设_(1),(X)_(2),(X)_(3),(X)_(4)是来自均值为_(1),(X)_(2),(X)_(3),(X)_(4)的泊松分布总体的样本，其中 _(1),(X)_(2),(X)_(3),(X)_(4) 未知，则下列估计量中是 _(1),(X)_(2),(X)_(3),(X)_(4) 的无偏估计量的是( )A._(1),(X)_(2),(X)_(3),(X)_(4)B._(1),(X)_(2),(X)_(3),(X)_(4)C._(1),(X)_(2),(X)_(3),(X)_(4)D._(1),(X)_(2),(X)_(3),(X)_(4)

步骤 1：计算每个选项的期望值
对于泊松分布，均值等于方差，即$E(X_i) = \lambda$，其中$X_i$是来自泊松分布总体的样本。
步骤 2：计算选项A的期望值
$E(I_1) = E(\dfrac{X_1}{2} + \dfrac{1}{9}(2X_2 + 3X_3 + 4X_4)) = \dfrac{1}{2}E(X_1) + \dfrac{1}{9}(2E(X_2) + 3E(X_3) + 4E(X_4)) = \dfrac{1}{2}\lambda + \dfrac{1}{9}(2\lambda + 3\lambda + 4\lambda) = \dfrac{1}{2}\lambda + \dfrac{9}{9}\lambda = \dfrac{1}{2}\lambda + \lambda = \dfrac{3}{2}\lambda$
步骤 3：计算选项B的期望值
$E(Y_2) = E(\dfrac{1}{4}(X_1 + X_2 + X_3 + X_4)) = \dfrac{1}{4}(E(X_1) + E(X_2) + E(X_3) + E(X_4)) = \dfrac{1}{4}(\lambda + \lambda + \lambda + \lambda) = \dfrac{1}{4}4\lambda = \lambda$
步骤 4：计算选项C的期望值
$E(\eta_3) = E(\dfrac{1}{6}(X_1 + X_2) + \dfrac{1}{4}(X_3 + X_4)) = \dfrac{1}{6}(E(X_1) + E(X_2)) + \dfrac{1}{4}(E(X_3) + E(X_4)) = \dfrac{1}{6}(\lambda + \lambda) + \dfrac{1}{4}(\lambda + \lambda) = \dfrac{1}{6}2\lambda + \dfrac{1}{4}2\lambda = \dfrac{1}{3}\lambda + \dfrac{1}{2}\lambda = \dfrac{5}{6}\lambda$
步骤 5：计算选项D的期望值
$E(I_1) = E(\dfrac{1}{6}(X_1 + 2X_2 + 3X_3 + 4X_4)) = \dfrac{1}{6}(E(X_1) + 2E(X_2) + 3E(X_3) + 4E(X_4)) = \dfrac{1}{6}(\lambda + 2\lambda + 3\lambda + 4\lambda) = \dfrac{1}{6}10\lambda = \dfrac{5}{3}\lambda$
步骤 6：判断无偏估计量
无偏估计量的期望值应该等于参数$\lambda$。根据步骤2到步骤5的计算结果，只有选项B的期望值等于$\lambda$，因此选项B是$\lambda$的无偏估计量。