大学数学基礎解説

こうやって教えてほしかったシリーズ①

統計学,基礎

121

はじめに

学部の1年がやる程度の統計学の公式を，ある程度級数や組み合わせ論の知識を持つ人に向けた解法で証明する．
以下の諸定義に注意せよ．

$k, n \in N^{+}$
$\frac{1}{(- n)!} := 0$
$p + q = 1 (0 < p, q < 1)$
降冪： $(x)_{n}$
昇冪： $[x]_{n}$
第二種スターリング数： ${\binom{n}{i}}$
確率変数 $X = i$ なる確率： $P (X = i)$
期待値： $E [X] : ＝ \sum_{i \in I} i P (X = i)$
分散： $V [X] := E [(X - E [X])^{2}] = E [X^{2}] - E [X]^{2}$

準備

モーメント $E [X^{k}]$

モーメント

E [X^{k}]

$E [X^{k}] = \sum_{j = 0}^{k} {\binom{k}{j}} E [(X)_{j}]$

$\begin{aligned} E [X^{k}] & = \sum_{i \in I} i^{k} P (X = i) \\ = \sum_{i \in I} \sum_{j = 0}^{k} {\binom{k}{j}} (i)_{j} P (X = i) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} \sum_{i \in I} (i)_{j} P (X = i) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} E [(X)_{j}] \end{aligned}$

二項分布

確率

$P (X = i) := (\binom{n}{i}) p^{i} q^{n - i}$

E [(X)_{k}]

$E [(X)_{k}] = (n)_{k} p^{k}$

$\begin{aligned} E [(X)_{k}] & = \sum_{i = 0}^{n} (i)_{k} (\binom{n}{i}) p^{i} q^{n - i} \\ = \sum_{i = 0}^{n} (n)_{k} (\binom{n - k}{i - k}) p^{i} q^{n - i} \\ = (n)_{k} p^{k} \sum_{i = 0}^{n} (\binom{n - k}{i - k}) p^{i - k} q^{(n - k) - (i - k)} \\ = (n)_{k} p^{k} (p + q)^{n - k} \\ = (n)_{k} p^{k} . \end{aligned}$

期待値と分散

$E [X] = n p, V [X] = n p q$

$E [X] = (n)_{1} p^{1} = n p .$

$\begin{aligned} E [X^{2}] & = E [(X)_{2}] + E [X] \\ = n (n - 1) p^{2} + n p \\ = (n p)^{2} + n p q \end{aligned}$

$V [X] = (n p)^{2} + n p q - (n p)^{2} = n p q .$

一応モーメントも載せておく．

モーメント

$E [X^{k}] = \sum_{j = 0}^{k} {\binom{k}{j}} (n)_{j} p^{j}$

$E [X^{k}] = \sum_{j = 0}^{k} {\binom{k}{j}} E [(X)_{j}] = \sum_{j = 0}^{k} {\binom{k}{j}} (n)_{j} p^{j} ．$

Poisson分布

確率

$P (X = i) := \frac{λ^{i}}{i!} e^{- λ}$

E [(X)_{k}]

$E [(X)_{k}] = λ^{k}$

$\begin{aligned} E [(X)_{k}] & = \sum_{i = 0}^{\infty} (i)_{k} \frac{λ^{i}}{i!} e^{- λ} \\ = λ^{k} e^{- λ} \sum_{i = 0}^{\infty} \frac{λ^{i - k}}{(i - k)!} \\ = λ^{k} e^{- λ} e^{λ} \\ = λ^{k} . \end{aligned}$

期待値と分散

$E [X] = λ, V [X] = λ$

$E [X] = λ^{1} = λ .$

$\begin{aligned} E [X^{2}] & = E [(X)_{2}] + E [X] \\ = λ^{2} + λ \end{aligned}$

$V [X] = λ^{2} + λ - λ^{2} = λ .$

一応モーメントも載せておく．

モーメント

$E [X^{k}] = \sum_{j = 0}^{k} {\binom{k}{j}} λ^{j}$

$E [X^{k}] = \sum_{j = 0}^{k} {\binom{k}{j}} E [(X)_{j}] = \sum_{j = 0}^{k} {\binom{k}{j}} λ^{j} ．$

幾何分布

確率

$P (X = i) := p q^{i - 1}$

E [(X)_{k}]

$E [(X)_{k}] = k! \frac{q^{k - 1}}{p^{k}}$

$\begin{aligned} E [(X)_{k}] & = \sum_{i = 0}^{\infty} (i)_{k} p q^{i - 1} \\ = p q^{k - 1} k! \sum_{i = 0}^{\infty} \frac{(k + (i - k))!}{(i - k)! k!} q^{i - k} \\ = p q^{k - 1} k! \sum_{i = 0}^{\infty} \frac{[k + 1]_{i - k}}{(i - k)!} q^{i - k} \\ = p q^{k - 1} \frac{k!}{(1 - q)^{k + 1}} \\ = k! \frac{q^{k - 1}}{p^{k}} \end{aligned}$