Theorem: Variance of Hat-Check Distribution

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:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
      | r = \sum_{k \mathop = 0}^n {k^2 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }
      | c = {{Defof|Hat-Check Distribution}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n {k^2 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }
      | c = as the $k = 0$ term vanishes
}}
{{eqn | r = \sum_{y \mathop = n - 1}^0 \paren {n - y }^2 \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}
      | c = Let $y = n - k$
}}
{{eqn | r = 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!}
      | c = 
}}
{{eqn | r = 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!}
      | c = Let $y = n - k$
}}
{{eqn | r = 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!}
      | c = adding $0$
}}
{{eqn | r = 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!}
      | c = 
}}
{{eqn | r = 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!}
      | c = as the $k = n$ term vanishes
}}
{{eqn | r = 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!}
      | c = canceling terms
}}
{{eqn | r = 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!}
      | c = Hat-Check Distribution Gives Rise to Probability Mass Function
}}
{{eqn | r = n^2 - 2n + n - \paren {n - 2}
      | c = Hat-Check Distribution Gives Rise to Probability Mass Function and Expectation of Hat-Check Distribution
}}
{{eqn | r = n^2 - 2n + 2
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \var X
      | r = \expect {X^2} - \paren {\expect X}^2
      | c = 
}}
{{eqn | r = n^2 - 2n + 2 - \paren {n - 1}^2
      | c = Expectation of Hat-Check Distribution: $\expect X = n - 1$
}}
{{eqn | r = \paren {n - 1}^2 + 1 - \paren {n - 1}^2
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
{{qed}}