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Expand around the expectation $\mu$:Thus variance as the minimum mean squared error of a constant givesThe finite second moment also guarantees the finite first absolute moment needed below.

For a probability density function $f$, splitting at $a$ givesLet $F$ be the cumulative distribution function. Since a density gives no mass at a singleton, differentiating under this integral yields $H^{\prime }(a)=F(a)-(1-F(a))=2F(a)-1$. This derivative is nondecreasing, so $H$ is convex andThis is the median minimizes expected absolute loss property. The minimizer need not be unique: a gap in the density can give an interval of medians, all with the same loss.

For every $a\in \mathbb{R}$,After squaring and taking expectations, the cross term vanishes because $\mathbb{E}[X-\mathbb{E}X]=0$. ThereforeThe second term is nonnegative and vanishes exactly at $a=\mathbb{E}X$, so the variance as the minimum mean squared error of a constant identity is