[1]
\(V[X]=\sigma^2\)なので、\(V[\bar{X}]=\displaystyle \frac{\sigma^2}{n}\)
[2]
\(E[(X_1-X_2)^3]\)
\(=E[\{(X_1-\mu)-(X_2-\mu)\}^3]\)
\(=E[(X_1-\mu)^3]\)\(-3E[(X_1-\mu)^2]E[X_2-\mu]\)\(+3E[X_1-\mu]E[(X_2-\mu)^2]\)\(-E[(X_2-\mu)^3]\)
\(E[(X_i-\mu)^3]=\tau,E[X_i-\mu]=0\)を用いて、
\(=0\)
[3]
\(E[Y_i^3]\)\(=E[(X_i-\mu)^3]\)\(=\tau\)
\(E[Y_i^2Y_j]\)\(=E[(X_i-\mu)^2]E[X_j-\mu]\)\(=0\)(\(E[X_j-\mu]=0\)のため)
\(E[Y_iY_jY_k]\)\(=E[X_i-\mu]E[X_j-\mu]E[X_k-\mu]\)\(=0\)
[4]
\(\displaystyle E[\sum^n_{i=1}(X_i-\bar{X})^3]\)
\(\displaystyle =\sum^n_{i=1} E[(X_i-\bar{X})^3]\)
\(\displaystyle =\sum^n_{i=1} E[\{(X_i-\mu)-(\bar{X}-\mu)\}^3]\)
\(\bar{Y}=\bar{X}-\mu\)として、
\(\displaystyle =\sum^n_{i=1} E[(Y_i-\bar{Y})^3]\)
\(\displaystyle =\sum^n_{i=1} E\left[\left(Y_i-\frac{Y_1+\cdots+Y_n}{n}\right)^3\right]\)
\(\displaystyle =\frac{1}{n^3}\sum^n_{i=1} E[(nY_i-(Y_1+\cdots+Y_n))^3]\)
\(\displaystyle =\frac{1}{n^3}\sum^n_{i=1} E[(-Y_1- \cdots \)\( +(n-1)Y_i-\cdots -\)\(Y_n)^3]\)
3乗の展開する際に[3]の結果を利用して、
\(\displaystyle =\frac{1}{n^3}\sum^n_{i=1}(-E[Y_1^3]-\cdots \)\(+(n-1)^3E[Y_i^3]-\cdots \)\(-E[Y_n^3])\)
\(\displaystyle =\frac{1}{n^3}\sum^n_{i=1}((n-1)^3\tau-(n-1)\tau)\)
\(\displaystyle =\frac{(n-1)(n-2)}{n}\tau\)
[5]
\(\displaystyle E\left[\left(\sum^n_{i=1}a_iX_i\right)^3\right]\)
\(\displaystyle =E\left[\left(\sum^n_{i=1}a_i(Y_i+\mu)\right)^3\right]\)
\(\displaystyle =E\left[\left(\sum^n_{i=1}a_iY_i+\mu\sum^n_{i=1}a_i\right)^3\right]\)
\(\displaystyle \mu\sum^n_{i=1}a_i=A\)として、[3]の結果を利用すると、
\(\displaystyle =E\left[\sum^n_{i=1}a_i^3Y_i^3\right.\)\(\displaystyle +3A\sum^n_{i=1}a_i^2Y_i^2\)\(\displaystyle \left. +3A^2\sum^n_{i=1}a_iY_i+A^3\right]\)
\(\displaystyle =\sum^n_{i=1}a_i^3E[Y_i^3]\)\(\displaystyle+3A\sum^n_{i=1}a_i^2E[Y_i^2]\)\(\displaystyle+3A^2\sum^n_{i=1}a_iE[Y_i]\)\(+A^3\)
1,2,3項目の期待値部分はそれぞれ、\(\tau,\sigma^2,0\)で、さらに\(A\)を元に戻して、
\(\displaystyle =\tau\sum^n_{i=1}a_i^3\)\(\displaystyle+3\mu\sigma^2\sum^n_{i=1}a_i\sum^n_{i=1}a_i^2\)\(\displaystyle + \mu^3\left(\sum^n_{i=1}a_i\right)^3\)
これが、未知パラメータに依存しないで\(\tau\)となるための条件は、\(\tau\)に関する恒等式と見て、
\(\displaystyle \sum^n_{i=1}a_i=0\)かつ\(\displaystyle \sum^n_{i=1}a_i^3=1\)
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