# Expectation 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 expectation of $X$ is given by:

$\expect X = \mu$

## 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 $\size t < \dfrac 1 s$.

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

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 $\ds$ $=$ $\ds \mu$

$\blacksquare$