# Variance as Expectation of Square minus Square of Expectation

From ProofWiki

## Contents

## Theorem

Let $X$ be a discrete random variable.

Then the variance of $X$ can be expressed as:

- $\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.

## Proof

We let $\mu = E \left({X}\right)$, and take the expression for variance:

- $\displaystyle \operatorname{var} \left({X}\right) := \sum_{x \mathop \in \operatorname{Im} \left({X}\right)} \left({x - \mu}\right)^2 \Pr \left({X = x}\right)$

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \operatorname{var} \left({X}\right)\) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_x \left({x^2 - 2 \mu x + \mu^2}\right) \Pr \left({X = x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_x x^2 \Pr \left({X = x}\right) - 2 \mu \sum_x x \Pr \left({X = x}\right) + \mu^2 \sum_x \Pr \left({X = x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_x x^2 \Pr \left({X = x}\right) - 2 \mu \sum_x x \Pr \left({X = x}\right) + \mu^2\) | \(\displaystyle \) | \(\displaystyle \) | Definition of Probability Mass Function: $\sum_x \Pr \left({X = x}\right) = 1$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_x x^2 \Pr \left({X = x}\right) - 2 \mu^2 + \mu^2\) | \(\displaystyle \) | \(\displaystyle \) | Definition of Expectation: $\sum_x x \Pr \left({X = x}\right) = \mu$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_x x^2 \Pr \left({X = x}\right) - \mu^2\) | \(\displaystyle \) | \(\displaystyle \) |

Hence the result, from $\mu = E \left({X}\right)$.

$\blacksquare$

## Comment

This is a significantly more convenient way of defining the variance than the first-principles version. In particular, it is far easier to program a computer to calculate this (you don't need to maintain a record of all the divergences). Therefore, this is by far the more usually encountered of the definitions for variance.

## Also see

## Sources

- Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*(1986)... (previous)... (next): $\S 2.4$: Expectation: $(22)$