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Question: Let $X_1,\dots,X_n$ be iid from a distribution with cdf $F(\cdot)$. Find the pdf of $\dfrac{1-F(X_{(2)})}{1-F(X_{(1)})}$.

I know that $F(X) \sim \mathcal{U}(0,1)$ for any r.v. $X$, so in this case, we're looking for the pdf of $\dfrac{1- U_{(2)}}{1-U_{(1)}}$, where $U_1,U_2,\dots,U_n$ are standard uniform.

From here, I'm sort of stuck. This is a test question, so it was meant to be reasonably quick. Some ideas:

  1. I can calculate $f_{U_{(1)}, U_{(2)}} (u,v) = n(n-1) \cdot \left[ 1 - v \right]^{n-2}\cdot \mathbb{1}[0 < u < v< 1] $. Then I suppose I could find $P \left( \frac{1-U_{(2)}}{1-U_{(1)}} \le x \right) $ by integrating and go from there.
  2. The hint was that for $U_1,\dots,U_n \stackrel{iid}{\sim} \mathcal{U}(0,1)$, we have the following fact: given $ U_{(n)} = u$, $(U_{(1)},\dots,U_{(n-1)})$ is conditionally distributed as order statistics on $[0,u]$. So I could condition on $U_{(3)}$? This is still a pain, though.

The above hint makes me think there's a typo, as it doesn't seem to help. Is there a clever way to do this problem?

user365239
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1 Answers1

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By symmetry $\frac{1-U_{(2)}}{1-U_{(1)}}$ has the same distribution as $\frac{U_{(n-1)}}{U_{(n)}}.$ Since $U_{(n-1)}$ is conditionally distributed like the maximum of $n-1$ uniforms on the interval $[0,U_{(n)}],$ the $U_{(n)}$ in the denominator just rescales the $[0,U_{(n)}]$ interval to unity. The result is that the answer is distributed like the maximum of $n-1$ independent uniforms on $[0,1]$.

spaceisdarkgreen
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