Variance as Expectation of Square minus Square of Expectation/Discrete
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Theorem
Let $X$ be a discrete random variable.
Then the variance of $X$ can be expressed as:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.
Proof
We let $\mu = \expect X$, and take the expression for variance:
- $\var X := \ds \sum_{x \mathop \in \Img X} \paren {x - \mu}^2 \map \Pr {X = x}$
Then:
\(\ds \var X\) | \(=\) | \(\ds \sum_x \paren {x^2 - 2 \mu x + \mu^2} \map \Pr {X = x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_x x^2 \map \Pr {X = x} - 2 \mu \sum_x x \map \Pr {X = x} + \mu^2 \sum_x \map \Pr {X = x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_x x^2 \map \Pr {X = x} - 2 \mu \sum_x x \map \Pr {X = x} + \mu^2\) | Definition of Probability Mass Function: $\ds \sum_x \map \Pr {X = x} = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_x x^2 \map \Pr {X = x} - 2 \mu^2 + \mu^2\) | Definition of Expectation: $\ds \sum_x x \map \Pr {X = x} = \mu$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_x x^2 \map \Pr {X = x} - \mu^2\) |
Hence the result, from $\mu = \expect X$.
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: $(22)$