cdots \int_{y_{n-2}}^{\infty}\prod_{i=1}^{n-3}f(y_{i})(1-F(y_{n-2}))^{2} \times \, dy_{n-2}\cdots dy_{m+2}\cdots \, dy_{k+2}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ =\ldots=\\ =n! \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\\ \cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\prod_{i=1}^{m+1}f(y_{i})\frac{(1-F(y_{m+1}))^{n-m-1}}{(n-m-1)!}\times \, dy_{k+3}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ =\frac{n!}{(n-m-1)!}(1-F(y_{m+1}))^{n-m-1}f(y_{m+1})f(y_{m})\cdots \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\\ \cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m-1}}^{y_{m}} \prod_{i=1}^{m-1}f(y_{i})\times \, dy_{k+3}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ =\frac{n!}{(n-m-1)!}(1-F(y_{m+1}))^{n-m-1}f(y_{m+1})f(y_{m})\cdots \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\\ \cdots \int_{-\infty}^{y_{2}}\int_{y_{k+1}}^{y_{m}}\int_{y_{k+2}}^{y_{m}}\cdots \int_{y_{m-1}}^{y_{m}} \prod_{i=1}^{m-1}f(y_{i})\times \, dy_{k+3}\, dy_{1}\cdots dy_{m-1}\cdots \, dy_{k-1}\\ =\frac{n!}{(n-m-1)!}(1-F(y_{m+1}))^{n-m-1}f(y_{m+1})f(y_{m}) \int_{-\infty}^{y_{k}}\int_{-\infty}^{y_{k-2}}\\ \cdots \int_{-\infty}^{y_{2}}(F(y_{m})-F(y_{k+1}))^{m-1-(k+1)}\frac{1}{(n-m-2)!}f(y_{k})f(y_{k+1})\Pi_{i=1}^{k-1}\\ =\frac{n!}{(n-m-1)!}\frac{1}{(n-m-2)!}(1-F(y_{m+1}))^{n-m-1}(F(y_{m})-F(y_{k+1}))^{m-k-2}\\ \times f(y_{m+1})f(y_{m})f(y_{k})f(y_{k+1})(F(y_{k}))^{k-1}\\ f_{X_{k},X_{k+1},X_{m},X_{m+1}}(y_{k},y_{k+1},y_{m},y_{m+1})=\frac{n!}{(n-m-1)!}\frac{1}{(n-m-2)!}(1-y_{m+1})^{n-m-1}\\ \times (y_{m}-y_{k+1})^{m-k-2}(y_{k})^{k-1}\cdot 1\cdot 1\cdot 1\cdot 1\\ U=\frac{X_{k}}{X_{k+1}}\qquad X_{k}=UZ\\ V=\frac{X_{m}}{X_{m+1}}\qquad X_{k+1}=Z\\ Z=X_{k+1} \qquad X_{m}=VQ\\ Q=X_{m+1}\qquad X_{m+1}=Q\\ |\mathbf{J}|= |\begin{bmatrix} \frac{\partial y_{k}}{\partial u}&\frac{\partial y_{k}}{\partial v}&\frac{\partial y_{k}}{\partial z}&\frac{\partial y_{k}}{\partial q}\\ \frac{\partial y_{k+1}}{\partial u}&\frac{\partial y_{k+1}}{\partial v}&\frac{\partial y_{k+1}}{\partial z}&\frac{\partial y_{k+1}}{\partial q}\\ \frac{\partial y_{m}}{\partial u}&\frac{\partial y_{m}}{\partial v}&\frac{\partial y_{m}}{\partial z}&\frac{\partial y_{m}}{\partial q}\\ \frac{\partial y_{m+1}}{\partial u}&\frac{\partial y_{m+1}}{\partial v}&\frac{\partial y_{m+1}}{\partial z}&\frac{\partial y_{m+1}}{\partial q} \end{bmatrix} |=|\begin{bmatrix} z&0&u&0\\ 0&0&1&0\\ 0&q&0&v\\ 0&0&0&1 \end{bmatrix}|\\ =|z\begin{bmatrix} 0&1&0\\ q&0&v\\ 0&0&1 \end{bmatrix}-0+u \begin{bmatrix} 0&0&0\\ 0&q&v\\ 0&0&1 \end{bmatrix}-0|\\ =|z(-(q\cdot 1\cdot 1))|=zq\\ \end{split} \end{equation*} \begin{equation*} \begin{split} f_{U,V,Z,Q}(u,v,z,q)=\frac{n!}{(n-m-1)!}\frac{1}{(n-m-2)!}(1-q)^{n-m-1}\\ \times (vq-z)^{m-k-2}(uz)^{k-1}\cdot zq\\ f_{U,V,Q}(u,v,q)=\frac{n!}{(n-m-1)!}\frac{1}{(n-m-2)!}(1-q)^{n-m-1}q\\ \times \int_{0}^{1}(vq-z)^{m-k-2}(uz)^{k-1}\cdot z\, dz \end{split} \end{equation*} At this step I am stuck, I have not fund a way to evaluate this integral. I want to find $f_{U,V}(u,v)$ and then the marginal distributions $f_{U}(u)$ and $f_{V}(v)$. Can it be done? Is this the best way to show independence between $U$ and $V$? Thanks in advance!

Answer 1

Rather than computing densities, you can work with moments: because the $X_{(k)}$ and the $V_j$ all take values in $(0,1)$, their joint moments fully determine their distributions. For example, it's easy to check that for a positive integer $p$, $$ \Bbb E[V_1^p]=2\int_{0<u<v<1}u^pv^{-p}\,du\,dv=2\int_0^1\int_0^v u^pv^{-p}\,du\,dv={1\over p+1}. $$ As these are also the moments of a random variable uniformly distributed on $(0,1)$, it follows that $V_1\sim U(0,1)$.

Now check that if $p_1,p_2, p_3$ are non-negative integers then $$ \Bbb E[X_{(1)}^{p_1}X_{(2)}^{p_2}X_{(3)}^{p_3}]=6\int_{u<v<w}u^{p_1}v^{p_2}w^{p_3}\,du\,dv\,dw={6\over(p_1+1)(p_1+p_2+2)(p_1+p_2+p_3+3)}. $$ From this it follows that for non-negative $q_1$ and $q_2$, $$ \Bbb E[V_1^{q_1}V_2^{q_2}]=\Bbb E[X_{(1)}^{q_1}X_{(2)}^{q_2-q_1}X_{(3)}^{-q_2}]={2\over (q_1+1)( q_2+2) }. $$ From this it follows that $V_1$ and $V_2$ are independent (the joint moments factor!) and that $V^2_2\sim U(0,1)$ (because $V_2^2$ has the correct moments!).