P.d.f of $X_{(1)}/X_{(n)}$

Question

Let $X_{i} \sim U(0, \theta) $ and $X=(X_1,\dots,X_n)$. Find the pdf of $$ \frac{X_{(1)}}{X_{(n)}}$$

I coulxnt find a way of doing it that looks convenient. Any idea?

P.s: $X_{(i)} $ are the order statistics of the vector

Answer 1

It's easy to see that it doesn't depend on $\theta$, so let's assume $\theta=1$. For $0<x<y<1$ we have: $$ \Pr(X_{(1)}>x \ \&\ X_{(n)}<y) = \Pr(\text{all $n$ random variables}\in(x,y)) = (y-x)^n. $$ And $$ y^n = \Pr(X_{(n)}<y)=\Pr(X_{(1)}>x \ \&\ X_{(n)}<y) + \Pr(X_{(1)}<x \ \&\ X_{(n)}<y). $$ Therefore $$ \Pr(X_{(1)}<x \ \&\ X_{(n)}<y) = y^n - \Pr(X_{(1)}>x \ \&\ X_{(n)}<y) = y^n - (y-x)^n. $$ This, then, is the joint c.d.f. for values of $x,y$ that satisfy $0<x<y<1$.

Use that to find the joint p.d.f. Then use that to find the conditional distribution of $X_{(1)}$ given $X_{(n)}$. Since it's given $X_{(n)}$, it follows that when you divide $X_{(1)}$ by $X_{(n)}$, you're just rescaling, and that gives you the conditional distribution of $X_{(1)}/X_{(n)}$ given $X_{(n)}$. You will find that – lo and behold – that conditional distribution does not depend on $X_{(n)}$. Therefore it's the same as the marginal (or "unconditional") distribution that you seek.

PS in response to comments: We have the c.d.f. $$ (x,y)\mapsto y^n - (y-x)^n \text{ if }0<x<y<1. $$ The values of the c.d.f. outside of the region $0<x<y<1$ need not concern us here since that is the support of the distribution. (One thing beginners sometimes fail to notice until it's pointed out is that unlike p.d.f.s, the values of joints c.d.f.s outside the support of the distribution are sometimes as complicated as within the support, or more so. But as I said, we won't need that here in this case since we're looking for the p.d.f.) Within the support the p.d.f. is $$ (x,y)\mapsto\frac{\partial^2}{\partial y\,\partial x}\Big( y^n - (y-x)^n\Big) = \frac{\partial}{\partial y} n(y-x)^{n-1} = n(n-1)(y-x)^{n-2}. $$ So we ask: What is the conditional density of $X_{(1)}$ given the event that $X_{(n)}=y$. It is just $$ x\mapsto c (y-x)^{n-2} \text{ for } 0<x<y. $$ (Here it says $x\mapsto\cdots$ rather than $(x,y)\mapsto\cdots$, indicating that it's a function of $x$ with $y$ held fixed, rather than a function of $x$ and $y$.) The normalizing constant $c$ is chosen so as to make the integral of this function equal to $1$: $$ \int_0^y c(y-x)^{n-2} \,dx = \left[\frac{-c(y-x)^{n-1}}{n-1}\right]_{x:=0}^{x:=y} = \frac{cy^{n-1}}{n-1} = 1, $$ so $c=(n-1)/y^{n-1}$. In other words $$ f_{X_{(1)}\mid X_{(n)}=y} (x) = \frac{(n-1)(y-x)^{n-2}}{y^{n-1}} \text{ for }0<x<y. \tag 1 $$ This of course depends on $y$, the value to which $X_{(n)}$ is conditioned. But now we rescale, finding the conditional density of $X_{(1)}/a$ given $X_{(n)}=y$: $$ f_{X_{(1)}/a\,\mid\, X_{(n)}=y} (x) = af_{X_{(1)}}(ax)\text{ for }0<x<\frac y a. $$ Now finally suppose $a=y$, so we're looking for the conditional distribution of $X_{(1)}/X_{(n)}$ given that $X_{(n)}=y$. We treat $X_{(n)}$ as a constant, like the "$a$" above, because it is conditioned to be equal to $y$. We get $$ f_{X_{(1)}/X_{(n)}\,\mid\, X_{(n)}=y} (x) = yf_{X_{(1)}}(yx)\text{ for }0<x<1. $$ Plug this into $(1)$ above: $$ f_{X_{(1)}/X_{(n)}\,\mid\, X_{(n)}=y} (x) = y\frac{(n-1)(y-yx)^{n-2}}{y^{n-1}}\text{ for }0<x<1. $$ This simplifies to $$ f_{X_{(1)}/X_{(n)}\,\mid\, X_{(n)}=y} (x) = (n-1)(1-x)^{n-2} \text{ for }0<x<1. $$ And you see that it does not depend on $y$.