21.设x1,x2,x3,x 4是来自均值为θ的指数分布总体的样本,其中θ未知,设有估计量-|||-_(1)=dfrac (1)(6)((x)_(1)+(x)_(2))+dfrac (1)(3)((x)_(3)+(x)_(4)),-|||-_(2)=dfrac (({x)_(1)+2(x)_(2)+3(x)_(3)+4(x)_(4))}(5)-|||-_(3)=dfrac (({x)_(1)+(x)_(2)+(x)_(3)+(x)_(4))}(4)-|||-(1)指出T1,T2,T3哪几个是θ的无偏估计量;-|||-(2)在上述θ的无偏估计中指出哪一个较为有效.

步骤 1：计算期望值
由于x1, x2, x3, x4是来自均值为θ的指数分布总体的样本，所以 $E(x_i) = \theta$，$D(x_i) = \theta^2$，其中i=1,2,3,4。我们首先计算每个估计量的期望值。
步骤 2：计算T1的期望值
$E(T_1) = \dfrac{1}{6}[E(x_1) + E(x_2)] + \dfrac{1}{3}[E(x_3) + E(x_4)] = \dfrac{1}{6}[\theta + \theta] + \dfrac{1}{3}[\theta + \theta] = \dfrac{1}{6} \cdot 2\theta + \dfrac{1}{3} \cdot 2\theta = \dfrac{1}{3}\theta + \dfrac{2}{3}\theta = \theta$
步骤 3：计算T2的期望值
$E(T_2) = \dfrac{1}{5}[E(x_1) + 2E(x_2) + 3E(x_3) + 4E(x_4)] = \dfrac{1}{5}[\theta + 2\theta + 3\theta + 4\theta] = \dfrac{1}{5} \cdot 10\theta = 2\theta$
步骤 4：计算T3的期望值
$E(T_3) = \dfrac{1}{4}[E(x_1) + E(x_2) + E(x_3) + E(x_4)] = \dfrac{1}{4}[\theta + \theta + \theta + \theta] = \dfrac{1}{4} \cdot 4\theta = \theta$
步骤 5：计算方差
由于x1, x2, x3, x4独立，我们计算每个估计量的方差。
步骤 6：计算T1的方差
$D(T_1) = \dfrac{1}{36}[D(x_1) + D(x_2)] + \dfrac{1}{9}[D(x_3) + D(x_4)] = \dfrac{1}{36}[\theta^2 + \theta^2] + \dfrac{1}{9}[\theta^2 + \theta^2] = \dfrac{1}{36} \cdot 2\theta^2 + \dfrac{1}{9} \cdot 2\theta^2 = \dfrac{1}{18}\theta^2 + \dfrac{2}{9}\theta^2 = \dfrac{5}{18}\theta^2$
步骤 7：计算T3的方差
$D(T_3) = \dfrac{1}{16}[D(x_1) + D(x_2) + D(x_3) + D(x_4)] = \dfrac{1}{16}[\theta^2 + \theta^2 + \theta^2 + \theta^2] = \dfrac{1}{16} \cdot 4\theta^2 = \dfrac{1}{4}\theta^2$
步骤 8：比较方差
$D(T_1) > D(T_3)$，所以T3较为有效。