The random variables $X_1,X_2,...X_n,Y_1,Y_2,...Y_n $ are independent and $U(0,a)$-distributed. Determine the distribution of

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

where $X_{(n)},Y_{(n)}$ denotes $n$th order variable.

This problem is giving me a headache. The answer is that $Z_n$ is $\text{Exp}(1)$-distributed. But i can't derive that conclusion. I'd like to show you my derivation, but its so long and contains so many integrals and fractions that it would take a while to write it here. But basically my approach was to derive the distribution of $M=\max(X_{(n)},Y_{(n)})$ and $L=\min(X_{(n)},Y_{(n)})$. I get the following cdf's:

$F_M=(F_X\cdot F_Y)^n=(F_X)^{2n}$ and $F_L=1-(1-(F_X)^n)\cdot (1-(F_Y)^n)=1-(1-(F_X)^n)^2$ ,

correct? $F_X $ stands for the cdf of $X_i$

Then i define the random variable $T=\frac{M}{L}$ and compute its cdf to finally be able to derive the distribution of $Z_n=n\cdot \log(T)$. But thats where i think everything goes wrong.

Any help appreciated, thanks.

  • $\begingroup$ Maybe one way to simplify: Note that $\displaystyle T = n\ln \frac {\max\{X_{(n)}, Y_{(n)}\} } {\min\{X_{(n)}, Y_{(n)}\} } = n\ln \max\left\{\frac {X_{(n)}} {Y_{(n)}}, \frac {Y_{(n)}} {X_{(n)}} \right\} = n \max\left\{\ln\frac {X_{(n)}} {Y_{(n)}}, \ln\frac {Y_{(n)}} {X_{(n)}} \right\} = n \left|\ln\frac {X_{(n)}} {Y_{(n)}}\right|$ Not sure if this can help $\endgroup$ – BGM Feb 13 '18 at 14:58
  • $\begingroup$ I'll check this, will take a moment to digest. Didnt think in this terms at all though. Thanks $\endgroup$ – JustANoob Feb 13 '18 at 15:56
  • $\begingroup$ Have a look at this post on CV:stats.stackexchange.com/questions/185227/…. $\endgroup$ – StubbornAtom Mar 26 '18 at 21:27

To go a step further from the comment,

$$\log \frac{X_{(n)}}{Y_{(n)}} = \log X_{(n)} - \log Y_{(n)},$$ hence we need only find the distribution of the maximum order statistic. Assuming without loss of generality that $a = 1$ hence $$F_{X_{(n)}}(x) = \prod_{i=1}^n \Pr[X_i \le x] = F_X(x)^n = x^n,$$ and it follows that $$f_{X_{(n)}}(x) = nx^{n-1}, \quad 0 \le x \le 1.$$ Note $Y_{(n)}$ also follows the same distribution. Continuing, $W = \log X_{(n)}$ has density $$f_W(w) = f_{X_{(n)}} (e^w) e^w = ne^{w(n-1)} e^w = ne^{nw}, \quad -\infty < w \le 0;$$ that is to say, $-W$ is exponential with rate $n$. The rest is quite straightforward and is left as an exercise.


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