Expectation 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 expectation of $X$ is given by:

$\expect X = \mu$


Lemma 1

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


Lemma 2

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


Lemma 3

$\forall k \in \N$:
$\ds \int_{\to 0}^{\to 1} \map {\ln^{2k + 1} } {\dfrac {1 - u} u} \rd u = 0$


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 the definition of the expected value of a continuous random variable:

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

So:

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


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
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 u - 1\) \(=\) \(\ds \paren {\map \exp {-\dfrac {\paren {x - \mu} } s} }\)
\(\ds \leadsto \ \ \) \(\ds \map \ln {\dfrac 1 u - 1}\) \(=\) \(\ds -\dfrac {\paren {x - \mu} } s\)
\(\ds \leadsto \ \ \) \(\ds -s \map \ln {\dfrac {1 - u} u} + \mu\) \(=\) \(\ds x\)


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:

\(\ds \expect X\) \(=\) \(\ds \int_{\to 0}^{\to 1} \paren {-s \map \ln {\dfrac {1 - u} u} + \mu} \rd u\)
\(\ds \) \(=\) \(\ds -s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + \mu \int_{\to 0}^{\to 1} \rd u\)
\(\ds \) \(=\) \(\ds -s \paren {\int_{\to 0}^{\to 1} \map \ln {1 - u} \rd u - \int_{\to 0}^{\to 1} \map \ln u \rd u} + \mu\) Integral of Constant and Difference of Logarithms
\(\ds \) \(=\) \(\ds -s \paren {\paren {-1} - \paren {-1} } + \mu\) Lemma 1 and Lemma 2
\(\ds \) \(=\) \(\ds \mu\)

$\blacksquare$


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 Moment in terms of Moment Generating Function:

$\expect X = \map { {M_X}'} 0$

From Moment Generating Function of Logistic Distribution: First Moment:

$\ds \map { {M_X}'} t = \map \exp {\mu t} \paren {\mu \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u - s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u}$

Hence setting $t = 0$:

\(\ds \map { {M_X}'} 0\) \(=\) \(\ds \mu \int_{\to 0}^{\to 1} \rd u - s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u\)
\(\ds \) \(=\) \(\ds \mu - s \paren {0 }\) Definite Integral of Constant and Lemma 3
\(\ds \) \(=\) \(\ds \mu\)

$\blacksquare$