Let [math]X[/math] be a random variable with [math]E(X) = \mu[/math] and [math]V(X) = \sigma^2[/math]. Show that the function [math]f(x)[/math] defined by

[[math]] f(x) = \sum_\omega (X(\omega) - x)^2 p(\omega) [[/math]]

has its minimum value when [math]x = \mu[/math].