Variance as Expectation of Square minus Square of Expectation

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

Then:

$$ $$ $$ $$ $$

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

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.