This is very much a homework question from my textbook though not an assignment, so caveats there
I know how to find a constant in the PDF of a random variable but I'm missing something in finding the constant in the PDF of a joint variable. So I'm given:
$$ f(x,y) = \begin{cases} kxy \; x \geq 0, y \geq 0, 20 \leq x + y \leq 30 \\ 0 \; otherwise \end{cases}$$
The book says the answer is 3/81250, I guess what I'm missing is how I set things up to find that value given what I know about x and y.
I think that I need to solve this by using the integral of f(x,y) which since we know it's a probability distribution function has to equal 1, so:
$$ \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} f(x,y) \cdot dx \cdot dy = 1$$
so we know in my case
$$ \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} kxy \cdot dx \cdot dy = 1$$
but I'm not sure how to set up the integral to solve it. My first thought is that we know x + y > 20 so we've got our lower bounds and we know that x + y < 30 so that should give us:
$$ \int_{20-x}^{30-x} \int_{20-y}^{30-y} kxy \cdot dx \cdot dy = 1$$
we then move k out and get it on the other side:
$$ \int_{20-x}^{30-x} \int_{20-y}^{30-y} xy \cdot dx \cdot dy = 1/k$$
so now I wind up with:
$$ \int_{20-x}^{30-x} \int_{20-y}^{30-y} x^2y^2 = 1/k$$
I substitute in my values to start solving for dx:
$$ \frac{(30-y)^3}{3} y^2 - \frac{(20-y)^3}{3} y^2$$
and this starts taking me in what feels like a weird direction. I end up with (apologies for skipping all the algebra):
$$ \int_{20-x}^{30-x} \frac{10y^6}{6} - \frac{5000y^5}{5} + \frac{6333y^4}{4}= 1/k$$
But then I'm putting 30-x and 20-x in for y and it starts getting really messy. Am I over-complicating things or doing something grossly incorrect?
Guide:
Start with making a picture of the region where the PDF takes value $kxy$ and convince yourself by means of that picture that:$$\int_{\infty}^{\infty}\int_{\infty}^{\infty}f\left(x,y\right)dxdy=\int_{0}^{20}\int_{20-y}^{30-y}kxydxdy+\int_{20}^{30}\int_{0}^{30-y}kxydxdy$$
The RHS of this equality equals $1$ and must be worked out in order to find $k$.
I will give you a start: $$\begin{aligned}1=\mathsf{RHS} & =k\int_{0}^{20}y\int_{20-y}^{30-y}xdxdy+k\int_{20}^{30}y\int_{0}^{30-y}xdxdy\\ & =k\int_{0}^{20}y\left[\frac{1}{2}x^{2}\right]_{20-y}^{30-y}dy+k\int_{20}^{30}y\left[\frac{1}{2}x^{2}\right]_{0}^{30-y}dy\\ & =k\int_{0}^{20}\frac{1}{2}y\left[\left(30-y\right)^{2}-\left(20-y\right)^{2}\right]dy+k\int_{20}^{30}\frac{1}{2}y\left(30-y\right)^{2}dy\\ & =\cdots \end{aligned} $$
I leave the rest to you.
Good luck!