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The demonstration of this property: $$ \begin{align}\mathsf E(XY) &= E(X)E(Y)\end{align}, $$ where $X$ and $Y$ are two independent random variables, is the following: $$ \begin{align}\mathsf E(XY) &= \int_\Bbb R \int_\Bbb R xy\,f_{X,Y}(x, y)\,\mathrm d x \mathrm d y\tag 1\\[1ex] &=\int_\Bbb R \int_\Bbb R xy\,f_{X}(x)f_{Y}(y)\,\mathrm d x\mathrm d y\tag 2\\[1ex] &= (\int_\Bbb R x\,f_{X}(x)\mathrm dx ) (\int_\Bbb R y\,f_{Y}(y)\mathrm dy).\tag 3\\[1ex] \end{align} $$

I don't understand how you can go from the second line to the last one. It might be a basic property of double integrals, but I have very little knowledge about that subject.

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  • $\begingroup$ Intuitively, you can think of it as $x,f_X$ being independent of $y$, and $y,f_Y$ being independent of $x$, so you can pull them out like constants. $\endgroup$ Jun 18, 2021 at 19:55
  • $\begingroup$ For independent variables, $f_{X,Y}(x,y)=f_X(x)\,f_Y(y)$. $\endgroup$
    – user65203
    Jun 19, 2021 at 17:03

1 Answer 1

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Let's talk about the following integral $$\int_\Bbb R \int_\Bbb R xy\,f_{X}(x)f_{Y}(y)\,dx\,dy$$ Since $X$ and $Y$ are independent, it is clear that $f_X(x)$ does not depend on $y$, and $f_Y(y)$ does not depend on $x$. How do you do double integrals? First, integrate with respect to one variable, and then with respect to the other. When you're integrating with respect to $x$, you can pull terms dependent on $y$ out of that integral - since they do not depend on $x$ - they are essentially "constants". So, we would rewrite the integral above as: $$\int_\Bbb R \int_\Bbb R xy\,f_{X}(x)f_{Y}(y)\,dx\,dy = \int_\Bbb R y f_Y(y) \left( \int_\Bbb R x f_X(x)\, dx\right)\,dy$$ $\left( \int_\Bbb R x f_X(x)\, dx\right)$ does not depend on $y$, so we can pull it out of the integral over $y$. We get $$\int_\Bbb R y f_Y(y) \left( \int_\Bbb R x f_X(x)\, dx\right)\,dy = \left( \int_\Bbb R x f_X(x)\, dx\right) \left( \int_\Bbb R y f_Y(y)\, dy\right)$$ which is what we wanted.

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