# Find $P(2Y_{(1)} < Y_{(2)})$ of a Uniformly Distributed Random Variable

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

Therefore, we determine

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

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.

• 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. – StubbornAtom Jan 2 '19 at 6:35

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)}
$$\mathsf{P}(2Y_2