题目
(3)已知总体X的数学期望为 (X)=0, 方差为 (X)=(sigma )^2, X1,X2,···,xn为总体x-|||-的简单随机样本, overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) . ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2, 则下列属于σ^2的无偏估计量的是-|||-() .-|||-A. ((overline {x))}^2+(S)^2 B. dfrac (1)(2)[ n((overline {X))}^2+(S)^2] C. dfrac (n)(3)((overline {X))}^2+(S)^2 D. dfrac (1)(4)[ n((overline {X))}^2+(S)^2]
题目解答
答案
解析
步骤 1：计算 $\overline {X}$ 的方差
由于 $E(X)=0$，则 $E(\overline {X})=0$，且 $\overline {X}$ 是 $X_1, X_2, \ldots, X_n$ 的简单随机样本的均值，因此 $\overline {X}$ 的方差为：
$$
D(\overline {X}) = D\left(\frac{1}{n}\sum_{i=1}^{n}X_i\right) = \frac{1}{n^2}\sum_{i=1}^{n}D(X_i) = \frac{1}{n^2} \cdot n \cdot \sigma^2 = \frac{\sigma^2}{n}
$$
步骤 2：计算 $n(\overline {X})^2$ 的期望
$$
E[n(\overline {X})^2] = nE[(\overline {X})^2] = n[D(\overline {X}) + (E(\overline {X}))^2] = n\left(\frac{\sigma^2}{n} + 0\right) = \sigma^2
$$
步骤 3：计算 $S^2$ 的期望
$S^2$ 是样本方差，对于简单随机样本，$S^2$ 是 $\sigma^2$ 的无偏估计量，因此：
$$
E(S^2) = \sigma^2
$$
步骤 4：计算各选项的期望
A. $E[n(\overline {X})^2 + S^2] = E[n(\overline {X})^2] + E(S^2) = \sigma^2 + \sigma^2 = 2\sigma^2$
B. $E\left[\frac{1}{2}[n(\overline {X})^2 + S^2]\right] = \frac{1}{2}E[n(\overline {X})^2 + S^2] = \frac{1}{2} \cdot 2\sigma^2 = \sigma^2$
C. $E\left[\frac{n}{3}(\overline {X})^2 + S^2\right] = \frac{n}{3}E[(\overline {X})^2] + E(S^2) = \frac{n}{3} \cdot \frac{\sigma^2}{n} + \sigma^2 = \frac{\sigma^2}{3} + \sigma^2 = \frac{4\sigma^2}{3}$
D. $E\left[\frac{1}{4}[n(\overline {X})^2 + S^2]\right] = \frac{1}{4}E[n(\overline {X})^2 + S^2] = \frac{1}{4} \cdot 2\sigma^2 = \frac{\sigma^2}{2}$
由于 $E(X)=0$，则 $E(\overline {X})=0$，且 $\overline {X}$ 是 $X_1, X_2, \ldots, X_n$ 的简单随机样本的均值，因此 $\overline {X}$ 的方差为：
$$
D(\overline {X}) = D\left(\frac{1}{n}\sum_{i=1}^{n}X_i\right) = \frac{1}{n^2}\sum_{i=1}^{n}D(X_i) = \frac{1}{n^2} \cdot n \cdot \sigma^2 = \frac{\sigma^2}{n}
$$
步骤 2：计算 $n(\overline {X})^2$ 的期望
$$
E[n(\overline {X})^2] = nE[(\overline {X})^2] = n[D(\overline {X}) + (E(\overline {X}))^2] = n\left(\frac{\sigma^2}{n} + 0\right) = \sigma^2
$$
步骤 3：计算 $S^2$ 的期望
$S^2$ 是样本方差，对于简单随机样本，$S^2$ 是 $\sigma^2$ 的无偏估计量，因此：
$$
E(S^2) = \sigma^2
$$
步骤 4：计算各选项的期望
A. $E[n(\overline {X})^2 + S^2] = E[n(\overline {X})^2] + E(S^2) = \sigma^2 + \sigma^2 = 2\sigma^2$
B. $E\left[\frac{1}{2}[n(\overline {X})^2 + S^2]\right] = \frac{1}{2}E[n(\overline {X})^2 + S^2] = \frac{1}{2} \cdot 2\sigma^2 = \sigma^2$
C. $E\left[\frac{n}{3}(\overline {X})^2 + S^2\right] = \frac{n}{3}E[(\overline {X})^2] + E(S^2) = \frac{n}{3} \cdot \frac{\sigma^2}{n} + \sigma^2 = \frac{\sigma^2}{3} + \sigma^2 = \frac{4\sigma^2}{3}$
D. $E\left[\frac{1}{4}[n(\overline {X})^2 + S^2]\right] = \frac{1}{4}E[n(\overline {X})^2 + S^2] = \frac{1}{4} \cdot 2\sigma^2 = \frac{\sigma^2}{2}$