こうやって教えてほしかったシリーズ①
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．
- $k , n \in \mathbb N^+$
- $\dfrac1{(-n)!}:=0$
- $p+q=1(0< p,q<1)$
- 降冪：$(x)_n$
- 昇冪：$[x]_n$
- 第二種スターリング数：$ \left\{n \atop i \right\}$
- 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$E[X^k]$
$$E[X^k]=\sum_{j=0}^k \left\{k\atop j\right\} E[(X)_j]$$
$$\begin{aligned} E[X^k] & = \sum_{i\in I} i^kP(X=i)\\ & = \sum_{i\in I} \sum_{j=0}^k \left\{k\atop j\right\}(i)_jP(X=i)\\ & = \sum_{j=0}^k \left\{k\atop j\right\} \sum_{i\in I} (i)_jP(X=i)\\ & = \sum_{j=0}^k \left\{k\atop j\right\} E[(X)_j] \end{aligned} $$
二項分布
$$P(X=i):=\binom{n}{i}p^iq^{n-i}\\$$
$$E[(X)_k]=(n)_k p^k$$
$$ \begin{aligned}E[(X)_k]&=\sum_{i=0}^n (i)_k\binom{n}{i}p^iq^{n-i}\\&=\sum_{i=0}^n (n)_k\binom{n-k}{i-k}p^iq^{n-i}\\&=(n)_k p^k\sum_{i=0}^n \binom{n-k}{i-k}p^{i-k}q^{(n-k)-(i-k)}\\&=(n)_k p^k (p+q)^{n-k}\\&=(n)_k p^k.\end{aligned} $$
$$E[X]=np,V[X]=npq$$
$$E[X]=(n)_1 p^1=np.$$
$$ \begin{aligned}E[X^2]&=E[(X)_2]+E[X]\\&=n(n-1)p^2+np\\&=(np)^2+npq\end{aligned} $$
$$ V[X]=(np)^2+npq-(np)^2=npq.$$
一応モーメントも載せておく．
$$ E[X^k]=\sum_{j=0}^k \left\{k\atop j\right\} (n)_j p^j$$
$$ E[X^k]= \sum_{j=0}^k \left\{k\atop j\right\} E[(X)_j]=\sum_{j=0}^k \left\{k\atop j\right\} (n)_j p^j．$$
Poisson分布
$$P(X=i):=\frac{\lambda^i}{i!}e^{-\lambda}$$
$$ E[(X)_k]=\lambda^{k} $$
$$ \begin{aligned} E[(X)_k] & = \sum_{i=0}^\infty (i)_k \frac{\lambda^i}{i!}e^{-\lambda}\\ & = \lambda^{k}e^{-\lambda}\sum_{i=0}^\infty \frac{\lambda^{i-k}}{(i-k)!}\\ & = \lambda^{k}e^{-\lambda}e^{\lambda}\\& = \lambda^{k}. \end{aligned} $$
$$E[X]=\lambda,V[X]=\lambda$$
$$E[X]=\lambda^1=\lambda.$$
$$ \begin{aligned}E[X^2]&=E[(X)_2]+E[X]\\&=\lambda^2+\lambda\\\end{aligned} $$
$$ V[X]=\lambda^2+\lambda-\lambda^2=\lambda.$$
一応モーメントも載せておく．
$$ E[X^k]=\sum_{j=0}^k\left\{k\atop j\right\}\lambda^{j}$$
$$ E[X^k]= \sum_{j=0}^k \left\{k\atop j\right\} E[(X)_j]=\sum_{j=0}^k \left\{k\atop j\right\} \lambda^{j}．$$
幾何分布
$$P(X=i):=pq^{i-1}$$
$$ E[(X)_k]=k!\frac{q^{k-1}}{p^{k}} $$
$$ \begin{aligned} E[(X)_k] & = \sum_{i=0}^\infty (i)_k pq^{i-1}\\ & = pq^{k-1}k!\sum_{i=0}^\infty \frac{(k+(i-k))!}{(i-k)!k!} q^{i-k}\\ & = pq^{k-1}k!\sum_{i=0}^\infty \frac{[k+1]_{i-k}}{(i-k)!}q^{i-k}\\ & = pq^{k-1}\frac{k!}{(1-q)^{k+1}}\\ & = k!\frac{q^{k-1}}{p^{k}}\\ \end{aligned} $$
$$E[X]=\frac1p,V[X]=\frac{q}{p^2}$$
$$E[X]=1!\frac{q^0}{p^{1}}=\frac1p.$$
$$ \begin{aligned} E[X^2] & = E[(X)_2]+E[X]\\ & = \frac{2q}{p^2}+\frac1p\\ \end{aligned} $$
$$ V[X]=\frac{2q}{p^2}+\frac1p-\frac1{p^2}=\frac{q}{p^2}.$$
一応モーメントも載せておく．
$$ E[X^k]=\sum_{j=0}^k \left\{k\atop j\right\}j!\frac{q^{j-1}}{p^{j}}$$
$$ E[X^k]= \sum_{j=0}^k \left\{k\atop j\right\} E[(X)_j]=\sum_{j=0}^k \left\{k\atop j\right\}j!\frac{q^{j-1}}{p^{j}}．$$