Variance of Logistic Distribution/Proof 2
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Theorem
Let $X$ be a continuous random variable which satisfies the logistic distribution:
- $X \sim \map {\operatorname {Logistic} } {\mu, s}$
The variance of $X$ is given by:
- $\var X = \dfrac {s^2 \pi^2} 3$
Proof
By Moment Generating Function of Logistic Distribution, the moment generating function of $X$ is given by:
- $\ds \map {M_X} t = \map \exp {\mu t} \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
for $\size t < \dfrac 1 s$.
From Variance as Expectation of Square minus Square of Expectation:
- $\ds \var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Logistic Distribution, we have:
- $\expect X = \mu$
From Moment Generating Function of Logistic Distribution: Second Moment:
- $\ds \map { {M_X}} t = \map \exp {\mu t} \paren {\mu^2 \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u}$
Hence setting $t = 0$:
\(\ds \map { {M_X}} 0\) | \(=\) | \(\ds \mu^2 \int_{\to 0}^{\to 1} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mu^2 - 2 s \mu \paren {0 } + s^2 \paren {\dfrac {\pi^2} 3}\) | Definite Integral of Constant, Expectation of Logistic Distribution: Lemma 3, and Lemma 4 | |||||||||||
\(\ds \) | \(=\) | \(\ds \mu^2 + \dfrac {s^2 \pi^2} 3\) |
So:
\(\ds \var X\) | \(=\) | \(\ds \mu^2 + \dfrac {s^2 \pi^2} 3 - \mu^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {s^2 \pi^2} 3\) |
$\blacksquare$