統計検定 1級過去問解答/解答例と解説
2015年11月29日 (日) 試験

統計数理問1 [3]

設問の要約

$\bar{X}$ の平均周りの $k$ 次モーメントを求めよ．

解答例

$X \sim N(\mu,~\sigma^2)$ の積率母関数 $M_{X}(t)$ は，

M_{X} (t) = \exp [μ t + \frac{σ^{2}}{2} t^{2}]

$\begin{equation} M_{X}(t) = \exp\left[\mu t + \frac{\sigma^2}{2} t^2\right] \end{equation}$

M_{X}^{(n)} (t) = E [X^{n} e^{t X}]

$\begin{equation} M_{X}^{(n)}(t) = E[X^{n}e^{tX}] \end{equation}$

M_{X}^{(n)} (0) = E [X^{n}]

$\begin{equation} M_{X}^{(n)}(0) = E[X^{n}] \end{equation}$

であるので，

\bar{X} \sim N (μ, \frac{σ^{2}}{n})

$\ds \bar{X} \sim N\left(\mu,~\frac{\sigma^2}{n}\right)$ の積率母関数

M_{\bar{X}} (t)

$M_{\bar{X}}(t)$ は，

M_{\bar{X}} (t) = \exp [μ t + \frac{σ^{2}}{2 n} t^{2}]

$\begin{equation} M_{\bar{X}}(t) = \exp\left[\mu t + \frac{\sigma^2}{2n} t^2\right] \end{equation}$

\bar{X}

$\bar{X}$ の平均周りの

k

$k$ 次のモーメントを求めるには，

\bar{X}

$\bar{X}$ の代わりに，

\bar{X} - μ

$\bar{X} - \mu$ の積率母関数を求め，

k

$k$ 階微分すると求めることができる．

M_{\bar{X} - μ}^{(k)} (0) = E [(\bar{X} - μ)^{k}]

$\begin{equation} M_{\bar{X} - \mu}^{(k)}(0) = E[(\bar{X} - \mu)^{k}] \end{equation}$

\bar{X} - μ

$\bar{X} - \mu$ の積率母関数は，

\begin{aligned} M_{\bar{X} - μ} (t) & = E [e^{(\bar{X} - μ) t}] \\ = E [e^{\bar{X} t} e^{- μ t}] \\ = E [e^{\bar{X} t}] e^{- μ t} \\ = M_{\bar{X}} (t) e^{- μ t} \\ = e^{μ t + \frac{σ^{2}}{2 n} t^{2}} e^{- μ t} \\ = e^{\frac{σ^{2}}{2 n} t^{2}} \end{aligned}

$\begin{align} M_{\bar{X} - \mu}(t) &= E\left[e^{(\bar{X} - \mu)t}\right] \\ &= E\left[e^{\bar{X}t}e^{- \mu t}\right] \\ &= E\left[e^{\bar{X}t}\right]e^{- \mu t} \\ &= M_{\bar{X}}(t)~e^{- \mu t} \\ &= e^{\mu t + \frac{\sigma^2}{2n} t^2}~e^{- \mu t} \\ &= e^{\frac{\sigma^2}{2n} t^2} \\ \end{align}$

マクローリン展開すると，

\begin{aligned} M_{\bar{X} - μ} (t) & = 1 + \frac{1}{1!} \frac{σ^{2}}{2 n} t^{2} + \frac{1}{2!} {(\frac{σ^{2}}{2 n} t^{2})}^{2} + \frac{1}{3!} {(\frac{σ^{2}}{2 n} t^{2})}^{3} + \dots \\ = 1 + \frac{1}{1!} \frac{σ^{2}}{2 n} t^{2} + \frac{1}{2!} \frac{σ^{4}}{2^{2} n^{2}} t^{4} + \frac{1}{3!} \frac{σ^{6}}{2^{3} n^{3}} t^{6} + \dots \\ = \sum_{ℓ = 0}^{\infty} \frac{1}{ℓ!} \frac{σ^{2 ℓ}}{2^{ℓ} n^{ℓ}} t^{2 ℓ} \end{aligned}

$\begin{align} M_{\bar{X} - \mu}(t) &= 1 + \frac{1}{1!}\frac{\sigma^2}{2n} t^2 + \frac{1}{2!}\left(\frac{\sigma^2}{2n} t^2\right)^2 + \frac{1}{3!}\left(\frac{\sigma^2}{2n} t^2\right)^3 + \cdots \\ &= 1 + \frac{1}{1!}\frac{\sigma^2}{2n} t^2 + \frac{1}{2!}\frac{\sigma^4}{2^2 n^2} t^4 + \frac{1}{3!}\frac{\sigma^6}{2^3 n^3} t^6 + \cdots \\ &= \sum_{\ell=0}^{\infty} \frac{1}{\ell!}\frac{\sigma^{2\ell}}{2^\ell n^\ell} t^{2\ell} \end{align}$

これより，

k

$k$ が奇数のとき，

M_{\bar{X} - μ}^{(k)} (0) = 0

$\begin{equation} M_{\bar{X} - \mu}^{(k)}(0) = 0 \end{equation}$

また，

k

$k$ が偶数のとき，

k = 2 m

$k = 2m$ と書くとすると，

\begin{aligned} M_{\bar{X} - μ}^{(k)} (t) & = \sum_{2 ℓ - k \geq 0} \frac{1}{ℓ!} \frac{σ^{2 ℓ}}{2^{ℓ} n^{ℓ}} 2 ℓ (2 ℓ - 1) (2 ℓ - 2) \dots (2 ℓ - k + 1) t^{2 ℓ - k} \\ = \sum_{ℓ = m}^{\infty} \frac{1}{ℓ!} \frac{σ^{2 ℓ}}{2^{ℓ} n^{ℓ}} 2 ℓ (2 ℓ - 1) (2 ℓ - 2) \dots (2 ℓ - 2 m + 1) t^{2 ℓ - 2 m} \end{aligned}

$\begin{align} M_{\bar{X} - \mu}^{(k)}(t) &= \sum_{2\ell - k \ge 0}\frac{1}{\ell!}\frac{\sigma^{2\ell}}{2^\ell n^\ell}2\ell\,(2\ell - 1)\,(2\ell - 2)\,\cdots\,(2\ell - k + 1)\,t^{2\ell - k} \\ &= \sum_{\ell=m}^{\infty} \frac{1}{\ell!}\frac{\sigma^{2\ell}}{2^\ell n^\ell}2\ell\,(2\ell - 1)\,(2\ell - 2)\,\cdots\,(2\ell -2m + 1)\,t^{2\ell - 2m} \\ \end{align}$

\begin{aligned} M_{\bar{X} - μ}^{(k)} (0) & = \frac{1}{m!} \frac{σ^{2 m}}{2^{m} n^{m}} 2 m (2 m - 1) (2 m - 2) \dots 2 \cdot 1 \\ = \frac{1}{m!} \frac{σ^{2 m}}{2^{m} n^{m}} (2 m)! \\ = \frac{(2 m)!}{m!} {(\frac{σ^{2}}{2 n})}^{m} \end{aligned}

$\begin{align} M_{\bar{X} - \mu}^{(k)}(0) &= \frac{1}{m!}\frac{\sigma^{2m}}{2^m n^m}2m\,(2m - 1)\,(2m - 2)\,\cdots\,2 \cdot 1\\ &= \frac{1}{m!}\frac{\sigma^{2m}}{2^m n^m}(2m)! \\ &= \frac{(2m)!}{m!}\left(\frac{\sigma^{2}}{2n}\right)^m \end{align}$

よって，

E [(\bar{X} - μ)^{k}] = E [(\bar{X} - μ)^{2 m}] = \frac{(2 m)!}{m!} {(\frac{σ^{2}}{2 n})}^{m}

$\begin{equation} E[(\bar{X} - \mu)^k] = E[(\bar{X} - \mu)^{2m}] = \frac{(2m)!}{m!}\left(\frac{\sigma^{2}}{2n}\right)^m \end{equation}$

補足1: $X \sim N(0,~\sigma^2)$ の $k$ 次モーメントの別の計算方法

X \sim N (0, σ^{2})

$\begin{equation} X \sim N(0, \sigma^2) \end{equation}$

E [X^{k}] = \int_{- \infty}^{\infty} x^{k} \frac{1}{\sqrt{2 π σ^{2}}} e^{- \frac{x^{2}}{2 σ^{2}}} d x

$\begin{equation} E[X^k] = \int_{-\infty}^{\infty}x^{k}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}\,dx \\ \end{equation}$

k = 2 m + 1 (m = 1, 2, \dots)

$k = 2m + 1~(m = 1, 2, \ldots)$ のとき，奇関数を積分することになるので，

E [X^{2 m + 1}] = 0

$\begin{equation} E[X^{2m+1}] = 0 \end{equation}$

k = 2 m (m = 1, 2, \dots)

$k = 2m~(m = 1, 2, \ldots)$ のとき，偶関数を積分することになるので，

\begin{aligned} E [X^{2 m}] & = \int_{- \infty}^{\infty} x^{2 m} \frac{1}{\sqrt{2 π σ^{2}}} e^{- \frac{x^{2}}{2 σ^{2}}} d x \\ = 2 \int_{0}^{\infty} x^{2 m} \frac{1}{\sqrt{2 π σ^{2}}} e^{- \frac{x^{2}}{2 σ^{2}}} d x \\ = \frac{2}{\sqrt{2 π σ^{2}}} \int_{0}^{\infty} x^{2 m} e^{- \frac{x^{2}}{2 σ^{2}}} d x \end{aligned}

$\begin{align} E[X^{2m}] &= \int_{-\infty}^{\infty}x^{2m}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}\,dx \\ &= 2\int_{0}^{\infty}x^{2m}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}\,dx \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\int_{0}^{\infty}x^{2m}e^{-\frac{x^2}{2\sigma^2}}\,dx \\ \end{align}$

ここで，

u = \frac{x^{2}}{2 σ^{2}}

$\begin{equation} u = \frac{x^2}{2\sigma^2} \end{equation}$

とおくと，

\frac{d u}{d x} = \frac{x}{σ^{2}}

$\begin{equation} \frac{du}{dx} = \frac{x}{\sigma^2} \end{equation}$

x d x = σ^{2} d u

$\begin{equation} x\,dx = \sigma^2\,du \end{equation}$

また，

x^{2} = 2 σ^{2} u

$\begin{equation} x^2 = 2\sigma^2 u \end{equation}$

x = \sqrt{2} σ u^{\frac{1}{2}}

$\begin{equation} x = \sqrt{2}\sigma u^{\frac{1}{2}} \end{equation}$

であるので，

\begin{aligned} E [X^{2 m}] & = \frac{2}{\sqrt{2 π σ^{2}}} \int_{0}^{\infty} x^{2 m - 1} e^{- \frac{x^{2}}{2 σ^{2}}} x d x \\ = \frac{2}{\sqrt{2 π σ^{2}}} \int_{0}^{\infty} {(\sqrt{2} σ u^{\frac{1}{2}})}^{2 m - 1} e^{- u} σ^{2} d u \\ = \frac{2}{\sqrt{2 π σ^{2}}} {(\sqrt{2} σ)}^{2 m - 1} σ^{2} \int_{0}^{\infty} {(u^{\frac{1}{2}})}^{2 m - 1} e^{- u} d u \\ = \frac{2}{\sqrt{2 π σ^{2}}} {(\sqrt{2} σ)}^{2 m - 1} σ^{2} \int_{0}^{\infty} u^{m - \frac{1}{2}} e^{- u} d u \\ = \frac{2}{\sqrt{2 π σ^{2}}} {(\sqrt{2} σ)}^{2 m - 1} σ^{2} \int_{0}^{\infty} u^{m + \frac{1}{2} - 1} e^{- u} d u \\ = \frac{2}{\sqrt{2 π σ^{2}}} {(\sqrt{2} σ)}^{2 m - 1} σ^{2} Γ (m + \frac{1}{2}) \\ = \frac{1}{\sqrt{π}} \frac{2 σ^{2}}{\sqrt{2 σ^{2}}} {(\sqrt{2 σ^{2}})}^{2 m - 1} Γ (m + \frac{1}{2}) \\ = \frac{1}{\sqrt{π}} {(\sqrt{2 σ^{2}})}^{2 m} Γ (m + \frac{1}{2}) \\ = \frac{1}{\sqrt{π}} {(2 σ^{2})}^{m} Γ (m + \frac{1}{2}) \\ = \frac{1}{Γ (\frac{1}{2})} {(2 σ^{2})}^{m} Γ (m + \frac{1}{2}) \\ = \frac{Γ (m + \frac{1}{2})}{Γ (\frac{1}{2})} {(2 σ^{2})}^{m} \end{aligned}

$\begin{align} E[X^{2m}] &= \frac{2}{\sqrt{2\pi\sigma^2}}\int_{0}^{\infty}x^{2m-1}e^{-\frac{x^2}{2\sigma^2}}\,x\,dx \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\int_{0}^{\infty}\left(\sqrt{2}\sigma u^{\frac{1}{2}}\right)^{2m-1}e^{-u}\sigma^2\,du \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\left(\sqrt{2}\sigma\right)^{2m-1}\sigma^2\int_{0}^{\infty}\left(u^{\frac{1}{2}}\right)^{2m-1}e^{-u}\,du \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\left(\sqrt{2}\sigma\right)^{2m-1}\sigma^2\int_{0}^{\infty}u^{m-\frac{1}{2}}e^{-u}\,du \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\left(\sqrt{2}\sigma\right)^{2m-1}\sigma^2\int_{0}^{\infty}u^{m+\frac{1}{2}-1}e^{-u}\,du \\ &= \frac{2}{\sqrt{2\pi\sigma^2}}\left(\sqrt{2}\sigma\right)^{2m-1}\sigma^2\varGamma\left(m+\frac{1}{2}\right) \\ &= \frac{1}{\sqrt{\pi}}\frac{2\sigma^2}{\sqrt{2\sigma^2}}\left(\sqrt{2\sigma^2}\right)^{2m-1}\varGamma\left(m+\frac{1}{2}\right) \\ &= \frac{1}{\sqrt{\pi}}\left(\sqrt{2\sigma^2}\right)^{2m}\varGamma\left(m+\frac{1}{2}\right) \\ &= \frac{1}{\sqrt{\pi}}\left(2\sigma^2\right)^{m}\varGamma\left(m+\frac{1}{2}\right) \\ &= \frac{1}{\varGamma\left(\frac{1}{2}\right)}\left(2\sigma^2\right)^{m}\varGamma\left(m+\frac{1}{2}\right) \\ &= \frac{\varGamma\left(m+\frac{1}{2}\right)}{\varGamma\left(\frac{1}{2}\right)}\left(2\sigma^2\right)^{m} \\ \end{align}$

または，

E [X^{k}] = \frac{Γ (\frac{1}{2} k + \frac{1}{2})}{Γ (\frac{1}{2})} 2^{\frac{1}{2} k} σ^{k}

$\begin{equation} E[X^{k}] = \frac{\varGamma\left(\frac{1}{2}k+\frac{1}{2}\right)}{\varGamma\left(\frac{1}{2}\right)}2^{\frac{1}{2}k} \sigma^k \end{equation}$

補足2: $X \sim N(\mu,~\sigma^2)$ の平均周りの $k$ 次モーメント

問題では $\bar{X}$ の平均周りの $k$ 次モーメントを計算したが， $X$ の平均周りの $k$ 次モーメントは， $\frac{\sigma}{\sqrt{n}}$ を $\sigma$ に置き換えれば良い．

\begin{aligned} E [(X - μ)^{k}] & = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ (k - 1)!! σ^{k} & k = 2 m (m = 1, 2, \dots) \end{cases} \\ = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ \frac{(2 m)!}{2^{m} m!} (σ^{2})^{m} & k = 2 m (m = 1, 2, \dots) \end{cases} \\ = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ \frac{Γ (m + \frac{1}{2})}{Γ (\frac{1}{2})} 2^{m} (σ^{2})^{m} & k = 2 m (m = 1, 2, \dots) \end{cases} \end{aligned}

$\begin{align} E[(X - \mu)^k] &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ (k - 1)!! \sigma^k & k = 2m~~(m=1, 2, \ldots) \end{cases} \\ &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ \dfrac{(2m)!}{2^m\,m!} (\sigma^2)^m & k = 2m~~(m=1, 2, \ldots) \end{cases} \\ &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ \dfrac{\varGamma\left(m+\frac{1}{2}\right)}{\varGamma\left(\frac{1}{2}\right)}2^m (\sigma^2)^m & k = 2m~~(m=1, 2, \ldots) \end{cases} \end{align}$

補足3: 二重階乗

$k \in N$ のとき，

k!! = {\begin{cases} 2^{m} m! & k = 2 m (m = 1, 2, \dots) \\ \frac{(2 m)!}{2^{m} m!} & k = 2 m - 1 (m = 1, 2, \dots) \end{cases}

$\begin{equation} k!! = \begin{cases} 2^m\,m! & k = 2m~~(m = 1, 2, \ldots) \\ \dfrac{(2m)!}{2^m\,m!} & k = 2m-1~~(m = 1, 2, \ldots) \end{cases} \end{equation}$

統計検定 1級 過去問 解答/解答例と解説 2015年11月29日 (日) 試験

統計数理 問1 [3]