Proof: Variance of constant is zero
Theorem: The variance of a constant is zero
\[\label{eq:var-const} a = \text{const.} \quad \Rightarrow \quad \mathrm{Var}(a) = 0\]and if the variance of $X$ is zero, then $X$ is a constant
\[\label{eq:var-zero} \mathrm{Var}(X) = 0 \quad \Rightarrow \quad X = \text{const.}\]Proof:
1) A constant is defined as a quantity that always has the same value. Thus, if understood as a random variable, the expected value of a constant is equal to itself:
\[\label{eq:mean-const} \mathrm{E}(a) = a \; .\]Plugged into the formula of the variance, we have
\[\label{eq:var-const-s1} \begin{split} \mathrm{Var}(a) &= \mathrm{E}\left[ (a-\mathrm{E}(a))^2 \right] \\ &= \mathrm{E}\left[ (a-a)^2 \right] \\ &= \mathrm{E}(0) \; . \end{split}\]Applied to the formula of the expected value, this gives
\[\label{eq:var-const-s2} \mathrm{E}(0) = \sum_{x=0} x \cdot f_X(x) = 0 \cdot 1 = 0 \; .\]Together, \eqref{eq:var-const-s1} and \eqref{eq:var-const-s2} imply \eqref{eq:var-const}.
2) The variance is defined as
\[\label{eq:var} \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right] \; .\]Because $(X-\mathrm{E}(X))^2$ is strictly non-negative, the only way for the variance to become zero is, if the squared deviation is always zero:
\[\label{eq:sqr-dev-zero} (X-\mathrm{E}(X))^2 = 0 \; .\]This, in turn, requires that $X$ is equal to its expected value
\[\label{eq:X-eq-E-X} X = \mathrm{E}(X)\]which can only be the case, if $X$ always has the same value:
\[\label{eq:X-const} X = \text{const.}\]- Wikipedia (2020): "Variance"; in: Wikipedia, the free encyclopedia, retrieved on 2020-06-27; URL: https://en.wikipedia.org/wiki/Variance#Basic_properties.
Metadata: ID: P124 | shortcut: var-const | author: JoramSoch | date: 2020-06-27, 06:44.