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I have $f_{X,Y}(x,y) = \lambda^2e^{-\lambda y}$ for 0 < x < y.

If I want to show that this is a joint PDF, I need to do a double integral and show that it is equal to 1. Do I set my integration limits up as 0,y and x,$\infty$ for x and y respectively?

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    $\begingroup$ Assuming that $f_{X,Y}(x,y) = 0$ if $0 < x < y$ does not hold, then yes. $\endgroup$
    – Jan Ladislav Dussek
    Nov 9, 2013 at 0:25

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There are two ways. One is: $$ \int_{0}^{+\infty} \int_{0}^{y} \lambda^2\exp(-\lambda\cdot y)\,dx\,dy$$ and the other is: $$ \int_{0}^{+\infty} \int_{x}^{+\infty} \lambda^2\exp(-\lambda\cdot y)\,dy\,dx.$$

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  • $\begingroup$ Can you please explain why my limits don't work? I assume that the domain is all points above the line y=x. $\endgroup$
    – AdronVenelucio
    Nov 9, 2013 at 0:32
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    $\begingroup$ It works. The order suggested by Sergio Parreiras is simplest, but you can also integrate $x=0$ to $y$, $y=0$ to $\infty$. $\endgroup$
    – André Nicolas
    Nov 9, 2013 at 0:36
  • $\begingroup$ @Whistlekins: yes this is the domain (assuming we don't look at negative values of $x$ or $y$) and you can describe it as: 1) for every positive $y$ you consider all $x$_s below it or 2) for every positive $x$ you consider all $y$_s above it. $\endgroup$
    – Sergio Parreiras
    Nov 9, 2013 at 0:45
  • $\begingroup$ @Andre Nicolas: this was his original domain: $\int_{0}^y \int_{x}^{+\infty} \lambda^2 \exp(-\lambda\cdot y) \,dy\,dx$ which clearly does not work. $\endgroup$
    – Sergio Parreiras
    Nov 9, 2013 at 0:46
  • $\begingroup$ True. Integrating to $y$ in the second integral didn't even make sense. $\endgroup$
    – André Nicolas
    Nov 9, 2013 at 0:49

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