# Math Help - another order stats question

1. ## another order stats question

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

Originally I thought it was:

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

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

$2(2y-y^2) < 2y$

$2y

but this doesn't seem to right. If anybody has any input, it would be greatly appreciated.

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

Originally I thought it was:

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

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

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

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

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

[snip]
The key is to know that if $Y_1$ and $Y_2$ are i.i.d. random variables with density function f(y) then the joint density function of $Y_{(1)}$ and $Y_{(2)}$ is

$g_{12}(y_1, \, y_2) = 2 f(y_1) f(y_2)$ for $y_1 \leq y_2$

and zero elsewhere. I can prove this result if you want, but you should have a copy of the proof somewhere.

For your problem $g_{12}(y_1, \, y_2) = 2 (1) (1) = 2$ for $y_1 \leq y_2$ and $0 \leq y_2 \leq 1$.

Then $\Pr(2Y_{(1)} < Y_{(2)}) = \Pr \left(Y_{(1)} < \frac{1}{2} Y_{(2)}\right) = \int_{y_2 = 0}^{y_2 = 1} \int_{y_1 = 0}^{y_1 = y_2/2} 2 \, dy_1 \, dy_2 = \frac{1}{2}$.