Consider $n$ independent uniform random variables $X_i \sim U(-\theta,\theta)$, and let $Y_1 = \min(X_1, \ldots, X_n)$ and $Y_n = \max(X_1, \ldots, X_n)$ .

What is distribution of $Z = \max (-Y_1,Y_n)$, given that the joint PDF of $Y_1$ and $Y_n$ is $$ f(y_1, y_n) = \frac{n(n-1)}{(2θ)^n} (y_n-y_1)^{n-2} , \quad -\theta < y_1 < y_n < \theta ? $$

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    $\begingroup$ Please double-check your equation for the joint PDF: what are you doing subtracting $\theta$ from it? $\endgroup$ – whuber Mar 16 '13 at 22:39
  • $\begingroup$ This looks like standard bookwork. If this is homework, or other work for some subject or just for the purpose of self study, you should mark it with the self-study tag. (Click the tag in this comment for details, and check the information relating to homework in the faq.) $\endgroup$ – Glen_b Mar 17 '13 at 0:41
  • $\begingroup$ possible duplicate of Likelihood Ratio of two-sample Uniform Distribution $\endgroup$ – Alecos Papadopoulos Jan 30 '14 at 0:35
  • $\begingroup$ The easy way to do this would be to look at $|Y|$ so you're back to a simple max problem $Z=max |Y_i|$. $\endgroup$ – Glen_b Jan 30 '14 at 1:34
  • $\begingroup$ Nice problem. Prima facie, it seems attractive to treat it, by symmetry, as the max of two separate univariate problems, or equivalently, as the pdf of the sample maximum given the parent $Uniform(-\theta, \theta)$ with sample size $2n$... but that will yield an incorrect solution, because it misses the crucial interdependency. And the interdependency is that the domain of support for $Z = max(-Y_1,Y_n)$ is not $(-\theta, \theta)$, but rather $(0,\theta)$, because (i) if $y_n > 0$, then $Z > 0$, and (ii) if $y_n < 0$, then $y_1 < 0$ ---> $Z>0$. The pdf of $Z$ is a Power Function. $\endgroup$ – wolfies Jan 30 '14 at 16:07

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