Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.