Expectation 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 expectation of $X$ is given by:

$\expect X = n - 1$


Proof

From the definition of expectation:

$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$

By definition of hat-check distribution:

$\ds \expect X = \sum_{k \mathop = 0}^n k \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}$

Then:

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

$\blacksquare$


Also see

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