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:
- $\expect X = \begin {cases} \dfrac {a b} {a - 1} & 1 < a \\ \text {does not exist} & 1 \ge a \end {cases}$
From Raw Moment of Pareto Distribution, we have:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by:
- $\expect {X^n} = \begin {cases} \dfrac {a b^n} {a - n} & n < a \\ \text {does not exist} & n \ge a \end {cases}$
Therefore, for $n = 2$ we have:
- $\expect {X^2} = \begin {cases} \dfrac {a b^2} {a - 2} & 2 < a \\ \text {does not exist} & 2 \ge a \end {cases}$
So:
\(\ds \var X\) | \(=\) | \(\ds \dfrac {a b^2} {a - 2} - \paren {\dfrac {a b} {a - 1} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a b^2 \paren {a - 1}^2 - a^2 b^2 \paren {a - 2} } {\paren {a - 2} \paren {a - 1}^2 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a b^2 } {\paren {a - 2} \paren {a - 1}^2 }\) |
$\blacksquare$