[1]
公式の解答を参照。
[2]
\(E[(\bar{X}-\mu)^3]\)
\(\displaystyle =E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i-\mu\right)^3\right]\)
\(\displaystyle =\frac{1}{n^3}E\left[\left(\sum_{i=1}^nX_i-n\mu\right)^3\right]\)
\(\displaystyle =\frac{1}{n^3}E\left[\left(\sum_{i=1}^n(X_i-\mu)\right)^3\right]\)
\(\displaystyle =\frac{1}{n^3}\left(E\left[\sum_{i=1}^n(X_i-\mu)^3\right]\right.\)\(\displaystyle +E\left[{}_3C_1\sum_{i\neq j}^n(X_i-\mu)^2(X_j-\mu)\right]\)\(\displaystyle +\left.E\left[3!\sum_{i<j<k}^n(X_i-\mu)(X_j-\mu)(X_k-\mu)\right]\right)\)
\(\displaystyle =\frac{1}{n^3}E\left[\sum_{i=1}^n(X_i-\mu)^3\right]\)
\(\displaystyle =\frac{1}{n^2}E\left[(X_i-\mu)^3\right]\)
\(\displaystyle =\frac{1}{n^2}\sigma^3\beta_1\)
よって、\(\bar{X}\)の歪度は、分散の3/2乗で割って、
\(\displaystyle \frac{\frac{1}{n^2}\sigma^3\beta_1}{\left(\frac{\sigma^2}{n}\right)^\frac{3}{2}}\)\(\displaystyle =\frac{\beta_1}{\sqrt{n}}\)
[3]
\(E[(\bar{X}-\mu)^4]\)
\(\displaystyle =E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i-\mu\right)^4\right]\)
\(\displaystyle =\frac{1}{n^4}E\left[\left(\sum_{i=1}^nX_i-n\mu\right)^4\right]\)
\(\displaystyle =\frac{1}{n^4}E\left[\left(\sum_{i=1}^n(X_i-\mu)\right)^4\right]\)
\(\displaystyle =\frac{1}{n^4}\left(E\left[\sum_{i=1}^n(X_i-\mu)^4\right]\right.\)\(\displaystyle +E\left[{}_4C_1\sum_{i \neq j}^n(X_i-\mu)^3(X_j-\mu)\right]\)\(\displaystyle +E\left[{}_4C_2\sum_{i < j}^n(X_i-\mu)^2(X_j-\mu)^2\right]\)\(\displaystyle +\left.E\left[\frac{4!}{2!}\sum_{i \neq j \neq k}^n(X_i-\mu)^2(X_j-\mu)(X_k-\mu)\right]\right)\)
\(\displaystyle =\frac{1}{n^4}\left(E\left[\sum_{i=1}^n(X_i-\mu)^4\right]\right.\)\(\displaystyle +\left.E\left[{}_4C_2\sum_{i < j}^n(X_i-\mu)^2(X_j-\mu)^2\right]\right)\)
\(\displaystyle =\frac{1}{n^4}(nE[(X_i-\mu)^4]\)\(\displaystyle +3n(n-1)E[(X_i-\mu)^2]\)\(E[(X_j-\mu)^2])\)
\(\displaystyle =\frac{1}{n^4}(n\sigma^4(\beta_2+3)\)\(\displaystyle +3n(n-1)\sigma^2\)\(\cdot \sigma^2)\)
\(\displaystyle =\frac{\sigma^4(\beta_2+3n)}{n^3}\)
よって、\(\bar{X}\)の歪度は、分散の2乗で割って3を引いて
\(\displaystyle \frac{\frac{\sigma^4(\beta_2+3n)}{n^3}}{\left(\frac{\sigma^2}{n}\right)^2}-3\)\(\displaystyle =\frac{\beta_2}{n}\)
[4]
\(n \to \infty\)として、歪度も尖度も0に近づく。
[5]
\(f(\boldsymbol{x})=\displaystyle \prod_{i=1}^n f(x_i)\)
\(\displaystyle =(2\pi\sigma^2)^{-\frac{n}{2}}\exp\left[-\frac{\sum_{i=1}^n(x_i-\mu)^2}{2\sigma^2}\right]\)
\(l=\ln f(\boldsymbol{x})\)
\(\displaystyle =-\frac{n}{2}\ln(2\pi\sigma^2)-\frac{\sum_{i=1}^n(x_i-\mu)^2}{2\sigma^2}\)
\(\mu\)既知の時、
\(\displaystyle \frac{dl}{d\sigma^2}\)\(\displaystyle =-\frac{n}{2\sigma^2}+\frac{\sum_{i=1}^n(x_i-\mu)^2}{2\sigma^4}\)\(=0\)
として、\(\sigma^2=T\)となる。
\(\mu\)未知の時、
\(\displaystyle \frac{\partial l}{\partial \mu}\)\(\displaystyle =\frac{1}{\sigma^2}\sum_{i=1}^n(x_i-\mu)\)\(=0\)
として、\(\mu=\bar{X}\)
\(\displaystyle \frac{dl}{d\sigma^2}=0\)によって得られた\(\sigma^2\)の\(\mu\)に\(\bar{X}\)を代入して、
\(\displaystyle \sigma^2=\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2\)
\(\displaystyle =\frac{n-1}{n}S^2\)
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