Skewness of Geometric Distribution/Formulation 1

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Theorem

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.

$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
$\map \Pr {X = k} = \paren {1 - p} p^k$


Then the skewness of $X$ is given by:

$\gamma_1 = \dfrac {1 + p} {\sqrt p}$


Proof

From the definition of skewness, we have:

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

where:

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

By Expectation of Geometric Distribution: Formulation 1, we have:

$\mu = \dfrac p {1 - p}$

By Variance of Geometric Distribution: Formulation 1, we have:

$\sigma = \dfrac {\sqrt p} {1 - p}$

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}\) Cube of Difference
\(\ds \) \(=\) \(\ds \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3}\) Expectation is Linear
\(\ds \) \(=\) \(\ds \frac {\expect {X^3} - 3 \paren {\dfrac p {1 - p} } \expect {X^2} + 2 \paren {\dfrac {p^3 } {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }\)


To calculate $\gamma_1$, we must calculate both $\expect {X^2}$ and $\expect {X^3}$.

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 Moment Generating Function of Geometric Distribution: Second Moment:

$\map { {M_X}''} t = p \paren {1 - p} e^t \paren {\dfrac {1 + p e^t} {\paren {1 - p e^t}^3} }$

From Moment Generating Function of Geometric Distribution: Third Moment:

$\map { {M_X}'''} t = p \paren {1 - p } e^t \paren {\dfrac {1 + 4p e^t + p^2 e^{2 t} } {\paren {1 - p e^t}^4 } }$

Setting $t = 0$ and from Exponential of Zero, we have:

$\map {M_X''} 0 = \dfrac {p \paren {1 + p} } {\paren {1 - p}^2}$
$\map {M_X'''} 0 = \dfrac {p \paren {1 + 4 p + p^2} } {\paren {1 - p}^3}$

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\expect {X^3} - 3 \paren {\dfrac p {1 - p} } \expect {X^2} + 2 \paren {\dfrac {p^3} {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }\)
\(\ds \) \(=\) \(\ds \frac {\dfrac {p \paren {1 + 4 p + p^2} } {\paren {1 - p}^3} - 3 \paren {\dfrac p {1 - p} } \dfrac {p \paren {1 + p} } {\paren {1 - p}^2} + 2 \paren {\dfrac {p^3} {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 + 4 p + p^2} - 3 p \paren {1 + p} + 2 p^2} {\sqrt p}\) $\dfrac p {\paren {1 - p}^3}$ cancels
\(\ds \) \(=\) \(\ds \dfrac {1 + p} {\sqrt p}\)

$\blacksquare$