京都大学 情報学研究科 システム科学専攻 2026年8月実施 確率統計 Author AnonTokyo #83056
Description 以下の問題において，E ( X ) E(X) E ( X ) X X X V ( X ) V(X) V ( X ) N ( μ , σ 2 ) N(\mu,\sigma^2) N ( μ , σ 2 ) μ \mu μ σ 2 \sigma^2 σ 2
問題 1
n n n X 1 , … , X n X_1,\ldots,X_n X 1 , … , X n X i ∼ N ( μ , 1 ) X_i\sim N(\mu,1) X i ∼ N ( μ , 1 ) i = 1 , … , n i=1,\ldots,n i = 1 , … , n μ \mu μ
X ˉ = 1 n ∑ i = 1 n X i \bar X=\frac{1}{n}\sum_{i=1}^n X_i X ˉ = n 1 i = 1 ∑ n X i とする。さらに，
R = X 1 − X ˉ R=X_1-\bar X R = X 1 − X ˉ とおく。なお，確率変数の組 ( Y , Z ) (Y,Z) ( Y , Z ) Y Y Y Z Z Z
(1) E ( R ) E(R) E ( R ) V ( R ) V(R) V ( R )
(2) R R R X ˉ \bar X X ˉ
(3) U = X 1 2 − 1 U=X_1^2-1 U = X 1 2 − 1 E ( U ) E(U) E ( U ) E ( U ∣ X ˉ = x ) E(U\mid\bar X=x) E ( U ∣ X ˉ = x )
問題 2
n n n a 1 , … , a n a_1,\ldots,a_n a 1 , … , a n ∑ i = 1 n a i 2 = 1 \sum_{i=1}^n a_i^2=1 ∑ i = 1 n a i 2 = 1 X X X X ∼ N ( 0 , σ 2 ) X\sim N(0,\sigma^2) X ∼ N ( 0 , σ 2 ) σ 2 > 0 \sigma^2>0 σ 2 > 0 X = x X=x X = x Y 1 , … , Y n Y_1,\ldots,Y_n Y 1 , … , Y n Y i ∼ N ( x a i , τ 2 ) Y_i\sim N(xa_i,\tau^2) Y i ∼ N ( x a i , τ 2 ) i = 1 , … , n i=1,\ldots,n i = 1 , … , n τ 2 > 0 \tau^2>0 τ 2 > 0 Z = ∑ i = 1 n a i Y i Z=\sum_{i=1}^n a_iY_i Z = ∑ i = 1 n a i Y i Y 1 = y 1 , … , Y n = y n Y_1=y_1,\ldots,Y_n=y_n Y 1 = y 1 , … , Y n = y n Z Z Z z = ∑ i = 1 n a i y i z=\sum_{i=1}^n a_iy_i z = ∑ i = 1 n a i y i
(1) X = x X=x X = x Z Z Z
(2) W = Z − X W=Z-X W = Z − X X = x X=x X = x W W W x x x W W W X X X Z Z Z
(3) Y 1 = y 1 , … , Y n = y n Y_1=y_1,\ldots,Y_n=y_n Y 1 = y 1 , … , Y n = y n X X X f ( x ∣ y 1 , … , y n ) f(x\mid y_1,\ldots,y_n) f ( x ∣ y 1 , … , y n ) f ( x ∣ y 1 , … , y n ) f(x\mid y_1,\ldots,y_n) f ( x ∣ y 1 , … , y n ) x x x x ^ \hat x x ^ z , σ 2 , τ 2 z,\sigma^2,\tau^2 z , σ 2 , τ 2
(4) 設問 (2) で求めた Z Z Z h ( z ; σ 2 ) h(z;\sigma^2) h ( z ; σ 2 ) Z = z Z=z Z = z h ( z ; σ 2 ) h(z;\sigma^2) h ( z ; σ 2 ) σ 2 \sigma^2 σ 2 σ ^ 2 \hat\sigma^2 σ ^ 2 z , τ 2 z,\tau^2 z , τ 2 σ 2 > 0 \sigma^2>0 σ 2 > 0 σ ^ 2 = 0 \hat\sigma^2=0 σ ^ 2 = 0
(5) 設問 (3) で求めた x ^ \hat x x ^ σ 2 \sigma^2 σ 2 σ ^ 2 \hat\sigma^2 σ ^ 2 z , τ 2 z,\tau^2 z , τ 2
Kai 問題 1
(1) R R R
E ( X 1 ) = E ( X ˉ ) = μ E(X_1)=E(\bar X)=\mu E ( X 1 ) = E ( X ˉ ) = μ
E ( R ) = 0 . \boxed{E(R)=0}. E ( R ) = 0 . また，V ( X 1 ) = 1 V(X_1)=1 V ( X 1 ) = 1 V ( X ˉ ) = 1 / n V(\bar X)=1/n V ( X ˉ ) = 1/ n Cov ( X 1 , X ˉ ) = 1 / n \operatorname{Cov}(X_1,\bar X)=1/n Cov ( X 1 , X ˉ ) = 1/ n
V ( R ) = V ( X 1 ) + V ( X ˉ ) − 2 Cov ( X 1 , X ˉ ) = 1 + 1 n − 2 n = 1 − 1 n . V(R)=V(X_1)+V(\bar X)-2\operatorname{Cov}(X_1,\bar X)
=1+\frac1n-\frac2n
=\boxed{1-\frac1n}. V ( R ) = V ( X 1 ) + V ( X ˉ ) − 2 Cov ( X 1 , X ˉ ) = 1 + n 1 − n 2 = 1 − n 1 . (2) 無相関の証明
Cov ( R , X ˉ ) = Cov ( X 1 − X ˉ , X ˉ ) = 1 n − 1 n = 0. \operatorname{Cov}(R,\bar X)
=\operatorname{Cov}(X_1-\bar X,\bar X)
=\frac1n-\frac1n=0. Cov ( R , X ˉ ) = Cov ( X 1 − X ˉ , X ˉ ) = n 1 − n 1 = 0. 従って R R R X ˉ \bar X X ˉ
(3) U U U
E ( X 1 2 ) = V ( X 1 ) + [ E ( X 1 ) ] 2 = 1 + μ 2 E(X_1^2)=V(X_1)+[E(X_1)]^2=1+\mu^2 E ( X 1 2 ) = V ( X 1 ) + [ E ( X 1 ) ] 2 = 1 + μ 2
E ( U ) = μ 2 . \boxed{E(U)=\mu^2}. E ( U ) = μ 2 . X 1 = X ˉ + R X_1=\bar X+R X 1 = X ˉ + R R R R X ˉ \bar X X ˉ
E ( U ∣ X ˉ = x ) = E [ ( x + R ) 2 − 1 ] = x 2 + 2 x E ( R ) + E ( R 2 ) − 1 = x 2 − 1 n . \begin{aligned}
E(U\mid\bar X=x)
&=E[(x+R)^2-1]\\
&=x^2+2xE(R)+E(R^2)-1\\
&=\boxed{x^2-\frac1n}.
\end{aligned} E ( U ∣ X ˉ = x ) = E [( x + R ) 2 − 1 ] = x 2 + 2 x E ( R ) + E ( R 2 ) − 1 = x 2 − n 1 . 問題 2
(1) Z ∣ X = x Z\mid X=x Z ∣ X = x
条件付きの Y i Y_i Y i Z Z Z ∑ i a i 2 = 1 \sum_i a_i^2=1 ∑ i a i 2 = 1
E ( Z ∣ X = x ) = ∑ i a i ( x a i ) = x , E(Z\mid X=x)=\sum_i a_i(xa_i)=x, E ( Z ∣ X = x ) = i ∑ a i ( x a i ) = x , V ( Z ∣ X = x ) = τ 2 ∑ i a i 2 = τ 2 . V(Z\mid X=x)=\tau^2\sum_i a_i^2=\tau^2. V ( Z ∣ X = x ) = τ 2 i ∑ a i 2 = τ 2 . ゆえに
Z ∣ X = x ∼ N ( x , τ 2 ) . \boxed{Z\mid X=x\sim N(x,\tau^2)}. Z ∣ X = x ∼ N ( x , τ 2 ) . (2) W W W Z Z Z
W = Z − X W=Z-X W = Z − X
W ∣ X = x ∼ N ( 0 , τ 2 ) W\mid X=x\sim N(0,\tau^2) W ∣ X = x ∼ N ( 0 , τ 2 ) であり，x x x W W W X X X W ∼ N ( 0 , τ 2 ) W\sim N(0,\tau^2) W ∼ N ( 0 , τ 2 ) Z = X + W Z=X+W Z = X + W
Z ∼ N ( 0 , σ 2 + τ 2 ) . \boxed{Z\sim N(0,\sigma^2+\tau^2)}. Z ∼ N ( 0 , σ 2 + τ 2 ) . (3) 条件付き密度の最大点
ベイズの定理より，x x x
f ( x ∣ y 1 , … , y n ) ∝ exp [ − x 2 2 σ 2 − 1 2 τ 2 ∑ i ( y i − a i x ) 2 ] ∝ exp [ − 1 2 ( 1 σ 2 + 1 τ 2 ) x 2 + z τ 2 x ] . \begin{aligned}
f(x\mid y_1,\ldots,y_n)
&\propto\exp\!\left[-\frac{x^2}{2\sigma^2}-\frac1{2\tau^2}\sum_i(y_i-a_ix)^2\right]\\
&\propto\exp\!\left[-\frac12\left(\frac1{\sigma^2}+\frac1{\tau^2}\right)x^2+\frac z{\tau^2}x\right].
\end{aligned} f ( x ∣ y 1 , … , y n ) ∝ exp [ − 2 σ 2 x 2 − 2 τ 2 1 i ∑ ( y i − a i x ) 2 ] ∝ exp [ − 2 1 ( σ 2 1 + τ 2 1 ) x 2 + τ 2 z x ] . 指数部を x x x
− ( 1 σ 2 + 1 τ 2 ) x + z τ 2 = 0. -\left(\frac1{\sigma^2}+\frac1{\tau^2}\right)x+\frac z{\tau^2}=0. − ( σ 2 1 + τ 2 1 ) x + τ 2 z = 0. よって
x ^ = σ 2 σ 2 + τ 2 z . \boxed{\hat x=\frac{\sigma^2}{\sigma^2+\tau^2}z}. x ^ = σ 2 + τ 2 σ 2 z . (4) σ 2 \sigma^2 σ 2
(2) より
h ( z ; σ 2 ) = 1 2 π ( σ 2 + τ 2 ) exp [ − z 2 2 ( σ 2 + τ 2 ) ] . h(z;\sigma^2)=\frac{1}{\sqrt{2\pi(\sigma^2+\tau^2)}}
\exp\!\left[-\frac{z^2}{2(\sigma^2+\tau^2)}\right]. h ( z ; σ 2 ) = 2 π ( σ 2 + τ 2 ) 1 exp [ − 2 ( σ 2 + τ 2 ) z 2 ] . 対数尤度の微分は
∂ log h ∂ σ 2 = z 2 − σ 2 − τ 2 2 ( σ 2 + τ 2 ) 2 . \frac{\partial\log h}{\partial\sigma^2}
=\frac{z^2-\sigma^2-\tau^2}{2(\sigma^2+\tau^2)^2}. ∂ σ 2 ∂ log h = 2 ( σ 2 + τ 2 ) 2 z 2 − σ 2 − τ 2 . z 2 > τ 2 z^2>\tau^2 z 2 > τ 2 σ 2 = z 2 − τ 2 \sigma^2=z^2-\tau^2 σ 2 = z 2 − τ 2 z 2 ≤ τ 2 z^2\le\tau^2 z 2 ≤ τ 2 σ 2 > 0 \sigma^2>0 σ 2 > 0
σ ^ 2 = max { z 2 − τ 2 , 0 } . \boxed{\hat\sigma^2=\max\{z^2-\tau^2,0\}}. σ ^ 2 = max { z 2 − τ 2 , 0 } . (5) 代入推定値
(3) の式に (4) の値を代入すると
x ^ p l u g - i n = { z − τ 2 z , ∣ z ∣ > τ , 0 , ∣ z ∣ ≤ τ . \boxed{
\hat x_{\mathrm{plug\text{-}in}}=
\begin{cases}
z-\dfrac{\tau^2}{z},&|z|>\tau,\\[5pt]
0,&|z|\le\tau.
\end{cases}} x ^ plug - in = ⎩ ⎨ ⎧ z − z τ 2 , 0 , ∣ z ∣ > τ , ∣ z ∣ ≤ τ . となる。z = 0 z=0 z = 0