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 $\size t < \dfrac 1 s$.
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \mu\) |
$\blacksquare$