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 $$ f(y_1, y_n) = \frac{n(n-1)}{(2θ)^n} (y_n-y_1)^{n-2} , \quad -\theta < y_1 < y_n < \theta ? $$

  • 1
    $\begingroup$ 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 -Reinstate Monica 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 -Reinstate Monica 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

Your Answer

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

Browse other questions tagged or ask your own question.