题目
设总体X服从泊松分布，即(X=k)=dfrac ({lambda )^k}(k!)(e)^-lambda (k=0,1,2,... )，(X=k)=dfrac ({lambda )^k}(k!)(e)^-lambda (k=0,1,2,... )为样本，证明(X=k)=dfrac ({lambda )^k}(k!)(e)^-lambda (k=0,1,2,... )是(X=k)=dfrac ({lambda )^k}(k!)(e)^-lambda (k=0,1,2,... )的无偏估计。
设总体X服从泊松分布，即
，
为样本，证明
是
的无偏估计。
题目解答
答案
解析
步骤 1：确定泊松分布的期望和方差
泊松分布的期望和方差都是$\lambda$，即$E(X)=\lambda$和$D(X)=\lambda$。
步骤 2：计算$\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}$的期望
由于$E({X}_{i}^{2})=D({X}_{i})+E({X}_{i})^{2}=\lambda +\lambda ^{2}$，所以$E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2})=\dfrac {1}{n}\sum _{i=1}^{n}E({X}_{i}^{2})=\dfrac {1}{n}\sum _{i=1}^{n}(\lambda +\lambda ^{2})=\lambda +\lambda ^{2}$。
步骤 3：计算$\overline {X}$的期望
$\overline {X}=\dfrac {1}{n}\sum _{i=1}^{n}X_{i}$，所以$E(\overline {X})=E(\dfrac {1}{n}\sum _{i=1}^{n}X_{i})=\dfrac {1}{n}\sum _{i=1}^{n}E(X_{i})=\dfrac {1}{n}\sum _{i=1}^{n}\lambda =\lambda$。
步骤 4：计算$\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}-\overline {X}$的期望
$E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}-\overline {X})=E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2})-E(\overline {X})=(\lambda +\lambda ^{2})-\lambda =\lambda ^{2}$。
泊松分布的期望和方差都是$\lambda$，即$E(X)=\lambda$和$D(X)=\lambda$。
步骤 2：计算$\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}$的期望
由于$E({X}_{i}^{2})=D({X}_{i})+E({X}_{i})^{2}=\lambda +\lambda ^{2}$，所以$E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2})=\dfrac {1}{n}\sum _{i=1}^{n}E({X}_{i}^{2})=\dfrac {1}{n}\sum _{i=1}^{n}(\lambda +\lambda ^{2})=\lambda +\lambda ^{2}$。
步骤 3：计算$\overline {X}$的期望
$\overline {X}=\dfrac {1}{n}\sum _{i=1}^{n}X_{i}$，所以$E(\overline {X})=E(\dfrac {1}{n}\sum _{i=1}^{n}X_{i})=\dfrac {1}{n}\sum _{i=1}^{n}E(X_{i})=\dfrac {1}{n}\sum _{i=1}^{n}\lambda =\lambda$。
步骤 4：计算$\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}-\overline {X}$的期望
$E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2}-\overline {X})=E(\dfrac {1}{n}\sum _{i=1}^{n}{{X}_{i}}^{2})-E(\overline {X})=(\lambda +\lambda ^{2})-\lambda =\lambda ^{2}$。