Variance of Logistic Distribution

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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:

$\var X = \dfrac {s^2 \pi^2} 3$


Lemma 1

$\ds \int_{\to 0}^{\to 1} \map {\ln^2} {1 - u} \rd u = 2$


Lemma 2

$\ds \int_{\to 0}^{\to 1} \map {\ln^2} u \rd u = 2$


Lemma 3

$\ds \int_{\to 0}^{\to 1} \map \ln u \map \ln {1 - u} \rd u = 2 - \dfrac {\pi^2} 6$


Lemma 4

$\ds \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u = \dfrac {\pi^2} 3$


Proof 1

From the definition of the logistic distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$

From Variance as Expectation of Square minus Square of Expectation:

$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$

So:

$\ds \var X = \frac 1 s \int_{-\infty}^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x - \mu^2$

let:

\(\ds u\) \(=\) \(\ds \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}\) Integration by Substitution
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-2} \paren {-\frac 1 s \map \exp {-\dfrac {\paren {x - \mu} } s} }\) Power Rule for Derivatives, Chain Rule for Derivatives and Derivative of Exponential Function: Corollary 1
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 u - 1\) \(=\) \(\ds \paren {\map \exp {-\dfrac {\paren {x - \mu} } s} }\)
\(\ds \leadsto \ \ \) \(\ds \map \ln {\dfrac 1 u - 1}\) \(=\) \(\ds -\dfrac {\paren {x - \mu} } s\)
\(\ds \leadsto \ \ \) \(\ds -s \map \ln {\dfrac {1 - u} u} + \mu\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \mu^2 -2 s \mu \map \ln {\dfrac {1 - u} u} + s^2 \map {\ln^2} {\dfrac {1 - u} u}\) \(=\) \(\ds x^2\)


and also:

\(\ds \lim_{x \mathop \to -\infty} \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}\) \(=\) \(\ds 0\)
\(\ds \lim_{x \mathop \to \infty} \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}\) \(=\) \(\ds 1\)


Then:

\(\ds \var X\) \(=\) \(\ds \int_{\to 0}^{\to 1} \paren {\mu^2 - 2 s \mu \map \ln {\dfrac {1 - u} u} + s^2 \map {\ln^2} {\dfrac {1 - u} u} } \rd u - \mu^2\)
\(\ds \) \(=\) \(\ds \mu^2 \int_{\to 0}^{\to 1} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u - \mu^2\)
\(\ds \) \(=\) \(\ds \mu^2 - 2 s \mu \paren {\int_{\to 0}^{\to 1} \map \ln {1 - u} \rd u - \int_{\to 0}^{\to 1} \map \ln u \rd u } + s^2 \paren {\int_{\to 0}^{\to 1} \map {\ln^2} {1 - u} \rd u - 2 \int_{\to 0}^{\to 1} \map \ln {1 - u} \map \ln u \rd u + \int_{\to 0}^{\to 1} \map {\ln^2} u \rd u } - \mu^2\) Definite Integral of Constant and Difference of Logarithms
\(\ds \) \(=\) \(\ds \mu^2 - 2 s \mu \paren {\paren {-1} - \paren {-1} } + s^2 \paren {2 - 2 \paren {2 - \dfrac {\pi^2} 6 } + 2 } - \mu^2\) Expectation of Logistic Distribution:Lemma 1, Expectation of Logistic Distribution:Lemma 2, Lemma 1, Lemma 2 and Lemma 3
\(\ds \) \(=\) \(\ds \dfrac {s^2 \pi^2} 3\)

$\blacksquare$


Proof 2

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:

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

$\blacksquare$


Proof 3

From the definition of the logistic distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$

From Variance as Expectation of Square minus Square of Expectation:

$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$

So:

\(\ds \var X\) \(=\) \(\ds \frac 1 s \int_{-\infty}^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x - \mu^2\)
\(\ds \) \(=\) \(\ds \frac 1 s \int_{-\infty}^\mu \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x + \frac 1 s \int_\mu^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x\ - \mu^2\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \frac 1 s \int_{-\infty}^0 \dfrac {\paren {sz + \mu}^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \paren {s\rd z } + \frac 1 s \int_0^\infty \dfrac {\paren {sz + \mu}^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \paren {s\rd z } - \mu^2\) $z = \dfrac {\paren {x - \mu} } s$; $s\rd z = \rd x$
\(\ds \) \(=\) \(\ds \int_{-\infty}^0 \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z\) cleaning up; s and $\mu^2$ cancel
\(\ds \) \(=\) \(\ds \int_0^\infty \dfrac {\paren {s^2z^2 - 2sz\mu} \map \exp {z} } {\paren {1 + \map \exp {z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z\) rewriting the first integral
\(\ds \) \(=\) \(\ds \int_0^\infty \dfrac {\paren {s^2z^2 - 2sz\mu} \map \exp {z} } {\map \exp {2 z} \paren {1 + \map \exp {-z} }^2} \rd z + \int_0^\infty \dfrac {\paren {s^2z^2 + 2sz\mu} \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z\) extracting $\map \exp {2 z}$ from the denominator of the first integral
\(\ds \) \(=\) \(\ds 2s^2 \int_0^\infty \dfrac {z^2 \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2} \rd z\) simplifying


From Sum of Infinite Geometric Sequence, we have:

$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \map \exp {-z}^n = \dfrac 1 {1 + \map \exp {-z} }$

Taking the derivative of both sides, we have:

\(\ds \sum_{n \mathop = 0}^\infty \paren {-n} \paren {-1}^n \map \exp {-z}^n\) \(=\) \(\ds \dfrac { \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2}\)
\(\ds \sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \map \exp {-z}^n\) \(=\) \(\ds \dfrac { \map \exp {-z} } {\paren {1 + \map \exp {-z} }^2}\) simplifying


Therefore:


\(\ds \var X\) \(=\) \(\ds 2 s^2 \paren {\int_0^\infty z^2 \sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \map \exp {-z}^n \rd z}\) substitution from above
\(\ds \) \(=\) \(\ds 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \int_0^\infty z^2 \map \exp {-n z} \rd z}\) Fubini's Theorem
\(\ds \) \(=\) \(\ds 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \bigintlimits {-\dfrac 1 n \map \exp {-nz} \paren {z^2 + \frac {2 z} n + \frac 2 {n^2} } } 0 \infty }\) Primitive of x squared by Exponential of a x
\(\ds \) \(=\) \(\ds 2 s^2 \paren {\sum_{n \mathop = 1}^\infty n \paren {-1}^{n + 1} \paren {0 + \dfrac 1 n \paren {0^2 + \frac 0 n + \frac 2 {n^2} } } }\)
\(\ds \) \(=\) \(\ds 4 s^2 \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac 1 {n^2} }\)
\(\ds \) \(=\) \(\ds 4 s^2 \paren {\dfrac {\pi^2} {12} }\) Sum of Reciprocals of Squares Alternating in Sign
\(\ds \) \(=\) \(\ds \dfrac {s^2 \pi^2} 3\)


$\blacksquare$