Skewness of Logistic Distribution

Theorem

Let $X$ be a continuous random variable which satisfies the logistic distribution:

$X \sim \map {\operatorname {Logistic} } {\mu, s}$

Then the skewness $\gamma_1$ of $X$ is equal to $0$.

Proof

From Skewness in terms of Non-Central Moments, we have:

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

where:

$\mu$ is the expectation of $X$.

$\sigma$ is the standard deviation of $X$.

By Expectation of Logistic Distribution we have:

$\mu = \mu$

By Variance of Logistic Distribution we have:

$\sigma = \dfrac {s \pi} {\sqrt 3}$

To calculate $\gamma_1$, we must calculate $\expect {X^3}$.

From Moment in terms of Moment Generating Function:

$\expect {X^n} = \map { {M_X}^{\paren n} } 0$

where $M_X$ is the moment generating function of $X$.

From Derivatives of Moment Generating Function of Logistic Distribution:

$\ds \map { {M_X}} 0 = \mu^3 \int_{\to 0}^{\to 1} \rd u - 3 \mu^2 s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + 3 \mu s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u - s^3 \int_{\to 0}^{\to 1} \map {\ln^3} {\dfrac {1 - u} u} \rd u $

Therefore:

\(\ds \expect {X^3}\)

\(=\)

\(\ds \mu^3 - 3 \mu^2 s \paren {0} + 3 \mu s^2 \paren {\dfrac {\pi^2} 3} - s^3 \paren {0}\)

Variance of Logistic Distribution/Lemma 4 and Expectation of Logistic Distribution/Lemma 3

\(\ds \)

\(=\)

\(\ds \mu^3 + \mu s^2 \pi^2\)

So:

\(\ds \gamma_1\)

\(=\)

\(\ds \frac {\mu^3 + \mu s^2 \pi^2 - 3 \mu \paren {\dfrac {s^2 \pi^2} 3} - \mu^3} {\paren {\dfrac {s \pi} {\sqrt 3} }^3}\)

\(\ds \)

\(=\)

\(\ds 0\)

$\blacksquare$