# Order statistic and independence

"If $Y_1, Y_2, ..., Y_n$ are independent, uniformly distributed random variables on the interval $[0, \theta]$, show that $U=Y_{(1)}/Y_{(n)}$ and $Y_{(n)}$ are independent."

I have already found $$f_{Y_{(1)}}(y) = \frac{n}{\theta}(1-\frac{1}{\theta}y)^{n-1}$$ and $$f_{Y_{(n)}}(y) = \frac{n}{\theta^n}y^{n-1}$$

I was thinking of showing independence by showing $f(u, y_{(n)}) = f_U(u)f_{Y_{(n)}}(y)$, but I'm not sure if this even makes sense?

Thanks for any help.

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What is the notation $Y_{(n)}$? –  Tunococ Nov 28 '12 at 5:05
Oh sorry, it's the $n$th order statistic of $Y$. –  quantum Nov 28 '12 at 5:21
It's perfectly fine to show the density factors as you have written, the left hand side being the joint law. As a possible hint, try conditioning on $Y_(n)$ –  Alex R. Nov 28 '12 at 5:25
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## 1 Answer

Approach 1: I would start with the joint distribution of $Y_{(1)}$ and $Y_{(n)}$. (Since you know how to find the individual order statistic distributions you can use a similar argument to get the joint distribution.) This is $$f_{Y_{(1)},Y_{(n)}}(y_1, y_n) = \frac{n(n-1)}{\theta^n}(y_n-y_1)^{n-2}, \:\:\:\: 0 < y_1 < y_n < \theta.$$

Then do a bivariate transformation to obtain $f_{U,Y_{(n)}}(u,y_n)$. The Jacobian of the transformation is just $Y_{(n)}$, and so you get

$$f_{U,Y_{(n)}}(u,y_n) = \frac{n(n-1)}{\theta^n}(y_n - uy_n)^{n-2} y_n, \:\:\: 0 < u < 1, \: 0 < y_n < \theta.$$ Since $f_{U,Y_{(n)}}(u,y_n)$ factors into a function of $u$ and a function of $y_n$, $U$ and $Y_{(n)}$ must be independent.

You can fill in the details, but this is the basic argument for this approach.

Approach 2: Because I can't stop myself, let me also give the argument described by did. :)

Obtain the conditional distribution $f_{Y_{(1)}|Y_{(n)}}(y_1|y_n)$ by dividing $f_{Y_{(1)},Y_{(n)}}(y_1,y_n)$ by the marginal distribution for $f_{Y_{(n)}}$. This yields $$f_{Y_{(1)}|Y_{(n)}}(y_1|y_n) = \frac{(n-1)(y_n-y_1)^{n-2}}{y_n^{n-1}}, \:\:\: 0 < y_1 < y_n.$$

Then calculate $P(U < u | Y_{(n)})$ from $f_{Y_{(1)}|Y_{(n)}}(y_1|y_n)$. This is \begin{align} P(U < u | Y_{(n)}) &= P(Y_{(1)} < u Y_{(n)} | Y_{(n)} = y_n) \\ &= \int_0^{u y_n} \frac{(n-1)(y_n-y_1)^{n-2}}{y_n^{n-1}} \, dy_1 \\ &= \frac{-1}{y_n^{n-1}} \left[(y_n - y_1)^{n-1} \right]_0^{u y_n} \\ &= \frac{-1}{y_n^{n-1}} \left[(y_n - uy_n)^{n-1} - y_n^{n-1} \right] \\ &= \frac{-1}{y_n^{n-1}} y_n^{n-1} \left[(1 - u)^{n-1} - 1\right]\\ &= 1 - (1-u^{n-1}). \end{align} Since $P(U < u | Y_{(n)})$ does not depend on $Y_{(n)}$, $U$ and $Y_{(n)}$ are independent.

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Thanks for your answer Mike. Is there also a way to do this without using the Jacobian? That's something I was not taught, so we are not expected to use that. –  quantum Nov 28 '12 at 6:57
@quantum Yes: use the density Mike indicates, to compute $P(U\leqslant u\mid Y_{(n)})$ for every $u$ in $[0,1]$. The result does not depend on $Y_{(n)}$ hence you are done. –  Did Nov 28 '12 at 8:13
Great, thanks guys! –  quantum Nov 28 '12 at 13:10
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