Equivalence of Definitions of Variance of Discrete Random Variable
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Theorem
Let $X$ be a discrete random variable.
Let $\mu = \expect X$ be the expectation of $X$.
The following definitions of the concept of Variance of Discrete Random Variable are equivalent:
Definition 1
- $\var X := \expect {\paren {X - \expect X}^2}$
That is: it is the expectation of the squares of the deviations from the expectation.
Definition 2
- $\ds \var X := \sum_{x \mathop \in \Omega_X} \paren {x - \mu^2} \map \Pr {X = x}$
where:
- $\mu := \expect X$ is the expectation of $X$
- $\Omega_X$ is the image of $X$
- $\map \Pr {X = x}$ is the probability mass function of $X$.
Definition 3
- $\var X := \expect {X^2} - \paren {\expect X}^2$
Proof
Definition 1 equivalent to Definition 2
Let $\var X$ be defined as:
- $\var X := \expect {\paren {X - \expect X}^2}$
Let $\mu = \expect X$.
Let $\map f X = \paren {X - \mu}^2$ be considered as a function of $X$.
Then by applying Expectation of Function of Discrete Random Variable:
- $\ds \expect {\map f X} = \sum_{x \mathop \in \Omega_X} \map f x \, \map \Pr {X = x}$
which leads to:
- $\ds \expect {\paren {X - \mu}^2} = \sum_{x \mathop \in \Omega_X} \paren {\paren {X - \mu}^2} \map \Pr {X = x}$
thus demonstrating the equality of Definition 1 to Definition 2.
$\Box$
Definition 2 equivalent to Definition 3
Let $\mu = \expect X$, and take the expression for variance:
- $\ds \var X := \sum_{x \mathop \in \Img X} \paren {x - \mu}^2 \map \Pr {X = x}$
Then from Variance as Expectation of Square minus Square of Expectation:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation