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Denote $Y_{(1)} = \min(Y_1,Y_2)$ and $Y_{(2)} = \max(Y_1,Y_2)$. Let $Y_1$ and $Y_2$ be independent and uniformly distributed over the interval $(0, 1)$. Find $P(2Y_{(1)} < Y_{(2)})$.

Attempted solution:

We know that

$$ f(y_i) = \begin{cases} 1 & 0<y_i<1\\ 0 & else \end{cases} $$

Therefore, we determine

$$ F(y_i) = \begin{cases} 0 & y_i < 0\\ y_i & 0<y_i<1\\ 1 & y_i > 1\\ \end{cases} $$

Using this, we can find the distribution functions for $Y_{(1)}$ and $Y_{(2)}$

$$ Y_{(1)} = 1 - (1-F(y))^2\\ Y_{(2)} = F(y)^2 $$

By differentiating, we get the density functions

$$ f_{Y_{(1)}}(y) = 2(1-y)\\ f_{Y_{(2)}}(y) = 2y $$

I'm not too sure where to go from here.

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  • $\begingroup$ Similar to the answer below, using total probability theorem, \begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1\ge Y_2) \\&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \\&=2\times P(2Y_1<Y_2) \end{align} If you can justify the steps above, the rest is straightforward. $\endgroup$ – StubbornAtom Jan 2 '19 at 6:35
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Since $a\vee b=(a+b)/2+|a-b|/2$ and $a\wedge b=(a+b)/2-|a-b|/2$,

\begin{align} \mathsf{P}(2Y_{(1)}<Y_{(2)})&=\mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \\ &=\mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+\mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \\ &=\mathsf{P}(2Y_2<Y_1)+\mathsf{P}(2Y_1<Y_2)=1/2. \end{align}


$$ \mathsf{P}(2Y_2<Y_1)=\int_0^1\mathsf{P}(Y_2<y/2)dy=\int_0^1(y/2)dy=1/4. $$

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  • $\begingroup$ Alright, that makes sense. I'm still not too sure how to continue, however. $\endgroup$ – Bryden C Jan 2 '19 at 2:57

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