Originally Posted by

**lllll** Let $\displaystyle Y_1$ and $\displaystyle Y_2$ be independent and uniformly distributed over the interval $\displaystyle (0, 1)$. Find $\displaystyle P(2Y_{(1)} < Y_{(2)})$. The solution is 0.5.

Originally I thought it was:

$\displaystyle g_{(2)} = n[1-F(y)]^{n-1} f(y)= 2y \times 1 = 2y$

$\displaystyle g_{(1)}=1-[1-F(y)]^n = 1-[1-y]^2 = 2y-y^2$ Mr F says: This is not the pdf for $\displaystyle {\color{red}Y_{(1)}}$, it's the cdf ..... The pdf is $\displaystyle {\color{red}n [1 - F(y)]^{n-1} f(y)}$ where n = 2 in this case.

Also, there should be a different symbol used for the variable in each pdf:

$\displaystyle {\color{red}g_{(2)}(y_2) = n[1-F(y_2)]^{n-1} f(y_2)= 2y_2 \times 1 = 2y_2}$

$\displaystyle {\color{red}g_{(1)}(y_1) = n [1-F(y_1)]^{n-1} f(y_1) = 2(1 - y_1)}$

[snip]