京都大学 情報学研究科 知能情報学専攻 2022年8月実施 専門科目 S-2 Author Isidore , 祭音Myyura
Description 大学公表の原題
設問1
確率変数 X X X f ( x ) f(x) f ( x )
f ( x ) = { − 1 2 x + 1 2 ( ∣ x ∣ ≤ 1 ) 0 ( ∣ x ∣ > 1 ) f(x) =
\left \{
\begin{aligned}
-\frac{1}{2}x + \frac{1}{2} \quad &(|x| \le 1) \\
0 \quad &(|x| > 1)
\end{aligned}
\right. f ( x ) = ⎩ ⎨ ⎧ − 2 1 x + 2 1 0 ( ∣ x ∣ ≤ 1 ) ( ∣ x ∣ > 1 ) X X X
設問2
独立な確率変数 X X X Y Y Y λ 1 , λ 2 \lambda_1,\lambda_2 λ 1 , λ 2 Z = X + Y Z = X + Y Z = X + Y ( λ 1 + λ 2 ) (\lambda_1 + \lambda_2) ( λ 1 + λ 2 ) λ \lambda λ
P ( X = k ) = λ k e − λ k ! P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} P ( X = k ) = k ! λ k e − λ 設問3
モデル y i = β x i + ϵ i ( i = 1 , 2 , … , n ; ϵ i は誤差項 ) y_i = \beta x_i + \epsilon_i(i = 1,2,\dots,n;\epsilon_i \text{は誤差項}) y i = β x i + ϵ i ( i = 1 , 2 , … , n ; ϵ i は誤差項 ) ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) (x_1,y_1),(x_2,y_2),\dots,(x_n,y_n) ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) y = β ^ x y = \hat{\beta}x y = β ^ x
(1) β ^ \hat{\beta} β ^
(2) ϵ i ^ = y i − β ^ x i \hat{\epsilon_i} = y_i - \hat{\beta}x_i ϵ i ^ = y i − β ^ x i ∑ i = 1 n x i ϵ i ^ = 0 \sum_{i = 1}^n x_i\hat{\epsilon_i} = 0 ∑ i = 1 n x i ϵ i ^ = 0
設問4
ある研究分野における全統計的仮説の中で、真の仮説 (帰無仮説が誤り) と偽の仮説 (帰無仮説が正しい) の数の比が R : 1 R:1 R : 1
(1) ある仮説について実験を実施し、有意水準 α \alpha α 1 − β 1 - \beta 1 − β R , α , β R,\alpha,\beta R , α , β R = 0.1 , α = 0.05 , β = 0.2 R = 0.1,\alpha = 0.05,\beta = 0.2 R = 0.1 , α = 0.05 , β = 0.2
(2) ある仮説について k k k k k k R , α , β , k R,\alpha,\beta,k R , α , β , k R = 0.1 , α = 0.05 , β = 0.2 , k = 2 R = 0.1,\alpha = 0.05,\beta = 0.2,k = 2 R = 0.1 , α = 0.05 , β = 0.2 , k = 2
题目描述
随机变量 X X X
f ( x ) = { − 1 2 x + 1 2 , ∣ x ∣ ≤ 1 , 0 , ∣ x ∣ > 1. f(x)=\begin{cases}
-\frac12x+\frac12,&|x|\le1,\\
0,&|x|>1.
\end{cases} f ( x ) = { − 2 1 x + 2 1 , 0 , ∣ x ∣ ≤ 1 , ∣ x ∣ > 1. 求 E [ X ] E[X] E [ X ] V [ X ] V[X] V [ X ]
独立随机变量 X , Y X,Y X , Y λ 1 , λ 2 \lambda_1,\lambda_2 λ 1 , λ 2 P ( X = k ) = λ k e − λ / k ! P(X=k)=\lambda^ke^{-\lambda}/k! P ( X = k ) = λ k e − λ / k ! Z = X + Y ∼ Poisson ( λ 1 + λ 2 ) Z=X+Y\sim\operatorname{Poisson}(\lambda_1+\lambda_2) Z = X + Y ∼ Poisson ( λ 1 + λ 2 )
数据由过原点模型
y i = β x i + ϵ i y_i=\beta x_i+\epsilon_i y i = β x i + ϵ i y = β ^ x y=\hat\beta x y = β ^ x
求 β ^ \hat\beta β ^
令残差 ϵ ^ i = y i − β ^ x i \hat\epsilon_i=y_i-\hat\beta x_i ϵ ^ i = y i − β ^ x i ∑ i x i ϵ ^ i = 0 \sum_i x_i\hat\epsilon_i=0 ∑ i x i ϵ ^ i = 0
某领域真假研究假设（真指零假设错误）的先验数量比为 R : 1 R:1 R : 1
一次检验显著性水平为 α \alpha α 1 − β 1-\beta 1 − β R = 0.1 , α = 0.05 , β = 0.2 R=0.1,\alpha=0.05,\beta=0.2 R = 0.1 , α = 0.05 , β = 0.2
对同一假设作 k k k k = 2 k=2 k = 2
Kai 設問1
E [ X ] = ∫ − 1 1 x ( − 1 2 x + 1 2 ) d x = − 1 3 E[X] = \int^{1}_{-1}x(-\frac{1}{2}x+\frac{1}{2})\mathrm{d}x = -\frac{1}{3} E [ X ] = ∫ − 1 1 x ( − 2 1 x + 2 1 ) d x = − 3 1 E [ X 2 ] = ∫ − 1 1 x 2 ( − 1 2 x + 1 2 ) d x = 1 3 E[X^2] = \int^{1}_{-1}x^2(-\frac{1}{2}x+\frac{1}{2})\mathrm{d}x = \frac{1}{3} \\ E [ X 2 ] = ∫ − 1 1 x 2 ( − 2 1 x + 2 1 ) d x = 3 1 Var [ X ] = E [ X 2 ] − E 2 [ X ] = 2 9 \text{Var}[X] = E[X^2] - E^2[X] = \frac{2}{9} Var [ X ] = E [ X 2 ] − E 2 [ X ] = 9 2 設問2
By the convolution Rule, we have
f Z ( z ) = P ( Z = z ) = ∑ i = 0 z f X ( i ) f Y ( z − i ) = ∑ i = 0 z λ 1 i e − λ 1 i ! λ 2 z − i e − λ 2 ( z − i ) ! = e − ( λ 1 + λ 2 ) z ! ∑ i = 0 z z ! ( z − i ) ! i ! λ 1 i λ 2 z − i \begin{align}
f_{Z}(z) &= P(Z=z) = \sum^{z}_{i=0}f_{X}(i)f_{Y}(z-i) \\
& = \sum^{z}_{i=0} \frac{\lambda_1^{i}e^{-\lambda_1}}{i!} \frac{\lambda_2^{z - i}e^{-\lambda_2}}{(z - i)!} \\
& = \frac{e^{-(\lambda_1 + \lambda_2)}}{z!}\sum^{z}_{i=0}\frac{z!}{(z-i)!i!}\lambda_1^{i}\lambda_2^{z - i}
\end{align} f Z ( z ) = P ( Z = z ) = i = 0 ∑ z f X ( i ) f Y ( z − i ) = i = 0 ∑ z i ! λ 1 i e − λ 1 ( z − i )! λ 2 z − i e − λ 2 = z ! e − ( λ 1 + λ 2 ) i = 0 ∑ z ( z − i )! i ! z ! λ 1 i λ 2 z − i By the Binomial Theorem, we can insert ∑ i = 0 z z ! ( z − i ) ! i ! λ 1 i λ 2 z − i = ( λ 1 + λ 2 ) z \sum^{z}_{i=0}\frac{z!}{(z-i)!i!}\lambda_1^{i}\lambda_2^{z - i} = (\lambda_1 + \lambda_2)^z ∑ i = 0 z ( z − i )! i ! z ! λ 1 i λ 2 z − i = ( λ 1 + λ 2 ) z
f Z ( z ) = e − ( λ 1 + λ 2 ) z ! ( λ 1 + λ 2 ) z f_{Z}(z) = \frac{e^{-(\lambda_1 + \lambda_2)}}{z!}(\lambda_1 + \lambda_2)^z f Z ( z ) = z ! e − ( λ 1 + λ 2 ) ( λ 1 + λ 2 ) z So is the PMF for a Poisson Distribution with the parameter ( λ 1 + λ 2 ) (\lambda_1 + \lambda_2) ( λ 1 + λ 2 )
PS : A easier solution is to use Moments Generating Function.
設問3
(1)
By the Least Square Method, we have the sum of square residuals,
S ( β ^ ) = ∑ i = 1 n ( y i − β ^ x i ) 2 S(\hat{\beta}) = \sum^{n}_{i=1}(y_i - \hat{\beta}x_i)^2 S ( β ^ ) = i = 1 ∑ n ( y i − β ^ x i ) 2 Calculate its derivative with the root
S ′ ( β ^ ) = ∑ i = 1 n ( − 2 x i ) ( y i − β ^ x i ) = 2 ( ∑ i = 1 n β ^ x i 2 − ∑ i = 1 n x i y i ) β ^ = ∑ i = 1 n x i y i ∑ i = 1 n x i 2 , ∑ i = 1 n x i 2 > 0. \begin{align}
S'(\hat{\beta}) &= \sum^{n}_{i=1}(-2x_i)(y_i-\hat{\beta}x_i) \\
&= 2(\sum^{n}_{i=1}\hat{\beta}x_i^2 - \sum^{n}_{i=1}x_iy_i) \\
\hat{\beta} &= \frac{\sum^{n}_{i=1}x_iy_i}{\sum^{n}_{i=1}x_i^2},\qquad \sum_{i=1}^n x_i^2>0.
\end{align} S ′ ( β ^ ) β ^ = i = 1 ∑ n ( − 2 x i ) ( y i − β ^ x i ) = 2 ( i = 1 ∑ n β ^ x i 2 − i = 1 ∑ n x i y i ) = ∑ i = 1 n x i 2 ∑ i = 1 n x i y i , i = 1 ∑ n x i 2 > 0. If x 1 = ⋯ = x n = 0 x_1=\cdots=x_n=0 x 1 = ⋯ = x n = 0 β ^ \hat\beta β ^ S S S
(2)
The normal equation directly gives
0 = ( ∑ i = 1 n x i y i − ∑ i = 1 n β ^ x i 2 ) = ∑ i = 1 n x i ( y i − β ^ x i ) = ∑ i = 1 n x i ϵ i ^ \begin{align}
0 &= (\sum^{n}_{i=1}x_iy_i-\sum^{n}_{i=1}\hat{\beta}x_i^2) \\
&= \sum^{n}_{i=1}x_i(y_i-\hat{\beta}x_i) \\
&= \sum^{n}_{i=1}x_i\hat{\epsilon_i}
\end{align} 0 = ( i = 1 ∑ n x i y i − i = 1 ∑ n β ^ x i 2 ) = i = 1 ∑ n x i ( y i − β ^ x i ) = i = 1 ∑ n x i ϵ i ^ 設問4
(1)
Denote the events below:
The hypothesis is true: T T T
The hypothesis is false: F F F
The test results in significance: S S S
Then, what is asked can be represented by the probability as P r [ T ∣ S ] Pr[T|S] P r [ T ∣ S ] R : 1 R:1 R : 1
P r [ T ] = R R + 1 ; P r [ F ] = 1 R + 1 Pr[T] = \frac{R}{R+1} ;\; Pr[F] = \frac{1}{R+1} P r [ T ] = R + 1 R ; P r [ F ] = R + 1 1 Assume each test has exact Type I error probability α \alpha α 1 − β 1-\beta 1 − β
P r [ S ∣ F ] = α ; P r [ S ∣ T ] = 1 − β Pr[S|F] = \alpha ;\; Pr[S|T] = 1 - \beta P r [ S ∣ F ] = α ; P r [ S ∣ T ] = 1 − β Therefore, with the Bayes' theorem, we have
P r [ T ∣ S ] = P r [ S ∩ T ] P r [ S ] = P r [ S ∣ T ] P r [ T ] P r [ S ∣ T ] P r [ T ] + P r [ S ∣ F ] P r [ F ] = ( 1 − β ) R α + ( 1 − β ) R \begin{align}
Pr[T|S] &= \frac{Pr[S\cap T]}{Pr[S]} \\
&= \frac{Pr[S|T]Pr[T]}{Pr[S|T]Pr[T]+Pr[S|F]Pr[F]} \\
&=\frac{(1-\beta)R}{\alpha+(1-\beta)R}
\end{align} P r [ T ∣ S ] = P r [ S ] P r [ S ∩ T ] = P r [ S ∣ T ] P r [ T ] + P r [ S ∣ F ] P r [ F ] P r [ S ∣ T ] P r [ T ] = α + ( 1 − β ) R ( 1 − β ) R Insert the values, the answer is
P r [ T ∣ S ] = 8 13 Pr[T|S] = \frac{8}{13} P r [ T ∣ S ] = 13 8 (2)
Conditional on whether the hypothesis is true or false, the k k k S k = S 1 ∩ ⋯ ∩ S k S^k=S_1\cap\cdots\cap S_k S k = S 1 ∩ ⋯ ∩ S k
P r [ T ∣ S k ] = P r [ S k ∣ T ] P r [ T ] P r [ S k ∣ T ] P r [ T ] + P r [ S k ∣ F ] P r [ F ] = P r k [ S ∣ T ] P r [ T ] P r k [ S ∣ T ] P r [ T ] + P r k [ S ∣ F ] P r [ F ] = ( 1 − β ) k R α k + ( 1 − β ) k R \begin{align}
Pr[T|S^k] &= \frac{Pr[S^k|T]Pr[T]}{Pr[S^k|T]Pr[T]+Pr[S^k|F]Pr[F]} \\
&= \frac{Pr^k[S|T]Pr[T]}{Pr^k[S|T]Pr[T]+Pr^k[S|F]Pr[F]} \\
&= \frac{(1-\beta)^kR}{\alpha^k+(1-\beta)^kR}
\end{align} P r [ T ∣ S k ] = P r [ S k ∣ T ] P r [ T ] + P r [ S k ∣ F ] P r [ F ] P r [ S k ∣ T ] P r [ T ] = P r k [ S ∣ T ] P r [ T ] + P r k [ S ∣ F ] P r [ F ] P r k [ S ∣ T ] P r [ T ] = α k + ( 1 − β ) k R ( 1 − β ) k R Insert the values, the answer is
P r [ T ∣ S 2 ] = 128 133 Pr[T|S^2] = \frac{128}{133} P r [ T ∣ S 2 ] = 133 128