京都大学 情報学研究科 システム科学専攻 2016年8月実施 専門科目 確率統計 Author uogxtc
Description 問題1
下記の確率密度関数にしたがう確率変数 X X X α > 0 \alpha > 0 α > 0 β > 0 \beta > 0 β > 0
f ( x ) = { α x α − 1 β α exp ( − ( x β ) α ) ( x > 0 ) 0 ( x ≤ 0 ) f(x) = \begin{cases}
\frac{\alpha x^{\alpha-1}}{\beta^\alpha} \exp \left( - \left( \frac{x}{\beta} \right)^\alpha \right) & (x > 0) \\
0 & (x \leq 0)
\end{cases} f ( x ) = { β α α x α − 1 exp ( − ( β x ) α ) 0 ( x > 0 ) ( x ≤ 0 ) (1) 確率変数 X X X
Γ ( θ ) = ∫ 0 ∞ x θ − 1 e − x d x ( θ > 0 ) \Gamma(\theta) = \int_0^\infty x^{\theta-1} e^{-x} dx \quad (\theta > 0) Γ ( θ ) = ∫ 0 ∞ x θ − 1 e − x d x ( θ > 0 ) (2) 確率密度関数 f ( x ) f(x) f ( x ) n n n
{ X 1 , X 2 , … , X n } \{X_1, X_2, \ldots, X_n\} { X 1 , X 2 , … , X n } が得られたとする。このとき、パラメータ α = α 0 \alpha = \alpha_0 α = α 0 β \beta β
問題2
以下の設問に答えよ。
(1) X 1 , X 2 , … , X n X_1, X_2, \ldots, X_n X 1 , X 2 , … , X n f ( x ) f(x) f ( x ) F ( x ) F(x) F ( x ) X 1 , X 2 , … , X n X_1, X_2, \ldots, X_n X 1 , X 2 , … , X n
Z = min ( X 1 , X 2 , … , X n ) Z = \min(X_1, X_2, \ldots, X_n) Z = min ( X 1 , X 2 , … , X n ) もまた確率変数となるが、その確率密度関数 g ( z ) g(z) g ( z ) f f f F F F
(2) 設問 (1) の X 1 , X 2 , … , X n X_1, X_2, \ldots, X_n X 1 , X 2 , … , X n [ 0 , b ] [0, b] [ 0 , b ] ( b > 0 ) (b > 0) ( b > 0 )
Z = min ( X 1 , X 2 , … , X n ) Z = \min(X_1, X_2, \ldots, X_n) Z = min ( X 1 , X 2 , … , X n ) の期待値を求めよ。
問題3
以下の設問に答えよ。
(1) 半径 a a a C C C A , B A,B A , B C C C A B AB A B R R R R 2 R^2 R 2
(2) 設問 (1) において、点 A A A A B AB A B R R R C C C
题目描述
随机变量 X X X
f ( x ) = { α x α − 1 β α exp [ − ( x β ) α ] ( x > 0 ) , 0 ( x ≤ 0 ) , f(x)=
\begin{cases}
\dfrac{\alpha x^{\alpha-1}}{\beta^\alpha}
\exp\!\left[-\left(\dfrac{x}{\beta}\right)^\alpha\right]
& (x>0),\\
0 & (x\leq0),
\end{cases} f ( x ) = ⎩ ⎨ ⎧ β α α x α − 1 exp [ − ( β x ) α ] 0 ( x > 0 ) , ( x ≤ 0 ) , 其中 α > 0 , β > 0 \alpha>0,\beta>0 α > 0 , β > 0
（1）使用 Gamma 函数
Γ ( θ ) = ∫ 0 ∞ x θ − 1 e − x d x ( θ > 0 ) \Gamma(\theta)=\int_0^\infty x^{\theta-1}e^{-x}\,dx
\qquad(\theta>0) Γ ( θ ) = ∫ 0 ∞ x θ − 1 e − x d x ( θ > 0 ) 和上述参数表示 X X X
（2）从该密度所规定的分布中得到容量为 n n n
{ X 1 , X 2 , … , X n } . \{X_1,X_2,\ldots,X_n\}. { X 1 , X 2 , … , X n } . 已知 α = α 0 \alpha=\alpha_0 α = α 0 β \beta β
回答下列问题。
（1）设 X 1 , … , X n X_1,\ldots,X_n X 1 , … , X n f ( x ) , F ( x ) f(x),F(x) f ( x ) , F ( x )
Z = min ( X 1 , X 2 , … , X n ) . Z=\min(X_1,X_2,\ldots,X_n). Z = min ( X 1 , X 2 , … , X n ) . 用 f , F f,F f , F Z Z Z g ( z ) g(z) g ( z )
（2）若第（1）问中的 X 1 , … , X n X_1,\ldots,X_n X 1 , … , X n [ 0 , b ] [0,b] [ 0 , b ] b > 0 b>0 b > 0 Z Z Z
回答下列问题。
（1）在半径为 a a a C C C A , B A,B A , B R R R E [ R 2 ] E[R^2] E [ R 2 ]
（2）在第（1）问中，以 A A A R = A B R=AB R = A B C C C
Kai 問題1
(Readers may refer to Weibull distribution.)
(1)
E [ X ] = ∫ 0 ∞ x f ( x ) d x = ∫ 0 ∞ α ( x β ) α e − ( x β ) α d x . \begin{aligned}
\mathbb{E}[X]&=\int_0^\infty xf(x)dx\\
&=\int_0^\infty\alpha\left(\frac x\beta\right)^\alpha e^{-\left(\frac x\beta\right)^\alpha}dx.
\end{aligned} E [ X ] = ∫ 0 ∞ x f ( x ) d x = ∫ 0 ∞ α ( β x ) α e − ( β x ) α d x . Let u = ( x β ) α u=\left(\frac x\beta\right)^\alpha u = ( β x ) α
β u 1 / α = x d x = β α u 1 α − 1 d u , \begin{aligned}
&\beta u^{1/\alpha}=x \\
&dx=\frac\beta\alpha u^{\frac1\alpha-1}du,
\end{aligned} β u 1/ α = x d x = α β u α 1 − 1 d u , Then
E [ X ] = β ∫ 0 ∞ u 1 α e − u d u = β Γ ( 1 α + 1 ) . \begin{aligned}
\mathbb{E}[X]&=\beta\int_{0}^{\infty}u^{\frac{1}{\alpha}}e^{-u}du\\
&=\beta\Gamma\left(\frac{1}{\alpha}+1\right).
\end{aligned} E [ X ] = β ∫ 0 ∞ u α 1 e − u d u = β Γ ( α 1 + 1 ) . (2)
The likelihood function is
L ( β ) = ∏ i = 1 n α 0 X i α 0 − 1 β α 0 e − ( x β ) α 0 , L(\beta)=\prod_{i=1}^n\frac{\alpha_0X_i^{\alpha_0-1}}{\beta^{\alpha_0}}e^{-\left(\frac{x}{\beta}\right)^{\alpha_0}}, L ( β ) = i = 1 ∏ n β α 0 α 0 X i α 0 − 1 e − ( β x ) α 0 , from which we know that the log-likelihood function is
log L = n log α 0 − n α 0 log β + ( α 0 − 1 ) ∑ i = 1 n log X i − ∑ i = 1 n ( X i β ) α 0 . \log L=n\log\alpha_{0}-n\alpha_{0}\log\beta+(\alpha_{0}-1)\sum_{i=1}^{n}\log X_{i}-\sum_{i=1}^{n}\left(\frac{X_{i}}{\beta}\right)^{\alpha_{0}}. log L = n log α 0 − n α 0 log β + ( α 0 − 1 ) i = 1 ∑ n log X i − i = 1 ∑ n ( β X i ) α 0 . By setting ∂ log L ∂ β = 0 \frac{\partial\log L}{\partial\beta}=0 ∂ β ∂ l o g L = 0
− n α 0 β + α 0 1 β α 0 + 1 ∑ i = 1 n X i α 0 = 0 ⇒ β ^ = ( ∑ i = 1 n X i α 0 n ) 1 / α 0 . \begin{aligned}
&-\:\frac{n\alpha_{0}}{\beta} + \alpha_0 \frac{1}{\beta^{\alpha_0 + 1}}\sum_{i=1}^n X_i^{\alpha_0} =0 \Rightarrow\quad\hat{\beta}=\left(\frac{\sum_{i=1}^{n}X_{i}^{\alpha_{0}}}{n}\right)^{1/\alpha_{0}}.
\end{aligned} − β n α 0 + α 0 β α 0 + 1 1 i = 1 ∑ n X i α 0 = 0 ⇒ β ^ = ( n ∑ i = 1 n X i α 0 ) 1/ α 0 . 問題2
(1)
(The problems of the CDF/PDF of min/max of i . i . d . i.i.d. i . i . d .
The cumulative distribution function (CDF, 累積分布関数) of Z Z Z
Pr ( Z < s ) = 1 − Pr ( Z ≥ s ) = 1 − Pr ( X 1 ≥ s , … , X n ≥ s ) = 1 − ∏ i = 1 n ( 1 − F ( s ) ) = 1 − ( 1 − F ( s ) ) n . \begin{aligned}
\Pr(Z<s)&=1-\Pr(Z\geq s)\\
&=1-\Pr(X_1\geq s,\ldots,X_n\geq s)\\
&=1-\prod_{i=1}^n\left(1-F(s)\right)\\
&=1-\left(1-F(s)\right)^n.
\end{aligned} Pr ( Z < s ) = 1 − Pr ( Z ≥ s ) = 1 − Pr ( X 1 ≥ s , … , X n ≥ s ) = 1 − i = 1 ∏ n ( 1 − F ( s ) ) = 1 − ( 1 − F ( s ) ) n . The probability density function (PDF,確率密度関数) is
f Z ( s ) = d Pr ( Z < s ) d s = n ( 1 − F ( s ) ) n − 1 f ( s ) . f_Z(s)=\frac{d\Pr(Z<s)}{ds}=n(1-F(s))^{n-1}f(s). f Z ( s ) = d s d Pr ( Z < s ) = n ( 1 − F ( s ) ) n − 1 f ( s ) . (2)
The PDF of X i X_i X i
f ( x ) = { 1 / b , if 0 < x < b 0 , otherwise. f(x)=\begin{cases}
1/b,&\text{if }\ 0<x<b \\
0,&\text{otherwise.}
\end{cases} f ( x ) = { 1/ b , 0 , if 0 < x < b otherwise. Then,
E [ Z ] = ∫ 0 b x f Z ( x ) d x = ∫ 0 b n b x ( 1 − x b ) n − 1 d x ( let y = 1 − x b ) = ∫ 0 1 n b ⋅ ( b − b y ) y n − 1 ( − b ) d y = b n + 1 . \begin{aligned}
\mathbb{E}[Z]&=\int_0^bxf_Z(x)dx\\
&=\int_0^b\frac nbx\left(1-\frac xb\right)^{n-1}dx\\
&(\text{let }y=1-\frac xb)\\
&=\int_0^1\frac nb\cdot(b-by)y^{n-1}(-b)dy\\
&=\frac b{n+1}.
\end{aligned} E [ Z ] = ∫ 0 b x f Z ( x ) d x = ∫ 0 b b n x ( 1 − b x ) n − 1 d x ( let y = 1 − b x ) = ∫ 0 1 b n ⋅ ( b − b y ) y n − 1 ( − b ) d y = n + 1 b . 問題3
(1)
Suppose that A A A ( r A , θ A ) (r_A, \theta _A) ( r A , θ A ) B B B ( r B , θ B ) (r_B,\theta_B) ( r B , θ B ) r A , r B ∼ Unif ( 0 , a ) r_A,r_B \sim \text{Unif}(0,a) r A , r B ∼ Unif ( 0 , a ) θ A , θ B ∼ Unif ( 0 , 2 π ) . \theta_A,\theta_B \sim \text{Unif} ( 0, 2\pi ). θ A , θ B ∼ Unif ( 0 , 2 π ) .
Then
R 2 = r A 2 + r B 2 − 2 r A r B cos ( θ A − θ B ) , R^2=r_A^2+r_B^2-2r_Ar_B\cos(\theta_A-\theta_B), R 2 = r A 2 + r B 2 − 2 r A r B cos ( θ A − θ B ) , E [ R 2 ] = 1 2 a 2 π ∫ 0 a ∫ 0 a ∫ 0 2 π ( r A 2 + r B 2 − 2 r A r B cos θ ) d θ ⋅ r A d r A ⋅ r B d r B = a 4 / 4. \begin{aligned}
\mathbb{E}[R^{2}]&=\frac{1}{2a^{2}\pi}\int_{0}^{a}\int_{0}^{a}\int_{0}^{2\pi}(r_{A}^{2}+r_{B}^{2}-2r_{A}r_{B}\cos\theta)\ d\theta \cdot r_{A}dr_{A}\cdot r_{B}dr_{B}\\
&=a^{4}/4.
\end{aligned} E [ R 2 ] = 2 a 2 π 1 ∫ 0 a ∫ 0 a ∫ 0 2 π ( r A 2 + r B 2 − 2 r A r B cos θ ) d θ ⋅ r A d r A ⋅ r B d r B = a 4 /4. (2)
The probability is
1 π a 2 ∫ 0 a 2 π r ⋅ ( a − r ) 2 a 2 d r . \frac{1}{\pi a^2}\int_0^a2\pi r\cdot \frac{(a-r)^2}{a^2}dr. π a 2 1 ∫ 0 a 2 π r ⋅ a 2 ( a − r ) 2 d r . Here 2 π r π a 2 \frac{2\pi r}{\pi a^2} π a 2 2 π r A A A C C C r r r
Suppose that the distance between A A A r r r B B B A A A ( a − r ) 2 (a-r)^2 ( a − r ) 2 ( a − r ) 2 a 2 \frac{(a-r)^{2}}{a^{2}} a 2 ( a − r ) 2 A , B A,B A , B