Skewness of Geometric Distribution/Formulation 2

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Theorem

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

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


Then the skewness of $X$ is given by:

$\gamma_1 = \dfrac {2 - p} {\sqrt {1 - p} }$


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 2, we have:

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

By Variance of Geometric Distribution: Formulation 2, we have:

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

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


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:

$\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^3} }$

From Moment Generating Function of Geometric Distribution: Third Moment:

$\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4} }$

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

$\map {M_X} 0 = \dfrac {\paren {1 - p} \paren {2 - p} } {p^2}$
$\map {M_X} 0 = \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3}$

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\expect {X^3} - 3 \paren {\dfrac {1 - p} p} \expect {X^2} + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } }\)
\(\ds \) \(=\) \(\ds \frac {\dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3} - 3 \paren {\dfrac {1 - p} p } \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2} {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } }\)
\(\ds \) \(=\) \(\ds \frac {\paren {6 - 6 p + p^2} - 3 \paren {1 - p} \paren {2 - p} + 2 \paren {1 - p}^2} {\sqrt {1 - p} }\) $\dfrac {1 - p} {p^3}$ cancels
\(\ds \) \(=\) \(\ds \frac {6 - 6 p + p^2 - 6 + 9 p - 3 p^2 + 2 - 4 p + 2 p^2} {\sqrt {1 - p} }\)
\(\ds \) \(=\) \(\ds \dfrac {2 - p} {\sqrt {1 - p} }\)

$\blacksquare$

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