Expectation of Logistic Distribution
Jump to navigation
Jump to search
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mu\) |
$\blacksquare$