Variance of Pareto Distribution

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Theorem

Let $X$ be a continuous random variable with the Pareto distribution with $a, b \in \R_{> 0}$.

Then the variance of $X$ is given by:

$\var X = \begin {cases} \dfrac {a b^2 } {\paren {a - 2} \paren {a - 1}^2 } & 2 < a \\ \text {does not exist} & 2 \ge a \end {cases}$


Proof

By Variance as Expectation of Square minus Square of Expectation, we have:

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

By Expectation of Pareto Distribution, we have:

$\ds \expect {X} = \begin {cases} \dfrac {a b} {a - 1} & 1 < a \\ \text {does not exist} & 1 \ge a \end {cases}$

From Raw Moment of Pareto Distribution, we have:

The $n$th raw moment $\expect {X^n}$ of $X$ is given by:

$\ds \expect {X^n} = \begin {cases} \dfrac {a b^n} {a - n} & n < a \\ \text {does not exist} & n \ge a \end {cases}$


Therefore, for $n = 2$ we have:

$\ds \expect {X^2} = \begin {cases} \dfrac {a b^2} {a - 2} & 2 < a \\ \text {does not exist} & 2 \ge a \end {cases}$

So:

\(\ds \var X\) \(=\) \(\ds \dfrac {a b^2} {a - 2} - \paren {\dfrac {a b} {a - 1} }^2\)
\(\ds \) \(=\) \(\ds \dfrac {a b^2 \paren {a - 1}^2 - a^2 b^2 \paren {a - 2} } {\paren {a - 2} \paren {a - 1}^2 }\)
\(\ds \) \(=\) \(\ds \dfrac {a b^2 } {\paren {a - 2} \paren {a - 1}^2 }\)

$\blacksquare$