Skewness of Hat-Check Distribution

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Theorem

Let $X$ be a discrete random variable with a Hat-Check distribution with parameter $n$. ($n \gt 2$)

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = -1$

Proof

From Skewness in terms of Non-Central Moments:

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

where:

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

We have, by Expectation of Hat-Check Distribution:

$\expect X = n - 1$

By Variance of Hat-Check Distribution:

$\var X = \sigma^2 = 1$

so:

$\sigma = \sqrt 1 = 1$


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

\(\ds \expect {X^3}\) \(=\) \(\ds \sum_{k \mathop = 0}^n {k^3 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }\) Definition of Hat-Check Distribution
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n {k^3 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }\) as the $k = 0$ term vanishes
\(\ds \) \(=\) \(\ds \sum_{y \mathop = n - 1}^0 \paren {n - y }^3 \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}\) Let $y = n - k$
\(\ds \) \(=\) \(\ds n^3 \sum_{y \mathop = 0}^{n - 1} \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{y \mathop = 0}^{n - 1} \dfrac y {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} + 3n \sum_{y \mathop = 0}^{n - 1} \dfrac {y^2} {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - \sum_{y \mathop = 0}^{n - 1} \dfrac {y^3} {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}\)
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) Let $y = n - k$
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 4 \dfrac {n^3} {n!} - 4 \dfrac {n^3} {n!}\) adding $0$ - re-index sums
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\)
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) as the $k = n$ term vanishes
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k} } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^2 } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) canceling terms
\(\ds \) \(=\) \(\ds n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n \sum_{k \mathop = 0}^{n - 1} \dfrac k {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 2n \sum_{k \mathop = 0}^{n - 1} \dfrac k {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {k^2 } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\)

To help complete the sum above, recall that:

\(\ds \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) \(=\) \(\ds 1\) Hat-Check Distribution Gives Rise to Probability Mass Function
\(\ds \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) \(=\) \(\ds 1\) Hat-Check Distribution Gives Rise to Probability Mass Function
\(\ds \sum_{k \mathop = 0}^{n - 1} \dfrac k {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) \(=\) \(\ds \paren {n - 1} - 1\) Expectation of Hat-Check Distribution
\(\ds \) \(=\) \(\ds \paren {n - 2}\)
\(\ds \sum_{k \mathop = 0}^{n - 1} \dfrac {k^2} {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) \(=\) \(\ds \paren{\paren {n - 1} - 1}^2 + 1\) Variance of Hat-Check Distribution
\(\ds \) \(=\) \(\ds \paren{n^2 -4n + 5}\)

Therefore:

\(\ds \) \(=\) \(\ds n^3 - 3n^2 + 3n^2 - 3n \paren {n - 2} - n^2 + 2n \paren {n - 2} - \paren {n^2 - 4n + 5}\)
\(\ds \) \(=\) \(\ds n^3 + \paren {- 3 + 3 - 3 - 1 + 2 - 1}n^2 + \paren {6 - 4 + 4}n - 5\)
\(\ds \) \(=\) \(\ds n^3 - 3n^2 + 6n - 5\)


So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\paren {n^3 - 3n^2 + 6n - 5 } - 3 \paren {n - 1} 1 - \paren {n - 1}^3} {1^{3/2} }\)
\(\ds \) \(=\) \(\ds \paren {n^3 - 3n^2 + 6n - 5 } - 3n + 3 - \paren {n^3 - 3n^2 + 3n - 1}\)
\(\ds \) \(=\) \(\ds -1\)

$\blacksquare$

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