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6.16 (難) 推定量の評価7 標本がiidでない場合の一致性の評価

Last updated at Posted at 2022-06-29

方針

大数法則 (LLN) は標本がiidである場合にのみ使うことができる.本問では,標本平均の中身をうまくiidの確率標本になるよう変形してLLNに持ち込む.

(3)の赤字になっているところはまだ分かっていない.N(0,sigma^2)の加重平均がなぜ0に確率収束するのか分かる人いたら教えてください.

(追記)
赤線部分ですが,以下の通り0に確率収束します.

$Z_i=(\mu_i-\bar{\mu_i})(X_i-\mu_i)$とおけば,

$$
\mathbb{E}[Z_i]=(\mu_i-\bar{\mu_i})\mathbb{E}[(X_i-\mu_i)]=0
$$

なので,LLNにより0に収束.

答案

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参考文献

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