# Thread: another bivariate stat question

1. ## another bivariate stat question

$f(x,y) = \left\{ \begin{array}{rcl}
k(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$

for $0 \leq x \leq y \leq 1$ is it $0 \leq x \leq 1 \ \mbox{and} \ x \leq y \leq 1$ which would give:

$\int_{0}^{1} \int_{x}^{1} k(1-y) \ dy \ dx$ ?

b) and does it matter if it's $0 \leq x \leq 1 \ \mbox{and} \ x \leq y \leq 1$ or $0 \leq y \leq \ 1 \ \mbox{and} \ 0 \leq x \leq y$ ?

c) and would $P(x \leq \frac{3}{4}, \ y \geq \frac{1}{2})$ be the same as $0 \leq x \leq \frac{3}{4}, \ \frac{3}{4} \leq y \leq 1$?

therefore: $\int_{0}^{0.75} \int_{0.75}^{1} k(1-y) \ dy \ dx$

2. Originally Posted by lllll
$f(x,y) = \left\{ \begin{array}{rcl}
k(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$

for $0 \leq x \leq y \leq 1$ is it $0 \leq x \leq \ \mbox{and} \ x \leq y \leq 1$ which would give:

$\int_{0}^{1} \int_{x}^{1} k(1-y) \ dy \ dx$ ?

and does it matter if it's $0 \leq x \leq 1 \ \mbox{and} \ x \leq y \leq 1$ or $0 \leq y \leq \ 1 \ \mbox{and} \ 0 \leq x \leq y$ ?
no, you are correct. x goes from 0 up to y, but y goes up to 1, so x goes up to 1. y goes from x up to 1. so you get the integral you posted