Supplemental Proof 1


\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

About dustinstansbury

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.

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