Proof of the formula for the variance as the mean of the square minus the square of the mean

Proof

We expand the square and use linearity of expectation.

1. Let $X$ be a real random variable with $E[X^2] < \infty$.
2. For each real $x$, $(\abs{x} - 1)^2 \ge 0$, so $\abs{x} \le (1 + x^2)/2$.
3. Hence $0 \le \abs{X} \le (1 + X^2)/2$. The integral of nonnegative functions is monotone, so the expected value $E[\abs{X}]$ is at most $(1 + E[X^2])/2 < \infty$.
4. So $X$ is integrable. Let $m = E[X]$, a real number. The variance of $X$ is $E[(X - m)^2]$.
5. Pointwise, $(X - m)^2 = X^2 - 2mX + m^2$.
6. The variables $X^2$, $X$ and the constant $m^2$ are integrable. By linearity of expectation, $E[(X - m)^2] = E[X^2] - 2m E[X] + m^2$.
7. Since $E[X] = m$, the right side is $E[X^2] - 2m^2 + m^2 = E[X^2] - m^2$.
8. So $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.