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...

  • $\begingroup$ 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 $\log(Z_i)$? $\endgroup$
    – whuber
    Dec 5 '15 at 22:38
  • $\begingroup$ @whuber my thought from your comment would be to set up a transform where I solve the density of n*log(Z$_i$) ? $\endgroup$
    – Susan
    Dec 5 '15 at 22:52
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Dec 6 '15 at 0:09
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    $\begingroup$ 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. ....... $\endgroup$
    – wolfies
    Dec 6 '15 at 7:45
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    $\begingroup$ 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). $\endgroup$
    – 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

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


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 ...


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 $f_{(n)}(x,y)$:

enter image description here

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:

enter image description here

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, $f_n(w)$:

enter image description here

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:

enter image description here

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:

enter image description here

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.


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

We are asked to find:

$$Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$$

where the denominator is min(xMax, yMax), ... not the minimum of all the $X$'s and $Y$'s.

  • $\begingroup$ Following your sketch, I understand how I misinterpreted the question. I understand how to calculate the joint pdf of the two sample maximums, but I am still not sure how we are to interpret the ratio of max/min. $\endgroup$
    – Susan
    Dec 6 '15 at 11:49
  • $\begingroup$ I've added an alternative derivation using order stats, which 'circumvents' the max/min. $\endgroup$
    – wolfies
    Dec 6 '15 at 13:36
  • $\begingroup$ If you had started with the logs of the data, Susan, then you would be looking at the differences of order statistics rather than ratios. $\endgroup$
    – whuber
    Dec 6 '15 at 14:59
  • $\begingroup$ I am not convinced using computer formal computations is the best way to explain the reason why the ratio is an Exp(1) random variable. $\endgroup$
    – Xi'an
    Dec 6 '15 at 15:38
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    $\begingroup$ Good point ... except the OP does not ask the reason ... but to show that it is Exp[1]. I am also unsure as to whether or not this is homework (or an assignment) ... and that is actually one nice advantage of using a computer: one provides the steps to follow, verifies the result, so that one has the right approach, but the mechanics are still left to the OP. Would be nice for someone to explore @whuber's suggestion of taking logs at the start. $\endgroup$
    – wolfies
    Dec 6 '15 at 16:34

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