# Determine K so that the function can serve as a joint probability function

I need to find k such that

$f(x,y) = \begin{cases} kx(x-y), & \text{if 0<$x$<1, -$x$<$y$<$x$} \\ 0, & \text{elsewhere} \end{cases}$

can serve as a joint probability density. I understand that I have to plug in values of $x$ and $y$, under the conditions given, and sum everything up and equal it to $1$. The problem I'm having is with the restrictions. I can plug in an infinite numbers of $X$ and $Y$ under that condition. I just don't know what to do. Any hint or suggestion would be greatly appreciated. Thank you!

• Hint: Read your book where it tells you about joint probability density functions, not where it tells you about joint probability mass functions. – Dilip Sarwate Oct 4 '14 at 17:05

For $f(x,y)$ to be a pdf, you need the integration of $f$ over $\mathbb{R}^2$ to be $1$. In other words, $$\iint_{\mathbb{R}^2} f(x,y)\, dy\,dx = \int_0^1 \int_{-x}^x kx(x-y)\,dy\,dx = 1$$ Note that the second integral is over the region $\{ 0 < x < 1, -x < y < x \}$ since $f$ is $0$ everywhere else. Evaluating the integral should give you a value for $k$.