第2講確率分布

標本分散

標本 $X_{1},~X_{2},~\ldots,X_{n}$ は同分布で， $E[X_i]=\mu,~V[X_i]=\sigma^2$ とする．

このとき，標本分散は

S^{2} = \frac{1}{n} {(X_{1} - \bar{X})^{2} + (X_{2} - \bar{X})^{2} + \dots + (X_{n} - \bar{X})^{2}}

$\begin{equation} S^2 = \frac{1}{n}\left\{(X_{1} - \bar{X})^2 + (X_{2} - \bar{X})^2 + \cdots + (X_{n} - \bar{X})^2\right\} \end{equation}$

ただし，

\bar{X} = \frac{X_{1} + X_{2} + \dots + X_{n}}{n}

$\begin{equation} \bar{X} = \frac{X_{1} + X_{2} + \cdots + X_{n}}{n} \end{equation}$

また，

\begin{aligned} E [S^{2}] & = E [\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - \bar{X})^{2}] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} {{X_{i}}^{2} - 2 \bar{X} X_{i} + (\bar{X})^{2}}] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} {X_{i}}^{2} - 2 \bar{X} \frac{1}{n} \sum_{i = 1}^{n} X_{i} + (\bar{X})^{2} \frac{1}{n} \sum_{i = 1}^{n} 1] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} {X_{i}}^{2} - 2 \bar{X} \cdot \bar{X} + (\bar{X})^{2}] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} {X_{i}}^{2} - (\bar{X})^{2}] \\ = \frac{1}{n} \sum_{i = 1}^{n} E [{X_{i}}^{2}] - E [(\bar{X})^{2}] \\ = \frac{1}{n} \sum_{i = 1}^{n} (V [{X_{i}}^{2}] + E [X_{i}]^{2}) - (V [\bar{X}] + E [\bar{X}]^{2}) \\ = \frac{1}{n} \sum_{i = 1}^{n} (σ^{2} + μ^{2}) - (\frac{σ^{2}}{n} + μ^{2}) \\ = \frac{1}{n} n (σ^{2} + μ^{2}) - (\frac{σ^{2}}{n} + μ^{2}) \\ = \frac{n - 1}{n} σ^{2} \end{aligned}

$\begin{align} E[S^2] &= E\left[\frac{1}{n}\sum_{i=1}^{n}(X_i - \bar{X})^2\right] \\ &= E\left[\frac{1}{n}\sum_{i=1}^{n}\{{X_i}^2 -2\bar{X}{X_i} + (\bar{X})^2\}\right] \\ &= E\left[\frac{1}{n}\sum_{i=1}^{n}{X_i}^2 -2\bar{X}\frac{1}{n}\sum_{i=1}^{n}{X_i} + (\bar{X})^2\frac{1}{n}\sum_{i=1}^{n}1\right] \\ &= E\left[\frac{1}{n}\sum_{i=1}^{n}{X_i}^2 -2\bar{X}\cdot\bar{X} + (\bar{X})^2\right] \\ &= E\left[\frac{1}{n}\sum_{i=1}^{n}{X_i}^2 - (\bar{X})^2\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}E\left[{X_i}^2\right] - E\left[(\bar{X})^2\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}\left(V[{X_i}^2] + E[X_i]^2\right) - \left(V[\bar{X}] + E[\bar{X}]^2\right) \\ &= \frac{1}{n}\sum_{i=1}^{n}\left(\sigma^2 + \mu^2\right) - \left(\frac{\sigma^2}{n} + \mu^2\right) \\ &= \frac{1}{n}n\left(\sigma^2 + \mu^2\right) - \left(\frac{\sigma^2}{n} + \mu^2\right) \\ &= \frac{n-1}{n}\sigma^2 \end{align}$

V [S^{2}] = \frac{(n - 1)^{2}}{n^{3}} {μ_{4} - \frac{n - 3}{n - 1} σ^{4}} (μ_{4} = E [(X - E [\bar{X}])^{4}])

$\begin{equation} V[S^2] = \frac{(n-1)^{2}}{n^3}\left\{\mu_4 - \frac{n-3}{n-1}\sigma^4\right\}~~~~\left(\mu_4 = E\left[(X - E[\bar{X}])^4\right]\right) \end{equation}$

ここで，

U^{2} = \frac{1}{n - 1} {(X_{1} - \bar{X})^{2} + (X_{2} - \bar{X})^{2} + \dots + (X_{n} - \bar{X})^{2}}

$\begin{equation} U^2 = \frac{1}{n-1}\left\{(X_{1} - \bar{X})^2 + (X_{2} - \bar{X})^2 + \cdots + (X_{n} - \bar{X})^2\right\} \end{equation}$

と定義すると，

E [U^{2}] = σ^{2}

$\begin{equation} E[U^2] = \sigma^2 \end{equation}$

となり，これを不偏分散という．

第2講 確率分布

標本分散

第2講確率分布