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:
- $\map \Pr {X = c} = 1$
$\blacksquare$