6
$\begingroup$

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 :)

$\endgroup$

1 Answer 1

10
$\begingroup$

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

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\,) $

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .