7.1.2 点估计例题

例7.1.1 
 
 设总体X的密度函数为 
 $$
f(x) = \begin{cases}
(\alpha + 1)x^\alpha, \text{ } 0 \lt x \lt1, \alpha \gt-1, \\
\textbf 0, \text{ } 其他
\end{cases}
$$ 
 其中 $\alpha$ 是未知参数,样本为($X_1$, $X_2$, .......,$X_n$),求$\alpha$的矩估计量。 
 解： 
 令$A_1= \overline{X} = \mu = E(x) = \frac{1}{n}\Sigma_{i=1}^{n}X_i $ ，则 
 $$
\begin{align}
\mu =E(x) \
= \int_{-\infty}^{+\infty} xf(x)dx \\
=\int_{0}^{1} x(\alpha + 1)x^\alpha dx \\
=\int_{0}^{1} (\alpha + 1)x^{\alpha+1}dx \\
= \int_{0}^{1} (\alpha*x^{\alpha + 1} + x^{\alpha + 1})dx \\
= \Big(\frac{\alpha}{\alpha+2} x ^{a+2} + \frac{1}{\alpha+2} x ^{a+2}\Big)\Big|^1_0 \\
= \frac{\alpha + 1}{\alpha +2}
\end{align}
$$ 
 所以: 
 $$
\overline{X} = \frac{\alpha + 1}{\alpha +2}
$$ 
 $$
\alpha = \frac{1-2\overline{x}}{\overline{x} - 1}
$$ 
 例 7.1.2 设总体X在[a,b]上均匀分布，a,b未知，$X_1, X_2, X_3,...,X_n$均是来自总体的样本，试求a，b的矩估计量。 
 解：由矩估计法，令$ \mu_1=E(x) = A_1, \mu_2 = E(x^2) =A_2$ 所以 
 由均匀分布的数学期望知 $E(X) = \frac{a+b}{2},$ 
 $E(x^2) = \frac{1}{b-a}\int_{a}^{b} x^2 dx =\frac{(b-a)^2}{12} + \frac{(a+b)^2}{4} $ 
 即 
 $$
\begin{cases}
a+b = 2\mu, \\
b-a = \sqrt{12(\mu_2-\mu_1^2)}
\end{cases}
$$ 
 解方程组得 
 $$ a = \mu_1 - \sqrt{3(\mu_2 - \mu_1^2)}, b = \mu_1 + \sqrt{3(\mu_2 - \mu_1^2)} $$ 
 令 $A_1 = \frac{1}{n}\Sigma_{i=1}^{n}X^1_i, A_2 = \frac{1}{n}\Sigma_{i=1}^{n}X^2_i,$ 
 又$ (\frac{1}{n}\Sigma_{i=1}^{n}X^2_i - \overline{X^2}) = \frac{1}{n}\Sigma_{i=1}^{n}(X_i-\overline{X})^2$ 
 得到a和b的矩估计值为 
 $$
a = A_1 -\sqrt{3(A_2-A_1^2)} = \overline{X} - \sqrt{3(\frac{1}{n}\Sigma_{i=1}^{n}(X_i - \overline{X})^2} = \overline{X}-\sqrt{3*\frac{n-1}{n}S^2} \\
b = A_1 + \sqrt{3(A_2-A_1^2)} = \overline{X} + \sqrt{3(\frac{1}{n}\Sigma_{i=1}^{n}(X_i - \overline{X})^2} = \overline{X}+ \sqrt{3*\frac{n-1}{n}S^2}
$$