We have two continuous random variables with joint pdf of $f(x,y) = \frac{3}{4}(2x+y^2)$ with $0 \le x \le 1$ and $0 \le y \le 1$.
We now need to compute $P(4X < Y)$.
The solution to this is for some reason the following:
$\int_0^{\frac{1}{4}}(\int_{4x}^{1} \frac{3}{4}(2x+y^2) dy)dx$
I'm confused about the following:
- Why are we integrating with respect to $y$ when we have $x$ in our bounds?
- Why are the bounds when integrating with respect to $x$ from $0$ to $\frac{1}{4}$ (Specifically why $\frac{1}{4}$.
Thanks in advance for any help!
You want to be integrating the density $f$ over the region $\{(x,y): 0\le x\le 1, 0\le y\le 1, 4x\le y\}$. To do this as an iterated integal you have two choices:
(a) For fixed $x\in[0,1]$ integrate with respect to $y$, taking the constraint $4x\le y$ into account; that is, integrate with respect to $y$ in the region $\{y: 0\le y\le 1, y\ge 4x\}$. This region is empty if $x>1/4$, while it's the interval $[4x,1]$ if $x\in[0,1/4]$. Having integrated out $y$ you then integrate with respect to $x$ — but notice that you'll only be integrating over $[0,1/4]$ because of the "emptiness" noted. (This choice corresponds to the iterated integral you have written.)
(b): For fixed $y\in[0,1]$ integrate with respect to $x$, taking the constraint $4x\le y$ into account; that is, integrate with respect to $x$ in the region $\{x: 0\le x\le 1, x\le y/4\}$. This region is simply the $x$-interval $[0,y/4]$. Having integrated out $x$ you then integrate with respect to $y\in[0,1]$.