# Supplemental Proof 1

Proof:

$\mathbb E[X^2] = \mathbb E[(X - \mathbb E[X])^2] + \mathbb E[X]^2$

Let’s start off with the following expression:

(1) $\mathbb E [(X - \mathbb E[X])^2]=$

(2) $\mathbb E[X^2-2X\mathbb E[X]+\mathbb E [X]^2]=$

(3) $\mathbb E[X^2] - 2\mathbb E[X]^2 + \mathbb E[X]^2=$

(here we take advantage of the notion that $\mathbb E[\mathbb E[X]]$ is a constant, namely $\mathbb E[X]$)

(4) $\mathbb E[X^2] - \mathbb E[X]^2$

Combining the terms in Line (1) and Line (4) and rearranging, we find that

(5) $\mathbb E[X^2] = \mathbb E[(X - \mathbb E[X])^2] + \mathbb E[X]^2$

I recently received my PhD from UC Berkeley where I studied computational neuroscience and machine learning.

Posted on April 21, 2013, in Proofs. Bookmark the permalink. 1 Comment.