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Uploaded by
Jun Iio
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不偏分散ではなぜNでなくN-1で割るのか
朝からゴリゴリ計算して疲れたので苦労を皆さんと共有すべくアップします
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不偏分散ではなぜNでなくN-1で割るのか
1.
n n−1
2.
s2 = 1 n Σn k=1(Xk − ¯X)2
3.
E[s2 ] = E[1/nΣ(Xk
− ¯X)2 ] Σ Σk(Xk − ¯X)2 = ( (n − 1)X1 − X2 − ⋯ − Xn n )2 + ⋯ = 1 n2 ((n − 1)2 X2 1 − 2(n − 1)X1(X2 + ⋯ + Xn) + (X2 + X3 + ⋯ + Xn)2 ) + ⋯ (X2 + X3 + ⋯ + Xn)2 = X2 2 + X2 3 + ⋯ + X2 n+ 2(X2X3 + X2X4 + ⋯ + Xn−1Xn)
4.
Σk(Xk − ¯X)2 =
( (n − 1)X1 − X2 − ⋯ − Xn n )2 + ⋯ = 1 n2 ((n − 1)2 X2 1 − 2nX1(X2 + ⋯ + Xn) + X2 2 + X2 3 + ⋯ + X2 n + 2(X1X2 + ⋯ + Xn−1Xn)) + ⋯ Xk = 1 n2 ((n − 1)2 X2 1 − 2nX1Σn k=2Xk + Σn k=2X2 k + 2Σi<jXiXj) + ⋯ = 1 n2 (Σ(n − 1)2 X2 k − 2n ⋅ 2Σi<jXiXj + (n − 1)ΣX2 k + 2nΣi<jXiXj)
5.
Σk(Xk − ¯X)2 = 1 n2 (Σ(n
− 1)2 X2 k − 2n ⋅ 2Σi<jXiXj + (n − 1)ΣX2 k + 2nΣi<jXiXj) Σk(Xk − ¯X)2 = 1 n2 (n(n − 1)ΣX2 k − 2nΣi<jXiXj) = 1 n ((n − 1)ΣX2 k − 2Σi<jXiXj) E[Σk(Xk − ¯X)2 ] = 1 n ((n − 1)ΣE[X2 k ] − 2Σi<jE[XiXj])
6.
V[X] = E[X2 ]
− E[X]2 E[Σk(Xk − ¯X)2 ] = 1 n ((n − 1)ΣE[X2 k ] − 2Σi<jE[XiXj]) Cov(X, Y) = E[XY] − E[X]E[Y] X Y 0 E[X2 ] = σ2 + μ2 E[XiXj] = μ2
7.
= n − 1 n (n(σ2 +
μ2 ) − nμ2 ) = (n − 1)σ2 E[Σk(Xk − ¯X)2 ] = 1 n ((n − 1)ΣE[X2 k ] − 2Σi<jE[XiXj]) = 1 n ((n − 1)n(σ2 + μ2 ) − 2 ⋅ 1 2 n(n − 1)μ2 ) Σ n Σ n(n-1)/2 E[ 1 n Σk(Xk − ¯X)2 ] = n − 1 n σ2
8.
(n−1)/n u2 u2 = n n − 1 s2 = 1 n
− 1 Σn k=1(Xk − ¯X)2
9.
n/ (n−1) E[u2 ]
= E[ n n − 1 s2 ] = n n − 1 E[s2 ] = n n − 1 n − 1 n σ2 = σ2
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