Skewness of Logistic Distribution
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Theorem
Let $X$ be a continuous random variable which satisfies the logistic distribution:
- $X \sim \map {\operatorname {Logistic} } {\mu, s}$
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:
\(\ds \gamma_1\) | \(=\) | \(\ds \frac {\mu^3 + \mu s^2 \pi^2 - 3 \mu \paren {\dfrac {s^2 \pi^2} 3} - \mu^3} {\paren {\dfrac {s \pi} {\sqrt 3} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
$\blacksquare$