3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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