0
$\begingroup$

I'm fairly new to joint probability density functions but I've taken a multivariable calculus course before to sort of understand what's going on. However, I just can't seem to figure out how to set up the integrals for the simplest of questions:

"Let X and Y be continuous random variables that have the following joint probability density function:

f(x,y) = A*e-(2x+y) for 0 < x < y < infinity, 0 otherwise.

$f(x,y) = \begin{cases} Ae^{-(2x+y)}, & \text{for 0 < x < y < ∞ } \\ 0, & \text{otherwise} \end{cases} $

Find the value of A."

What I know so far is that the integrated pdf is always equal to 1, which is how you find the constant value A. However, the part that's holding me up is what exactly to put as the bounds on those integrals and which variable to anti-differentiate first.

Any ideas/tips on how to do this or how to get started?

Cheers

$\endgroup$
  • $\begingroup$ Start by figuring out what the region of integration looks like. Once you know that, you should be able to decide which variable to integrate out first. $\endgroup$ – SplitInfinity Mar 15 '17 at 19:38
0
$\begingroup$

Draw a picture of the region of points $(x,y)$ that satisfy $0<x<y<\infty$.

If you want to think of the region as many vertical strips, then note that $x$ is in the range $(0,\infty)$, and then for a fixed $x$, the strip in that region consists of all $y$ in the range $(x,\infty)$. This produces the integral $\int_0^\infty \int_x^\infty Ae^{-(2x+y)} \mathop{dy}\mathop{dx}$.

If instead you want to think of the region as many horizontal strips, then note that $y$ is in the range $(0,\infty)$, and then for a fixed $y$, the strip in the region consists of $x$ in the range $(0,y)$. This produces $\int_0^\infty \int_0^y Ae^{-(2x+y)} \mathop{dx} \mathop{dy}$.

$\endgroup$
0
$\begingroup$

I would highly recommend drawing a picture, and it doesn't matter which variable you integrate with respect to first.

If you integrate $dxdy$, $0<x<y$, and $0< y < \infty$

If you integrate $dydx$, $x<y<\infty$, and $0<x<\infty$

$\endgroup$

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.