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Suppose a random sample of size $2$ is obtained from a distribution that has pdf $f(x) = 2(1-x), 0 < x < 1$ and zero elsewhere. We want to find the probability that one sample observation is at least twice as large as the other

this question doesnt seem too complicated but I'm having a hard time setting double integral of this question. I, suppose we have two samples, $Y_1, Y_2$ and let $2Y_1 < Y_2$. so (my guess is that)we want to find $P(2Y_1 < Y_2) = \int \int 2(1-y_1)(1-y_2)dy_1dy_2$?? How would you construct integral?

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    $\begingroup$ I think this is a difficult way to look at it. You have to make sure you have the right densities inside the integral and get the right limits of integration. $\endgroup$ Dec 18, 2016 at 21:04
  • $\begingroup$ what would be the right integral? im not sure.. $\endgroup$
    – Allie
    Dec 18, 2016 at 21:31

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(Note: I am treating this question as self study.)

Expanding on the comment by Michael Chernick, perhaps this will help to understand the appropriate limits of integration:

domain

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You are on the right track. The joint pdf $g(y_1,y_2)=n!f(y_1)f(y_2)$. So for this problem, you would want to set it up as the following: $P(2Y_1<Y_2)=\int_{0}^{1} \int_{0}^{\frac{Y_2}{2}}2!2(1-y_1)2(1-y_2)dy_1dy_2$. Evaluate that closely and you'll arrive at the correct answer.

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