はじめに
「入門・演習 数理統計」の演習問題の自作解答を紹介します。
|
間違い等発見されましたらご指摘ください。
他の解答はこちらから。
なお、問題文は(必要がない限り)掲載しておりません。テキストを横に置いてご覧ください。
また、スマートフォン等では数式が画面からはみ出る場合があります。数式部分は横スクロールできます。
スポンサーリンク
$\newcommand{\lnl}{\\[8pt]}$ $\newcommand{\Lnl}{\\[18pt]}$ $\newcommand{\delt}{\mathrm{d}}$ $\newcommand{\comb}{\mathrm{C}}$ $\DeclareMathOperator*{\ssum}{\Sigma}$ $\DeclareMathOperator*{\sprod}{\Pi}$
ex4.2.3
(i)
$Y_i = (X_i – \mu)$とおく.
\begin{align}
(\overline{X}_n – \mu)^3 &= \left\{\left(\frac{1}{n}\sum_{i=1}^n X_i\right) – \mu\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n (X_i – \mu )\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n Y_i\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n {Y_i}^3 + \underline{ 3\sum_{i\neq j} Y_i {Y_j}^2 + 6\sum_{i\neq j \neq k \neq i}Y_i Y_j Y_k}\right\}
\end{align}
(\overline{X}_n – \mu)^3 &= \left\{\left(\frac{1}{n}\sum_{i=1}^n X_i\right) – \mu\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n (X_i – \mu )\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n Y_i\right\}^3\lnl
&= \frac{1}{n^3}\left\{\sum_{i=1}^n {Y_i}^3 + \underline{ 3\sum_{i\neq j} Y_i {Y_j}^2 + 6\sum_{i\neq j \neq k \neq i}Y_i Y_j Y_k}\right\}
\end{align}
ここで, $E(Y_i) = E(X_i-\mu) = 0$であることから,上記の期待値をとると下線部は$0$となるので,($Y_i$の表記を$(X_i-\mu)$に戻して)
\begin{align}
E(\overline{X}_n – \mu)^3 &= \frac{1}{n^3} \cdot E\left(\sum_{i=1}^n (X_i-\mu)^3\right)\lnl
&= \frac{1}{n^3} \cdot \sum_{i=1}^n E((X_i – \mu)^3)\lnl
&= \frac{1}{n^3} \cdot \sum_{i=1}^n \mu_3\lnl
&= \frac{\mu_3}{n^2}
\end{align}
E(\overline{X}_n – \mu)^3 &= \frac{1}{n^3} \cdot E\left(\sum_{i=1}^n (X_i-\mu)^3\right)\lnl
&= \frac{1}{n^3} \cdot \sum_{i=1}^n E((X_i – \mu)^3)\lnl
&= \frac{1}{n^3} \cdot \sum_{i=1}^n \mu_3\lnl
&= \frac{\mu_3}{n^2}
\end{align}
よって示された.
(ii)
はじめに以下計算しておく,(少し長いが)
\begin{align}
E({\overline{X}_n}^2) &= E\left\{\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\right\}\lnl
&= \frac{1}{n^2}E\left(\sum_{i=1}^n {X_i}^2 + \sum_{i\neq j}X_i X_j\right)\lnl
&= \frac{1}{n}E(X^2) + \frac{n-1}{n}\mu^2\Lnl
(\overline{X}_n – \mu)^3 &= {\overline{X}_n}^3 – 3\mu{\overline{X}_n}^2 + 3\mu^2 {\overline{X}_n} – \mu^3\Lnl
\therefore E((\overline{X}_n – \mu)^3) = \frac{\mu_3}{n^2} &= E({\overline{X}_n}^3) – 3\mu E({\overline{X}_n}^2) + 2\mu^3\lnl
&= E({\overline{X}_n}^3) – 3\mu \left( \frac{1}{n}E(X^2) + \frac{n-1}{n} \mu^2\right) + 2\mu^3\lnl
&= E({\overline{X}_n}^3) – \frac{3}{n}\mu E(X^2) – \left(1-\frac{3}{n}\right)\mu^3\Lnl
\therefore E({\overline{X}_n}^3) &= \frac{\mu_3}{n^2}+ \frac{3}{n}\mu E(X^2) +\left(1-\frac{3}{n}\right)\mu^3\label{eq-barx3}\Lnl
E({S_n}^2) &= E\left(\frac{1}{n}\sum_{i=1}^n {X_i}^2 – {\overline{X}_n}^2\right)\lnl
&= E(X^2) – E({\overline{X}_n}^2)\lnl
&= E(X^2) – \left(\frac{1}{n}E(X^2) + \frac{n-1}{n}\mu^2\right) \lnl
&= \frac{n-1}{n}(E(X^2) – \mu^2) \label{eq-sn}\Lnl
E((X-\mu)^3) = \mu_3 &= E(X^3 – 3\mu X^2 + 3\mu^2 X – \mu^3)\lnl
&= E(X^3) -3\mu E(X^2) + 2\mu^3\Lnl
\therefore E(X^3) &= \mu_3 + 3\mu E(X^2) -2\mu^3\label{eq-x3}
\end{align}
E({\overline{X}_n}^2) &= E\left\{\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\right\}\lnl
&= \frac{1}{n^2}E\left(\sum_{i=1}^n {X_i}^2 + \sum_{i\neq j}X_i X_j\right)\lnl
&= \frac{1}{n}E(X^2) + \frac{n-1}{n}\mu^2\Lnl
(\overline{X}_n – \mu)^3 &= {\overline{X}_n}^3 – 3\mu{\overline{X}_n}^2 + 3\mu^2 {\overline{X}_n} – \mu^3\Lnl
\therefore E((\overline{X}_n – \mu)^3) = \frac{\mu_3}{n^2} &= E({\overline{X}_n}^3) – 3\mu E({\overline{X}_n}^2) + 2\mu^3\lnl
&= E({\overline{X}_n}^3) – 3\mu \left( \frac{1}{n}E(X^2) + \frac{n-1}{n} \mu^2\right) + 2\mu^3\lnl
&= E({\overline{X}_n}^3) – \frac{3}{n}\mu E(X^2) – \left(1-\frac{3}{n}\right)\mu^3\Lnl
\therefore E({\overline{X}_n}^3) &= \frac{\mu_3}{n^2}+ \frac{3}{n}\mu E(X^2) +\left(1-\frac{3}{n}\right)\mu^3\label{eq-barx3}\Lnl
E({S_n}^2) &= E\left(\frac{1}{n}\sum_{i=1}^n {X_i}^2 – {\overline{X}_n}^2\right)\lnl
&= E(X^2) – E({\overline{X}_n}^2)\lnl
&= E(X^2) – \left(\frac{1}{n}E(X^2) + \frac{n-1}{n}\mu^2\right) \lnl
&= \frac{n-1}{n}(E(X^2) – \mu^2) \label{eq-sn}\Lnl
E((X-\mu)^3) = \mu_3 &= E(X^3 – 3\mu X^2 + 3\mu^2 X – \mu^3)\lnl
&= E(X^3) -3\mu E(X^2) + 2\mu^3\Lnl
\therefore E(X^3) &= \mu_3 + 3\mu E(X^2) -2\mu^3\label{eq-x3}
\end{align}
これらを用いて, $\mathrm{Cov}(\overline{X}_n,{S_n}^2) = E(\overline{X}_n {S_n}^2) – E(\overline{X}_n)E( {S_n}^2) $を計算する.
\begin{align}
\overline{X}_n {S_n}^2 &= \overline{X}_n\left(\frac{1}{n}\sum_{i=1}^n {X_i}^2 – {\overline{X}_n}^2\right)\lnl
&= \frac{1}{n^2}\left(\sum_{j=1}^n X_j\right)\left(\sum_{i=1}^n {X_i}^2\right) – {\overline{X}_n}^3\lnl
&=\frac{1}{n^2}\left(\sum_{j=1}^n {X_j}^3 + \sum_{i\neq j} X_j {X_i}^2\right) – {\overline{X}_n}^3\Lnl
\therefore E(\overline{X}_n {S_n}^2) &= \frac{1}{n^2}\Big(nE(X^3) + n(n-1)\mu E(X^2)\Big) – E({\overline{X}_n}^3)\label{eq-a1}\lnl
&= \left(\frac{n-1}{n^2}\right)\mu_3 + \frac{n-1}{n}\left(\mu E(X^2) – \mu^3 \right)
\label{eq-a2}\end{align}
\overline{X}_n {S_n}^2 &= \overline{X}_n\left(\frac{1}{n}\sum_{i=1}^n {X_i}^2 – {\overline{X}_n}^2\right)\lnl
&= \frac{1}{n^2}\left(\sum_{j=1}^n X_j\right)\left(\sum_{i=1}^n {X_i}^2\right) – {\overline{X}_n}^3\lnl
&=\frac{1}{n^2}\left(\sum_{j=1}^n {X_j}^3 + \sum_{i\neq j} X_j {X_i}^2\right) – {\overline{X}_n}^3\Lnl
\therefore E(\overline{X}_n {S_n}^2) &= \frac{1}{n^2}\Big(nE(X^3) + n(n-1)\mu E(X^2)\Big) – E({\overline{X}_n}^3)\label{eq-a1}\lnl
&= \left(\frac{n-1}{n^2}\right)\mu_3 + \frac{n-1}{n}\left(\mu E(X^2) – \mu^3 \right)
\label{eq-a2}\end{align}
$\eqref{eq-a1}$から$\eqref{eq-a2}$への変形は,$\eqref{eq-barx3} , \eqref{eq-x3}$を$E({\overline{X}_n}^3) , E(X^3)$に代入し整理した.
\begin{align}
E(\overline{X}_n)E({S_n}^2) &= \mu \cdot \frac{n-1}{n}\Big(E(X^2) – \mu^2\Big) \qquad(\because \eqref{eq-sn})\lnl
&=\frac{n-1}{n}\left(\mu E(X^2) -\mu^3\right)
\end{align}
E(\overline{X}_n)E({S_n}^2) &= \mu \cdot \frac{n-1}{n}\Big(E(X^2) – \mu^2\Big) \qquad(\because \eqref{eq-sn})\lnl
&=\frac{n-1}{n}\left(\mu E(X^2) -\mu^3\right)
\end{align}
以上より,
\begin{align}
\mathrm{Cov}(\overline{X}_n,{S_n}^2) &= E(\overline{X}_n {S_n}^2) – E(\overline{X}_n)E( {S_n}^2) \lnl
&=\frac{n-1}{n^2}\mu_3
\end{align}
\mathrm{Cov}(\overline{X}_n,{S_n}^2) &= E(\overline{X}_n {S_n}^2) – E(\overline{X}_n)E( {S_n}^2) \lnl
&=\frac{n-1}{n^2}\mu_3
\end{align}
となり示された.