Random Variable has Zero Variance iff Almost Surely Constant

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Theorem

Let $X$ be a random variable such that $\expect {X^2}$ exists.

Then $\var X = 0$ if and only if there exists $c \in \R$ with $\map \Pr {X = c} = 1$.

That is, $X$ is almost surely constant.


Proof

Sufficient Condition

Suppose that there exists some $c \in \R$ with $\map \Pr {X = c} = 1$.

From Expectation of Almost Surely Constant Random Variable:

$\expect X = c$

Let $\map \supp X$ be the support of $X$.

Since $\map \Pr {X = c} = 1$, we have:

$\map \supp X = \set c$

We therefore have:

\(\ds \expect {X^2}\) \(=\) \(\ds \sum_{x \mathop \in \map \supp X} x^2 \map \Pr {X = x}\)
\(\ds \) \(=\) \(\ds c^2 \map \Pr {X = c}\)
\(\ds \) \(=\) \(\ds c^2\)

Hence, from Variance as Expectation of Square minus Square of Expectation:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\)
\(\ds \) \(=\) \(\ds c^2 - c^2\)
\(\ds \) \(=\) \(\ds 0\)

$\Box$

Necessary Condition

Suppose now that $\var X = 0$.

By definition of variance:

$\expect {\paren {X - \expect X}^2} = 0$

From Condition for Expectation of Non-Negative Random Variable to be Zero:

$\map \Pr {\paren {X - \expect X}^2 = 0} = 1$

That is:

$\map \Pr {X - \expect X = 0} = 1$

giving:

$\map \Pr {X = \expect X} = 1$

Setting $c = \expect X \in \R$, we see that we have $c \in \R$ such that:

$\map \Pr {X = c} = 1$

$\blacksquare$