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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

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

  • 0
    But your function is defined everywhere! It's $0$ whenever we are not in the interval $02011-12-11
  • 0
    sure, but I only want to integrate in the given interval to see if the double integral results equal 12011-12-11
  • 0
    I suggest you edit "..as the function is only defined on the interval.." to "..as the function is non-zero only on the interval..".2011-12-11
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    Indeed, it was a bad translation because if this function outside the range defined by taking the value zero. Excuse my english2011-12-11
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    The set $0\lt x\lt y\lt2$ is not an interval.2011-12-11
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    You would find this problem a _lot_ easier if you were to sketch a diagram of the plane with $x$ and $y$ axes, and figure out the _region_ of the plane where the joint probability distribution is nonzero.2011-12-11

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Consider the interval $0

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

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.