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!

  • 3
    $\begingroup$ Hint: Read your book where it tells you about joint probability density functions, not where it tells you about joint probability mass functions. $\endgroup$ – 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 $k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.