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My book says the following:

If $X$ and $Y$ are continuous random variables with joint density $f(x, y)$, and if $g$ is any function of the two variables $X$ and $Y$, then:

$$E(g(X,Y)) = \int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f(x,y)dydx$$

My question is: is $E(g(X,Y)) = \int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f(x,y)dxdy$ equivalent to the above? I think it is, but I just want to make sure and if it is, I would appreciate an explanation the reason. Is it cause with double integrals it doesn't matter whether you integrate with respect to $y$ or $x$ first due to fubini's theorem?

Thank You!

Sebastiano
  • 7,649
  • There is something called Fubini/Tonelli theorem that takes care of this question. It is possible to interchange the integrals provided the absolute value of the integral is finite, further positive functions such as the absolute value can be interchanged. – Dole Mar 30 '21 at 18:32
  • And of course... for random variables it is quite often assumed that $Var(X)<\infty$. – Dole Mar 30 '21 at 18:38
  • I see, so it doesn't matter cause of fubini's theorem like I thought. Thank you so much –  Mar 30 '21 at 19:26

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