Horvitz–Thompson推定量
Horvitz-Thompson推定量の期待値 E ( X ̄ n H T ) {\displaystyle \mathbb {E} \left({\bar {X}}_{n}^{\mathrm {HT} }\right)} を評価することで、Horvitz-Thompson推定量の不偏性を示すことができる。
E ( X ̄ n H T ) = E ( 1 N ∑ i = 1 n X I i π I i ) = E ( 1 N ∑ i = 1 N X i π i 1 i ∈ D n ) = ∑ b = 1 B P ( D n ( b ) ) [ 1 N ∑ i = 1 N X i π i 1 i ∈ D n ( b ) ] = 1 N ∑ i = 1 N X i π i ∑ b = 1 B 1 i ∈ D n ( b ) P ( D n ( b ) ) = 1 N ∑ i = 1 N ( X i π i ) π i = 1 N ∑ i = 1 N X i {\displaystyle {\begin{aligned}\mathbb {E} \left({\bar {X}}_{n}^{\mathrm {HT} }\right)&=\mathbb {E} \left({\frac {1}{N}}\sum _{i=1}^{n}{\frac {\mathbf {X} _{I_{i}}}{\pi _{I_{i}}}}\right)\\&=\mathbb {E} \left({\frac {1}{N}}\sum _{i=1}^{N}{\frac {X_{i}}{\pi _{i}}}1_{i\in D_{n}}\right)\\&=\sum _{b=1}^{B}P(D_{n}^{(b)})\left[{\frac {1}{N}}\sum _{i=1}^{N}{\frac {X_{i}}{\pi _{i}}}1_{i\in D_{n}^{(b)}}\right]\\&={\frac {1}{N}}\sum _{i=1}^{N}{\frac {X_{i}}{\pi _{i}}}\sum _{b=1}^{B}1_{i\in D_{n}^{(b)}}P(D_{n}^{(b)})\\&={\frac {1}{N}}\sum _{i=1}^{N}\left({\frac {X_{i}}{\pi _{i}}}\right)\pi _{i}\\&={\frac {1}{N}}\sum _{i=1}^{N}X_{i}\\\end{aligned}}}