統計検定 1級 2015年 統計数理 問1 解答 解説

[1]

\(E[\bar{X}^2]\)\(=V[\bar{X}]+E[\bar{X}]^2\)

\(=\displaystyle \frac{\sigma^2}{n}+\mu^2\)

[2]

\(E[S^2]\)\(=\displaystyle E\left[\frac{1}{n-1}\sum^n_{i=1}(X_i-\bar{X})^2\right]\)

\(=\displaystyle \frac{1}{n-1}E\left[\sum^n_{i=1}(X_i-\bar{X})^2\right]\)

\(=\displaystyle \frac{1}{n-1}E\left[\sum^n_{i=1}((X_i-\mu)-(\bar{X}-\mu))^2\right]\)

\(=\displaystyle \frac{1}{n-1}E\left[\sum^n_{i=1}(X_i-\mu)^2\right . \)\(\displaystyle -2\sum^n_{i=1}(X_i-\mu)(\bar{X}-\mu) \)\(\displaystyle \left . +\sum^n_{i=1}(\bar{X}-\mu)^2\right]\)

\(=\displaystyle \frac{1}{n-1}E\left[\sum^n_{i=1}(X_i-\mu)^2\right . \)\(\displaystyle -2n(\bar{X}-\mu)^2 \)\(\displaystyle \left . +n(\bar{X}-\mu)^2\right]\)

\(=\displaystyle \frac{1}{n-1}E\left[\sum^n_{i=1}(X_i-\mu)^2 -n(\bar{X}-\mu)^2 \right]\)

\(=\displaystyle \frac{1}{n-1}(n\sigma^2 – \sigma^2)\)

\(=\sigma^2\)

[3]

\(E[(\bar{X}-\mu)^k]\)\(\displaystyle =\int^\infty_{-\infty} (\bar{X}-\mu)^k \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}e^{-\frac{(\bar{X}-\mu)^2}{2\sigma^2/n}}d\bar{X}\)

\(\bar{X}-\mu=Y\)とすると、

\(\displaystyle =\int^\infty_{-\infty}y^k \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}e^{-\frac{y^2}{2\sigma^2/n}}dy\)

\(k\)が奇数の時、積分内は奇関数になるので、\(0\)。

\(k\)が偶数の時、積分内は偶関数になるので、

\(\displaystyle =2\int^\infty_0y^k \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}e^{-\frac{y^2}{2\sigma^2/n}}dy\)

\(y^2=t\)と変数変換を行うと、

\(\displaystyle =2\int^\infty_0 t^{\frac{k}{2}} \frac{1}{\sqrt{2\pi}\sigma/\sqrt{n}}e^{-\frac{t}{2\sigma^2/n}}\frac{1}{2\sqrt{t}}dt\)

\(\displaystyle =\frac{\sqrt{n}}{\sqrt{2\pi}\sigma}\int^\infty_0 t^{\frac{k-1}{2}}e^{-\frac{t}{2\sigma^2/n}}dt\)

\(\displaystyle =\frac{\sqrt{n}}{\sqrt{2\pi}\sigma}\Gamma \left(\frac{k+1}{2}\right)\left(\frac{2\sigma^2}{n}\right)^{\frac{k+1}{2}}\)

\(\displaystyle =\frac{1}{\sqrt{\pi}}\Gamma \left(\frac{k+1}{2}\right)\left(\frac{2\sigma^2}{n}\right)^{\frac{k}{2}}\)

[4]

チェビシェフの不等式により、

\(P(|S^2-\sigma^2| \geq \varepsilon) \leq \displaystyle \frac{V[S^2]}{\varepsilon^2}\)

\(\displaystyle \frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\)なので、\(\displaystyle V\left[\frac{(n-1)S^2}{\sigma^2}\right]\)\(=2(n-1)\)となり、\(V[S^2]\)\(\displaystyle =\frac{2\sigma^4}{n-1}\)

\(P(|S^2-\sigma^2| \geq \varepsilon) \leq \displaystyle \frac{2\sigma^4}{\varepsilon^2(n-1)}\)\(\overset{n \to \infty}{\longrightarrow}\)\(0\)

よって一致推定量。

詳細

\(P(|S^2-\sigma^2| \geq \varepsilon) \)\(\overset{n \to \infty}{\longrightarrow}\)\(0\)

より、

\(\displaystyle \lim_{n \to \infty} P(|S^2-\sigma^2| \geq \varepsilon)=0\)

\(\iff \displaystyle \lim_{n \to \infty} P(|S^2-\sigma^2| < \varepsilon)\)\(=1\)

\(\iff \)「任意の\(\varepsilon\)に対して、\(n\)を限りなく大きくとれば、\(|S^2-\sigma^2| < \varepsilon\)が成り立つ」

\(\iff \)「\( \forall\varepsilon>0\)\(, \exists N \in \mathbb{N}\)\(;|S_N^2-\sigma^2| < \varepsilon\)」(\(\varepsilon\)-論法)

\(\iff\) \(\displaystyle \lim_{n \to \infty} S^2 =\sigma^2\)

[5]

\(MSE[cS^2]=E[(cS^2-\sigma^2)^2]\)

\(\displaystyle = V[cS^2-\sigma^2]+E[cS^2-\sigma^2]^2\)

\(\displaystyle = c^2V[S^2]+(cE[S^2]-\sigma^2)^2\)

\(\displaystyle = c^2 \frac{2\sigma^4}{n-1}+(c\sigma^2-\sigma^2)^2\)

\(\displaystyle =\left \{\frac{2c^2}{n-1}+(c-1)^2\right\}\sigma^4\)

\(\{\}\)内を\(f(c)\)とすると、

\(f'(c) = \displaystyle \frac{4}{n-1}c +2(c-1)\)\(=0\)

として、\(\displaystyle c = \frac{n-1}{n+1}\)となり、この時、\(MSE = \displaystyle \frac{2\sigma^4}{n+1}\)

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