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I'm trying to do an exercise of joint probability distributions in which I ask to show that the pdf is a valid pdf

$$ f(x,y) =\left\{\begin{array}{cl} \frac{3(x^2+y^2)}{16} & \text{if }0<x<y<2 \\\\ 0 & \text{else} \\\\ \end{array}\right.$$

to demonstrate that it is a valid pdf I have to integrate all $\mathbb R$, however as the function is non-zero only on the interval $0 <x <y <2$. As would be the limits? I have serious doubt that. there any way to get those limits?

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But your function is defined everywhere! It's $0$ whenever we are not in the interval $0<x<y<2$. – matt Dec 11 '11 at 4:12
sure, but I only want to integrate in the given interval to see if the double integral results equal 1 – franvergara66 Dec 11 '11 at 4:15
I suggest you edit " the function is only defined on the interval.." to " the function is non-zero only on the interval..". – matt Dec 11 '11 at 4:19
Indeed, it was a bad translation because if this function outside the range defined by taking the value zero. Excuse my english – franvergara66 Dec 11 '11 at 4:25
The set $0\lt x\lt y\lt2$ is not an interval. – Did Dec 11 '11 at 8:21
up vote 2 down vote accepted

Consider the interval $0<x<y<2$. Now $x$ can take any value between $0$ and $y$. Thus we obtain:

$$\int_0^y \frac{3(x^2+y^2)}{16} dx$$

Now $y$ can take any value between $x$ and $2$. Moreover, since $x$ can take any value greater than $0$ we have that $y$ can take any value between $0$ and $2$. Thus we obtain:

$$\int_0^2\int_0^y \frac{3(x^2+y^2)}{16} dxdy$$

Evaluting this integral gives $1$ as desired.

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