Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Variance ▷ Non-negativity

Theorem: The variance is always non-negative, i.e.

$\label{eq:var-nonneg} \mathrm{Var}(X) \geq 0 \; .$

Proof: The variance of a random variable is defined as

$\label{eq:var} \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right] \; .$

1) If $X$ is a discrete random variable, then, because squares and probabilities are stricly non-negative, all the addends in

$\label{eq:var-disc} \mathrm{Var}(X) = \sum_{x \in \mathcal{X}} (x-\mathrm{E}(X))^2 \cdot f_X(x)$

are also non-negative, thus the entire sum must be non-negative.

2) If $X$ is a continuous random variable, then, because squares and probability densities are strictly non-negative, the integrand in

$\label{eq:var-cont} \mathrm{Var}(X) = \int_{\mathcal{X}} (x-\mathrm{E}(X))^2 \cdot f_X(x) \, \mathrm{d}x$

is always non-negative, thus the term on the right-hand side is a Lebesgue integral, so that the result on the left-hand side must be non-negative.

Sources:

Metadata: ID: P123 | shortcut: var-nonneg | author: JoramSoch | date: 2020-06-06, 07:29.