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:

\(\ds \var X\) \(=\) \(\ds \mu^2 + \dfrac {s^2 \pi^2} 3 - \mu^2\)
\(\ds \) \(=\) \(\ds \dfrac {s^2 \pi^2} 3\)

$\blacksquare$