Variance of Hat-Check Distribution

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Theorem

Let $X$ be a discrete random variable with the Hat-Check distribution with parameter $n$.


Then the variance of $X$ is given by:

$\var X = 1$


Proof

From the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:

$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$


So:

\(\ds \expect {X^2}\) \(=\) \(\ds \sum_{k \mathop = 0}^n {k^2 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }\) Definition of Hat-Check Distribution
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n {k^2 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }\) as the $k = 0$ term vanishes
\(\ds \) \(=\) \(\ds \sum_{y \mathop = n - 1}^0 \paren {n - y }^2 \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}\) Let $y = n - k$
\(\ds \) \(=\) \(\ds n^2 \sum_{y \mathop = 0}^{n - 1} \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - 2n \sum_{y \mathop = 0}^{n - 1} \dfrac y {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} + \sum_{y \mathop = 0}^{n - 1} \dfrac {y^2} {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}\)
\(\ds \) \(=\) \(\ds n^2 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 2n \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) Let $y = n - k$
\(\ds \) \(=\) \(\ds n^2 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 2n \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 2 \dfrac {n^2} {n!} - 2 \dfrac {n^2} {n!}\) adding $0$
\(\ds \) \(=\) \(\ds n^2 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 2n \sum_{k \mathop = 0}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \sum_{k \mathop = 0}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\)
\(\ds \) \(=\) \(\ds n^2 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 2n \sum_{k \mathop = 0}^{n - 1} \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) as the $k = n$ term vanishes
\(\ds \) \(=\) \(\ds n^2 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 2n \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k} } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) canceling terms
\(\ds \) \(=\) \(\ds n^2 - 2n + n \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac k {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}\) Hat-Check Distribution Gives Rise to Probability Mass Function
\(\ds \) \(=\) \(\ds n^2 - 2n + n - \paren {n - 2}\) Hat-Check Distribution Gives Rise to Probability Mass Function and Expectation of Hat-Check Distribution
\(\ds \) \(=\) \(\ds n^2 - 2n + 2\)


Then:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\)
\(\ds \) \(=\) \(\ds n^2 - 2n + 2 - \paren {n - 1}^2\) Expectation of Hat-Check Distribution: $\expect X = n - 1$
\(\ds \) \(=\) \(\ds \paren {n - 1}^2 + 1 - \paren {n - 1}^2\)
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Also see

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