# Transforming Order Statistics

Assume random variables $X_1, ... , X_n$ and $Y_1, ..., Y_n$ are independent and $U(0,a)$-distributed. Show that $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$ has an $\text{Exp}(1)$ distribution.

I've started this problem by setting $\{X_1,...,X_n,Y_1,...Y_n\} = \{Z_1,...,Z_n\}$ Then the $\max(Y_n,X_n)= Z_{(2n)}$ would be distributed as $(\frac{z}{a})^{2n}$ and $\min(Y_n,X_n)= Z_{(1)}$ would be distributed as $1 - (1 - \frac{z}{a})^{2n}$ The densities can be found easily as $f_{Z_{1}}(z) = (2n)(1-\frac{z}{a})^{2n-1}\frac{1}{a}$ and $f_{Z_{(2n)}}(z) = (2n)(\frac{z}{a})^{2n-1} \frac{1}{a}$

This is where I'm having a hard time knowing where to go next now that these are calculated. I'm thinking it has to do something with a transformation, but I'm unsure...

• Surely you need to assume in addition that not only are the $X_i$ and $Y_i$ iid, but also the $X_i$ are independent of the $Y_j$. Given that, have you thought of working directly with the $\log(Z_i)$? – whuber Dec 5 '15 at 22:38
• @whuber my thought from your comment would be to set up a transform where I solve the density of n*log(Z$_i$) ? – Susan Dec 5 '15 at 22:52
• I did a little reformatting (especially turning $log$ and $min$ into $\log$ and $\min$) but if you don't like it how it is, you can roll back to the previous version (by clicking the "edited <x> ago" link above my gravatar at the bottom of your post) and then clicking the "roll back" link above your previous version. – Glen_b -Reinstate Monica Dec 6 '15 at 0:09
• Susan, you appear to have misinterpreted/misread the question. The question seeks the ratio of $$\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$$ The denominator refers to $\min(Y_{(n)},X_{(n)})$: where $Y_{(n)}$ is the maximum order statistic of the $Y$s, and $X_{(n)}$ is the maximum order statistic of the $X$s. In other words, ${\min(Y_{(n)},X_{(n)})}$ seeks min( maxX, maxY), NOT the minimum of all the $X$s and $Y$s, so you cannot use your Z trick to flatten / combine all the X and Y values. ....... – wolfies Dec 6 '15 at 7:45
• In any event, and as a separate matter, there is no point (as you have done) calculating the density of $Z_{(1)}$,and separately the density of $Z_{(2n)}$, because the different order stats are not generally independent. To find the ratio of $Z_{(2n)}/Z_{(1)}$, one would need to first find the joint pdf of $(Z_{(1)},Z_{(2n)})$, if that was the problem at hand (which it is not). – wolfies Dec 6 '15 at 7:51

This problem can be solved from the definitions alone: the only advanced calculation is the integral of a monomial.

### Preliminary observations

Let's work with the variables $X_i/a$ and $Y_i/a$ throughout: this does not change $Z_n$ but it makes $(X_1, \ldots, Y_n)$ iid with Uniform$(0,1)$ distributions, eliminating all distracting appearances of $a$ in the calculations. Thus we may assume $a=1$ without any loss of generality.

Note that the independence of the $Y_i$ and their uniform distribution imply that for any number $y$ for which $0\le y \le 1$,

$$\Pr(y \ge Y_{(n)}) = \Pr(y \ge Y_1 , \ldots, y \ge Y_n) = \Pr(y \ge Y_1)\cdots \Pr(y \ge Y_n) = y^n,$$

with an identical result holding for $X_{(n)}$. For future reference, this allows us to compute

$$\mathbb{E}(2X_{(n)}^n) = \int_0^1 2x^n\mathrm{d}(x^n) = \int_0^1 2nx^{2n-1}\mathrm{d}x = 1.$$

### Solution

Let $t$ be a positive real number. To find the distribution of $Z_n$, substitute its definition and simplify the resulting inequality:

\eqalign{ \Pr(Z_n \gt t) &= \Pr(Z_n / n \gt t/n) = \Pr\left(\exp(Z_n/n) \gt e^{t/n}\right) \\ &=\Pr\left(\frac{\max(X_{(n)}, Y_{(n)})}{\min(X_{(n)}, Y_{(n)})} \gt e^{t/n}\right) \\ &= \Pr\left(e^{-t/n}{\max(X_{(n)}, Y_{(n)})} \gt {\min(X_{(n)}, Y_{(n)})}\right). }

This event breaks into two equiprobable cases, depending on whether $X_{(n)}$ or $Y_{(n)}$ is the smaller of the two (and their intersection, with zero probability, can be ignored). Thus we need only compute the chance of one of these cases (say where $Y_{(n)}$ is the smaller) and double it. Since $t\ge 0$, $0 \le e^{-t/n}X_{(n)} \le 1$, allowing us (upon letting $e^{-t/n}X_{(n)}$ to play the role of $y$) to apply the computations in the preliminary section:

$$\Pr(Z_n \gt t) =2\Pr\left(e^{-t/n}X_{(n)} \gt {Y_{(n)}}\right) =2 \mathbb{E}\left[\left(e^{-t/n}{X_{(n)}}\right)^n\right] = e^{-t} \mathbb{E}\left[2{X_{(n)}^n}\right] = e^{-t}.$$

That's what it means for $Z_n$ to have an Exp$(1)$ distribution.

I will sketch the solution, here using a computer algebra system to do the nitty gritties ...

Solution

If $X_1, ... , X_n$ is a sample of size $n$ on parent $X \sim \text{Uniform}(0,a)$, then the pdf of the sample maximum is: $$f_{n}(x) = \frac{n}{a^n} x^{n-1}$$ and similarly for $Y$.

Approach 1: Find the joint pdf of $(X_{(n)},Y_{(n)})$

Since $X$ and $Y$ are independent, the joint pdf of the 2 sample maximums $(X_{(n)},Y_{(n)})$ is simply the product of the 2 pdf's, say $f_{(n)}(x,y)$:

Given $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$. Then, the cdf of $Z_n$ is $P(Z_n < z)$ is:

where I am using the Prob function from the mathStatica package for Mathematica to automate. Differentiating the cdf wrt $z$ yields the pdf of $Z_n$ as standard Exponential.

Approach 2: Order statistics

We can use order statistics to 'by-pass' the mechanics of having to deal with the Max and Min functions.

Once again: If $X_1, ... , X_n$ is a sample of size $n$ on parent $X \sim \text{Uniform}(0,a)$, then the pdf of the sample maximum $W = X_{(n)}$ is, say, $f_n(w)$:

The sample maximums $X_{(n)}$ and $Y_{(n)}$ are just two independent drawings from this distribution of $W$; i.e. the $1^{st}$ and $2^{nd}$ order statistics of $W$ (in a sample of size 2) are just what we are looking for:

• $W_{(1)} = \min(Y_{(n)},X_{(n)})$

• $W_{(2)} = \max(Y_{(n)},X_{(n)})$

The joint pdf of $(W_{(1)}, W_{(2)})$, in a sample of size 2, say $g(.,.)$, is:

Given $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$. Then, the cdf of $Z_n$ is $P(Z_n < z)$ is:

The advantage of this approach is that the probability calculation no longer involves the max/min functions, which may make the derivation (especially by hand) somewhat easier to express.

Other

As per my comment above, it appears you have misinterpreted the question ...

$$Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$$
where the denominator is min(xMax, yMax), ... not the minimum of all the $X$'s and $Y$'s.