Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Variance ▷ Variance of a constant

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:

$\label{eq:X-const} X = \text{const.}$
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Metadata: ID: P124 | shortcut: var-const | author: JoramSoch | date: 2020-06-27, 06:44.