# 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