Expected squared prediction error I'm reading about statistical decision theory and on one point in my book the author defines the expected squared prediction error by: $$EPE = E(Y-g(X))^2 = \int(y -g(x))^2Pr(dx, dy)$$ I like to write this with the density function so that it stays more precise: $$EPE = \int\int(y-g(x))^2f(x,y)\;dx\;dy$$ Now on the other part the author says that by conditioning on $X$, $EPE$ can be written as: $$EPE = E_XE_{Y|X}([Y-g(X)]^2\;|\;X)$$ For some reason this notation confuses me...could someone write this conditional notation of $EPE$ more precisely, i.e. so that it would include the joint density function of random variables $X$ and $Y$ etc.? Just to be sure: $X$ is the variable we use to predict $Y$ and $g(X)$ is the function we are trying to solve, which minimizes $EPE$. Thank you for any help :)

$EPE = \int\int {(y-g(x))^2f(x,y)dxdy}$

By Bayes' Theorem $f(x,y)=f(y\,|\,x)\,f(x)$ we have:

$EPE = \int\int {(y-g(x))^2f(y\,|\,x)\,f(x)dxdy}$

Rearranging gives:

$EPE = \int f(x)\;\left(\,\int (y-g(x))^2f(y\,|\,x)dy\,\right)\;dx$

Using definition of $E_x$ we get:

$EPE = E_x(\;\int (y-g(x))^2f(y\,|\,x)dy\;) $

Using definition of $E_{Y\,|\,X}$ we get:

$EPE = E_x(\;\;\;E_{Y\,|\,X}(\,(Y-g(X))^2\,|\,X\,)\;\;\;) $

Or an even shorter notation:

$EPE = E_xE_{Y\,|\,X}(\,[Y-g(X)]^2\,|\,X\,) $