当前位置:网站首页>[the Nine Yang Manual] 2019 Fudan University Applied Statistics real problem + analysis
[the Nine Yang Manual] 2019 Fudan University Applied Statistics real problem + analysis
2022-07-06 13:30:00 【Elder martial brother statistics】
Catalog
The real part
One 、(15 branch ) There is n − 1 n-1 n−1 A white ball 1 A black ball , There is n n n A white ball , Take one ball from each of the two bags for Exchange , Seek exchange N N N The probability that the black ball will still be in the armour pocket after times .
Two 、(15 branch ) f ( x , y ) = A e − ( 2 x + 3 y ) I [ x > 0 , y > 0 ] , f(x, y)=A e^{-(2 x+3 y)} I[x>0, y>0], f(x,y)=Ae−(2x+3y)I[x>0,y>0], seek
(1)(3 branch ) A A A;
(2)(3 branch ) P ( X < 2 , Y < 1 ) P(X<2, Y<1) P(X<2,Y<1);
(3)(3 branch ) X X X Marginal density of ;
(4)(3 branch ) P ( X < 3 ∣ Y < 1 ) P(X<3 \mid Y<1) P(X<3∣Y<1);
(5)(3 branch ) f ( x ∣ y ) f(x \mid y) f(x∣y).
3、 ... and 、(10 branch ) X 1 , X 2 , X_{1}, X_{2}, X1,X2,i.i.d ∼ N ( μ , σ 2 ) , \sim N\left(\mu, \sigma^{2}\right), ∼N(μ,σ2), seek E max { X 1 , X 2 } E \max \left\{X_{1}, X_{2}\right\} Emax{ X1,X2}.
Four 、(20 branch ) E X = 0 , Var ( X ) = σ 2 , E X=0, \operatorname{Var}(X)=\sigma^{2}, EX=0,Var(X)=σ2, Prove for any ε > 0 , \varepsilon>0, ε>0, Yes
(1)(10 branch ) P ( ∣ X ∣ > ε ) ≤ σ 2 ε 2 P(|X|>\varepsilon) \leq \frac{\sigma^{2}}{\varepsilon^{2}} P(∣X∣>ε)≤ε2σ2;
(2)(10 branch ) P ( X > ε ) ≤ σ 2 σ 2 + ε 2 P(X>\varepsilon) \leq \frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}} P(X>ε)≤σ2+ε2σ2.
5、 ... and 、(10 branch ) X 1 , X 2 , i . i . d ∼ N ( 0 , 1 ) , X_{1}, X_{2}, i . i . d \sim N(0,1), X1,X2,i.i.d∼N(0,1), seek X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 The distribution of .
6、 ... and 、(20 branch ) X 1 , X 2 , … , X n , i . i . d ∼ N ( μ , σ 2 ) , μ X_{1}, X_{2}, \ldots, X_{n}, i . i . d \sim N\left(\mu, \sigma^{2}\right), \mu X1,X2,…,Xn,i.i.d∼N(μ,σ2),μ It is known that , prove :
(1)(10 branch ) 1 n ∑ i = 1 n ( X i − μ ) 2 \frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} n1∑i=1n(Xi−μ)2 yes σ 2 \sigma^{2} σ2 Effective estimation of ;
(2)(10 branch ) 1 n π 2 ∑ i = 1 n ∣ X i − μ ∣ \frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right| n12π∑i=1n∣Xi−μ∣ yes σ \sigma σ Unbiased estimation of , But it's not effective .
7、 ... and 、(20 branch ) The overall distribution function F ( x ) F(x) F(x) Continuous single increase , X ( 1 ) , X ( 2 ) , … , X ( n ) X_{(1)}, X_{(2)}, \ldots, X_{(n)} X(1),X(2),…,X(n) Is the order statistic of random samples from the population , Y i = F ( X ( i ) ) , Y_{i}=F\left(X_{(i)}\right), Yi=F(X(i)), seek
(1)(10 branch ) E Y i , Var ( Y i ) E Y_{i}, \operatorname{Var}\left(Y_{i}\right) EYi,Var(Yi);
(2)(10 branch ) ( Y 1 , Y 2 , … , Y n ) T \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)^{T} (Y1,Y2,…,Yn)T The covariance matrix of .
8、 ... and 、(20 branch ) Some come from the general U ( θ , 2 θ ) U(\theta,2\theta) U(θ,2θ) A random sample of X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn, seek θ \theta θ Moment estimation sum of MLE, And verify the unbiasedness and consistency .
Nine 、(20 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn Is from N ( μ , 1 ) N(\mu,1) N(μ,1) A random sample of , Consider the hypothesis test problem H 0 : μ = 1 v s H 1 : μ = 2 H_0:\mu = 1 \quad \mathrm{vs} \quad H_1:\mu=2 H0:μ=1vsH1:μ=2 Given denial domain W = { X ˉ > 1.6 } W=\{\bar{X}>1.6\} W={ Xˉ>1.6}, Answer the following questions :
(1)(10 branch ) Find the probability of making two kinds of mistakes α \alpha α, β \beta β;
(2)(10 branch ) Request the second kind of error β ≤ 0.01 \beta\le0.01 β≤0.01, Find the value range of the sample size .
The analysis part
One 、(15 branch ) There is n − 1 n-1 n−1 A white ball 1 A black ball , There is n n n A white ball , Take one ball from each of the two bags for Exchange , Seek exchange N N N The probability that the black ball will still be in the armour pocket after times .
Solution:
set up p k p_{k} pk For exchange k k k The probability that the black ball will still be in the armour pocket after times , Then according to the full probability formula ,
p k + 1 = p k ⋅ n − 1 n + ( 1 − p k ) ⋅ 1 n = ( 1 − 2 n ) p k + 1 n , p_{k+1}=p_{k} \cdot \frac{n-1}{n}+\left(1-p_{k}\right) \cdot \frac{1}{n}=\left(1-\frac{2}{n}\right) p_{k}+\frac{1}{n}, pk+1=pk⋅nn−1+(1−pk)⋅n1=(1−n2)pk+n1, There are p N − 1 2 = ( 1 − 2 n ) ( p N − 1 − 1 2 ) = ⋯ = ( 1 − 2 n ) N ( p 0 − 1 2 ) , p_{N}-\frac{1}{2}=\left(1-\frac{2}{n}\right)\left(p_{N-1}-\frac{1}{2}\right)=\cdots=\left(1-\frac{2}{n}\right)^{N}\left(p_{0}-\frac{1}{2}\right), pN−21=(1−n2)(pN−1−21)=⋯=(1−n2)N(p0−21), namely p N = 1 2 + 1 2 ( 1 − 2 n ) N . p_{N}=\frac{1}{2}+\frac{1}{2}\left(1-\frac{2}{n}\right)^{N} \text {. } pN=21+21(1−n2)N.
Two 、(15 branch ) f ( x , y ) = A e − ( 2 x + 3 y ) I [ x > 0 , y > 0 ] , f(x, y)=A e^{-(2 x+3 y)} I[x>0, y>0], f(x,y)=Ae−(2x+3y)I[x>0,y>0], seek
(1)(3 branch ) A A A;
(2)(3 branch ) P ( X < 2 , Y < 1 ) P(X<2, Y<1) P(X<2,Y<1);
(3)(3 branch ) X X X Marginal density of ;
(4)(3 branch ) P ( X < 3 ∣ Y < 1 ) P(X<3 \mid Y<1) P(X<3∣Y<1);
(5)(3 branch ) f ( x ∣ y ) f(x \mid y) f(x∣y).
Solution:
(1) By the regularity of probability , 1 = ∫ R 2 f ( x , y ) d x d y = A ∫ 0 + ∞ e − 2 x d x ∫ 0 + ∞ e − 3 y d y = A 6 ⇒ A = 6 1=\int_{R^{2}} f(x, y) d x d y=A \int_{0}^{+\infty} e^{-2 x} d x \int_{0}^{+\infty} e^{-3 y} d y=\frac{A}{6} \Rightarrow A=6 1=∫R2f(x,y)dxdy=A∫0+∞e−2xdx∫0+∞e−3ydy=6A⇒A=6.
(2) P ( X < 2 , Y < 1 ) = 6 ∫ 0 2 e − 2 x d x ∫ 0 1 e − 3 y d y = 6 ( 1 − e − 4 ) ( 1 − e − 3 ) P(X<2, Y<1)=6 \int_{0}^{2} e^{-2 x} d x \int_{0}^{1} e^{-3 y} d y=6\left(1-e^{-4}\right)\left(1-e^{-3}\right) P(X<2,Y<1)=6∫02e−2xdx∫01e−3ydy=6(1−e−4)(1−e−3).
(3) f X ( x ) = ∫ 0 + ∞ 6 e − 2 x − 3 y d y = 2 e − 2 x , x > 0 f_{X}(x)=\int_{0}^{+\infty} 6 e^{-2 x-3 y} d y=2 e^{-2 x}, x>0 fX(x)=∫0+∞6e−2x−3ydy=2e−2x,x>0.
(4) because f X ( x ) = 2 e − 2 x , f Y ( y ) = 3 e − 3 y f_{X}(x)=2 e^{-2 x}, f_{Y}(y)=3 e^{-3 y} fX(x)=2e−2x,fY(y)=3e−3y, so X , Y X, Y X,Y Are independent of each other , therefore P ( X < 3 ∣ Y < 1 ) = P ( X < 3 ) = 1 − e − 6 . P(X<3 \mid Y<1)=P(X<3)=1-e^{-6} . P(X<3∣Y<1)=P(X<3)=1−e−6.(5) f ( x ∣ y ) = f ( x , y ) f Y ( y ) = 2 e − 2 x , x > 0 , y > 0 f(x \mid y)=\frac{f(x, y)}{f_{Y}(y)}=2 e^{-2 x}, x>0, y>0 f(x∣y)=fY(y)f(x,y)=2e−2x,x>0,y>0.
3、 ... and 、(10 branch ) X 1 , X 2 , X_{1}, X_{2}, X1,X2,i.i.d ∼ N ( μ , σ 2 ) , \sim N\left(\mu, \sigma^{2}\right), ∼N(μ,σ2), seek E max { X 1 , X 2 } E \max \left\{X_{1}, X_{2}\right\} Emax{ X1,X2}.
Solution:
Make Y i = X i − μ σ , i = 1 , 2 Y_{i}=\frac{X_{i}-\mu}{\sigma}, i=1,2 Yi=σXi−μ,i=1,2, so max { Y 1 , Y 2 } = max { X 1 , X 2 } − μ σ \max \left\{Y_{1}, Y_{2}\right\}=\frac{\max \left\{X_{1}, X_{2}\right\}-\mu}{\sigma} max{ Y1,Y2}=σmax{ X1,X2}−μ.
and E max { Y 1 , Y 2 } = E Y 1 I [ Y 1 ≥ Y 2 ] + E Y 2 I [ Y 1 < Y 2 ] = 2 E Y 1 I [ Y 1 ≥ Y 2 ] , E \max \left\{Y_{1}, Y_{2}\right\}=E Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]}+E Y_{2} I_{\left[Y_{1}<Y_{2}\right]}=2 E Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]}, Emax{ Y1,Y2}=EY1I[Y1≥Y2]+EY2I[Y1<Y2]=2EY1I[Y1≥Y2], E [ Y 1 I [ Y 1 ≥ Y 2 ] ] = 1 2 π ∫ − ∞ + ∞ ∫ x + ∞ y e − x 2 + y 2 2 d x d y = 1 2 π ∫ π 4 5 π 4 sin θ d θ ∫ 0 + ∞ r 2 e − r 2 2 d r E[ Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]}]=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} \int_{x}^{+\infty} y e^{-\frac{x^{2}+y^{2}}{2}} d x d y=\frac{1}{2 \pi} \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \sin \theta d \theta \int_{0}^{+\infty} r^{2} e^{-\frac{r^{2}}{2}} d r E[Y1I[Y1≥Y2]]=2π1∫−∞+∞∫x+∞ye−2x2+y2dxdy=2π1∫4π45πsinθdθ∫0+∞r2e−2r2dr
among ∫ 0 + ∞ r 2 e − r 2 2 d r = 2 ∫ 0 + ∞ r 2 2 e − r 2 2 d ( r 2 2 ) = 2 Γ ( 3 2 ) = π 2 \int_{0}^{+\infty} r^{2} e^{-\frac{r^{2}}{2}} d r=\sqrt{2} \int_{0}^{+\infty} \sqrt{\frac{r^{2}}{2}} e^{-\frac{r^{2}}{2}} d\left(\frac{r^{2}}{2}\right)=\sqrt{2} \Gamma\left(\frac{3}{2}\right)=\sqrt{\frac{\pi}{2}} ∫0+∞r2e−2r2dr=2∫0+∞2r2e−2r2d(2r2)=2Γ(23)=2π, so E max { Y 1 , Y 2 } = 1 π , E max { X 1 , X 2 } = μ + σ π E \max \left\{Y_{1}, Y_{2}\right\}=\frac{1}{\sqrt{\pi}}, E \max \left\{X_{1}, X_{2}\right\}=\mu+\frac{\sigma}{\sqrt{\pi}} Emax{ Y1,Y2}=π1,Emax{ X1,X2}=μ+πσ.
Four 、(20 branch ) E X = 0 , Var ( X ) = σ 2 , E X=0, \operatorname{Var}(X)=\sigma^{2}, EX=0,Var(X)=σ2, Prove for any ε > 0 , \varepsilon>0, ε>0, Yes
(1)(10 branch ) P ( ∣ X ∣ > ε ) ≤ σ 2 ε 2 P(|X|>\varepsilon) \leq \frac{\sigma^{2}}{\varepsilon^{2}} P(∣X∣>ε)≤ε2σ2;
(2)(10 branch ) P ( X > ε ) ≤ σ 2 σ 2 + ε 2 P(X>\varepsilon) \leq \frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}} P(X>ε)≤σ2+ε2σ2.
Solution:
(1) remember I [ ∣ X ∣ > ε ] = { 1 , ∣ X ∣ > ε , 0 , ∣ X ∣ ≤ ε . I_{[|X|>\varepsilon]}=\left\{\begin{array}{ll}1, & |X|>\varepsilon, \\ 0, & |X| \leq \varepsilon .\end{array}\right. I[∣X∣>ε]={ 1,0,∣X∣>ε,∣X∣≤ε. It's a collection { ω : ∣ X ( ω ) ∣ > ε } \{\omega:|X(\omega)|>\varepsilon\} { ω:∣X(ω)∣>ε} The indicative function of , It can be seen that : I [ ∣ X ∣ > ε ] ≤ ∣ X ∣ 2 ε 2 I_{[|X|>\varepsilon]} \leq \frac{|X|^{2}}{\varepsilon^{2}} I[∣X∣>ε]≤ε2∣X∣2 so P ( ∣ X ∣ > ε ) = E I [ ∣ X ∣ > ε ] ≤ E ∣ X ∣ 2 ε 2 = σ 2 ε 2 P(|X|>\varepsilon)=E I_{[|X|>\varepsilon]} \leq \frac{E|X|^{2}}{\varepsilon^{2}}=\frac{\sigma^{2}}{\varepsilon^{2}} P(∣X∣>ε)=EI[∣X∣>ε]≤ε2E∣X∣2=ε2σ2.
(2) By Markov inequality , Yes P ( X > ε ) = P ( X + a > ε + a ) ≤ E ( X + a ) 2 ( ε + a ) 2 = σ 2 + a 2 ( ε + a ) 2 , P(X>\varepsilon)=P(X+a>\varepsilon+a) \leq \frac{E(X+a)^{2}}{(\varepsilon+a)^{2}}=\frac{\sigma^{2}+a^{2}}{(\varepsilon+a)^{2}}, P(X>ε)=P(X+a>ε+a)≤(ε+a)2E(X+a)2=(ε+a)2σ2+a2, take a = σ 2 ε a=\frac{\sigma^{2}}{\varepsilon} a=εσ2, Then there happens to be σ 2 + a 2 ( ε + a ) 2 = σ 2 + σ 4 ε 2 ( ε + σ 2 ε ) 2 = σ 2 σ 2 + ε 2 , \frac{\sigma^{2}+a^{2}}{(\varepsilon+a)^{2}}=\frac{\sigma^{2}+\frac{\sigma^{4}}{\varepsilon^{2}}}{\left(\varepsilon+\frac{\sigma^{2}}{\varepsilon}\right)^{2}}=\frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}}, (ε+a)2σ2+a2=(ε+εσ2)2σ2+ε2σ4=σ2+ε2σ2, therefore P ( X > ε ) ≤ σ 2 σ 2 + ε 2 P(X>\varepsilon) \leq \frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}} P(X>ε)≤σ2+ε2σ2.
5、 ... and 、(10 branch ) X 1 , X 2 X_{1}, X_{2} X1,X2, i.i.d. ∼ N ( 0 , 1 ) , \sim N(0,1), ∼N(0,1), seek X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 The distribution of .
Solution:
Due to denominator X 2 X_{2} X2 The distribution of is about 0 symmetry , therefore X 1 ∣ X 2 ∣ \frac{X_{1}}{\left|X_{2}\right|} ∣X2∣X1 And X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 Homodistribution , And obviously N ( 0 , 1 ) χ 2 ( 1 ) 1 \frac{N(0,1)}{\sqrt{\frac{\chi^{2}(1)}{1}}} 1χ2(1)N(0,1) It's a The degree of freedom is 1 Of t t t Distribution , therefore X 1 ∣ X 2 ∣ \frac{X_{1}}{\left|X_{2}\right|} ∣X2∣X1 Also, the degree of freedom is 1 Of t t t Distribution , Its probability density is f ( x ) = Γ ( 1 ) π Γ ( 1 2 ) ( x 2 + 1 ) − 1 = 1 π ⋅ 1 1 + x 2 , − ∞ < x < + ∞ , f(x)=\frac{\Gamma(1)}{\sqrt{\pi} \Gamma\left(\frac{1}{2}\right)}\left(x^{2}+1\right)^{-1}=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}},-\infty<x<+\infty, f(x)=πΓ(21)Γ(1)(x2+1)−1=π1⋅1+x21,−∞<x<+∞, The standard Cauchy distribution .
6、 ... and 、(20 branch ) X 1 , X 2 , … , X n , i . i . d ∼ N ( μ , σ 2 ) , μ X_{1}, X_{2}, \ldots, X_{n}, i . i . d \sim N\left(\mu, \sigma^{2}\right), \mu X1,X2,…,Xn,i.i.d∼N(μ,σ2),μ It is known that , prove :
(1)(10 branch ) 1 n ∑ i = 1 n ( X i − μ ) 2 \frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} n1∑i=1n(Xi−μ)2 yes σ 2 \sigma^{2} σ2 Effective estimation of ;
(2)(10 branch ) 1 n π 2 ∑ i = 1 n ∣ X i − μ ∣ \frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right| n12π∑i=1n∣Xi−μ∣ yes σ \sigma σ Unbiased estimation of , But it's not effective .
Solution:
(1) To calculate σ 2 \sigma^{2} σ2 Of Fisher The amount of information , According to the definition I ( σ 2 ) = E [ ∂ ln f ( X ; σ 2 ) ∂ σ 2 ] 2 = 1 4 σ 4 E [ ( X − μ σ ) 2 − 1 ] 2 , I\left(\sigma^{2}\right)=E\left[\frac{\partial \ln f\left(X ; \sigma^{2}\right)}{\partial \sigma^{2}}\right]^{2}=\frac{1}{4 \sigma^{4}} E\left[\left(\frac{X-\mu}{\sigma}\right)^{2}-1\right]^{2}, I(σ2)=E[∂σ2∂lnf(X;σ2)]2=4σ41E[(σX−μ)2−1]2, just E [ ( X − μ σ ) 2 − 1 ] 2 E\left[\left(\frac{X-\mu}{\sigma}\right)^{2}-1\right]^{2} E[(σX−μ)2−1]2 yes χ 2 \chi^{2} χ2 (1) The variance of , so I ( σ 2 ) = 1 2 σ 4 I\left(\sigma^{2}\right)=\frac{1}{2 \sigma^{4}} I(σ2)=2σ41. therefore , σ 2 \sigma^{2} σ2 Of C-R The lower bound is 1 n I ( σ 2 ) = 2 n σ 4 . \frac{1}{n I\left(\sigma^{2}\right)}=\frac{2}{n} \sigma^{4} . nI(σ2)1=n2σ4. Calculate again 1 n ∑ i = 1 n ( X i − μ ) 2 \frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} n1∑i=1n(Xi−μ)2 The expected variance of , because 1 σ 2 ∑ i = 1 n ( X i − μ ) 2 ∼ χ 2 ( n ) \frac{1}{\sigma^{2}} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} \sim \chi^{2}(n) σ21∑i=1n(Xi−μ)2∼χ2(n), Expectation is n n n, The variance is 2 n 2 n 2n, therefore E [ 1 n ∑ i = 1 n ( X i − μ ) 2 ] = σ 2 , Var [ 1 n ∑ i = 1 n ( X i − μ ) 2 ] = 2 n σ 4 , E\left[\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]=\sigma^{2}, \quad \operatorname{Var}\left[\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]=\frac{2}{n} \sigma^{4}, E[n1i=1∑n(Xi−μ)2]=σ2,Var[n1i=1∑n(Xi−μ)2]=n2σ4, It is an unbiased estimate , The variance just reaches C-R Lower bound , Therefore, it is an effective estimate .
(2) First , Make g ( x ) = x g(x)=\sqrt{x} g(x)=x, be σ \sigma σ Of C − R \mathrm{C}-\mathrm{R} C−R The lower bound is [ g ′ ( σ 2 ) ] 2 n I ( σ 2 ) = 1 2 n σ 2 \frac{\left[g^{\prime}\left(\sigma^{2}\right)\right]^{2}}{n I\left(\sigma^{2}\right)}=\frac{1}{2 n} \sigma^{2} nI(σ2)[g′(σ2)]2=2n1σ2, Let's calculate 1 n π 2 ∑ i = 1 n ∣ X i − μ ∣ \frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right| n12π∑i=1n∣Xi−μ∣ The expected variance of : 1 σ E ∣ X 1 − μ ∣ = ∫ − ∞ + ∞ ∣ x ∣ 1 2 π e − x 2 2 d x = 2 π ∫ 0 + ∞ x e − x 2 2 d x = 2 π ⇒ E ∣ X 1 − μ ∣ = 2 π σ , \frac{1}{\sigma} E\left|X_{1}-\mu\right|=\int_{-\infty}^{+\infty}|x| \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} d x=\sqrt{\frac{2}{\pi}} \int_{0}^{+\infty} x e^{-\frac{x^{2}}{2}} d x=\sqrt{\frac{2}{\pi}} \Rightarrow E\left|X_{1}-\mu\right|=\sqrt{\frac{2}{\pi}} \sigma, σ1E∣X1−μ∣=∫−∞+∞∣x∣2π1e−2x2dx=π2∫0+∞xe−2x2dx=π2⇒E∣X1−μ∣=π2σ, E ∣ X 1 − μ ∣ 2 = σ 2 , so Var ( ∣ X 1 − μ ∣ ) = σ 2 − 2 π σ 2 = ( 1 − 2 π ) σ 2 , E\left|X_{1}-\mu\right|^{2}=\sigma^{2} \text {, so } \operatorname{Var}\left(\left|X_{1}-\mu\right|\right)=\sigma^{2}-\frac{2}{\pi} \sigma^{2}=\left(1-\frac{2}{\pi}\right) \sigma^{2} \text {, } E∣X1−μ∣2=σ2, so Var(∣X1−μ∣)=σ2−π2σ2=(1−π2)σ2, so E [ 1 n π 2 ∑ i = 1 n ∣ X i − μ ∣ ] = σ , Var [ 1 n π 2 ∑ i = 1 n ∣ X i − μ ∣ ] = ( π 2 − 1 ) n σ 2 E\left[\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right|\right]=\sigma, \operatorname{Var}\left[\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right|\right]=\frac{\left(\frac{\pi}{2}-1\right)}{n} \sigma^{2} E[n12π∑i=1n∣Xi−μ∣]=σ,Var[n12π∑i=1n∣Xi−μ∣]=n(2π−1)σ2, It is an unbiased estimate , but Don't reach C-R Lower bound , Not a valid estimate .
7、 ... and 、(20 branch ) The overall distribution function F ( x ) F(x) F(x) Continuous single increase , X ( 1 ) , X ( 2 ) , … , X ( n ) X_{(1)}, X_{(2)}, \ldots, X_{(n)} X(1),X(2),…,X(n) Is the order statistic of random samples from the population , Y i = F ( X ( i ) ) , Y_{i}=F\left(X_{(i)}\right), Yi=F(X(i)), seek
(1)(10 branch ) E Y i , Var ( Y i ) E Y_{i}, \operatorname{Var}\left(Y_{i}\right) EYi,Var(Yi);
(2)(10 branch ) ( Y 1 , Y 2 , … , Y n ) T \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)^{T} (Y1,Y2,…,Yn)T The covariance matrix of .
Solution:
(1) because U 1 = F ( X 1 ) , U 2 = F ( X 2 ) , ⋯ , U n = F ( X n ) U_{1}=F\left(X_{1}\right), U_{2}=F\left(X_{2}\right), \cdots, U_{n}=F\left(X_{n}\right) U1=F(X1),U2=F(X2),⋯,Un=F(Xn) Independence and obedience [ 0 , 1 ] [0,1] [0,1] Evenly distributed , so Y i = U ( i ) Y_{i}=U_{(i)} Yi=U(i) It happens to be the order statistics of uniform distribution , According to the definition of order statistics , Yes :
f i ( y ) = n ! ( i − 1 ) ! ( n − i ) ! f U ( y ) P i − 1 { U ≤ y } P n − i { U ≥ y } , Insert one by one , Yes f i ( y ) = n ! ( i − 1 ) ! ( n − i ) ! y i − 1 ( 1 − y ) n − i , 0 ≤ y ≤ 1 , Exactly Beta ( i , n − i + 1 ) . \begin{gathered} f_{i}(y)=\frac{n !}{(i-1) !(n-i) !} f_{U}(y) P^{i-1}\{U \leq y\} P^{n-i}\{U \geq y\}, \text { Insert one by one , Yes } \\ f_{i}(y)=\frac{n !}{(i-1) !(n-i) !} y^{i-1}(1-y)^{n-i}, 0 \leq y \leq 1 \text {, Exactly } \operatorname{Beta}(i, n-i+1) . \end{gathered} fi(y)=(i−1)!(n−i)!n!fU(y)Pi−1{ U≤y}Pn−i{ U≥y}, Insert one by one , Yes fi(y)=(i−1)!(n−i)!n!yi−1(1−y)n−i,0≤y≤1, Exactly Beta(i,n−i+1). According to the nature of beta distribution E Y i = i n + 1 , Var ( Y i ) = i ( n − i + 1 ) ( n + 1 ) 2 ( n + 2 ) E Y_{i}=\frac{i}{n+1}, \operatorname{Var}\left(Y_{i}\right)=\frac{i(n-i+1)}{(n+1)^{2}(n+2)} EYi=n+1i,Var(Yi)=(n+1)2(n+2)i(n−i+1).
(2) We consider the ( Y i , Y j ) , i < j \left(Y_{i}, Y_{j}\right), i<j (Yi,Yj),i<j The joint distribution of , According to the definition of order statistics , Yes
f i , j ( x , y ) = n ! ( i − 1 ) ! ( j − i − 1 ) ! ( n − j ) ! f U ( x ) f U ( y ) P i − 1 { U ≤ x } P j − i − 1 { x ≤ U ≤ y } P n − j { U ≥ y } f_{i, j}(x, y)=\frac{n !}{(i-1) !(j-i-1) !(n-j) !} f_{U}(x) f_{U}(y) P^{i-1}\{U \leq x\} P^{j-i-1}\{x \leq U \leq y\} P^{n-j}\{U \geq y\} fi,j(x,y)=(i−1)!(j−i−1)!(n−j)!n!fU(x)fU(y)Pi−1{ U≤x}Pj−i−1{ x≤U≤y}Pn−j{ U≥y} namely f i , j ( x , y ) = n ! ( i − 1 ) ! ( j − i − 1 ) ! ( n − j ) ! x i − 1 ( y − x ) j − i − 1 ( 1 − y ) n − j , 0 ≤ x ≤ y ≤ 1. f_{i, j}(x, y)=\frac{n !}{(i-1) !(j-i-1) !(n-j) !} x^{i-1}(y-x)^{j-i-1}(1-y)^{n-j}, 0 \leq x \leq y \leq 1. fi,j(x,y)=(i−1)!(j−i−1)!(n−j)!n!xi−1(y−x)j−i−1(1−y)n−j,0≤x≤y≤1.
The covariance problem is reduced to the calculation of integral ∫ 0 1 ∫ 0 y x y ⋅ x i − 1 ( y − x ) j − i − 1 ( 1 − y ) n − j d x d y \int_{0}^{1} \int_{0}^{y} x y \cdot x^{i-1}(y-x)^{j-i-1}(1-y)^{n-j} d x d y ∫01∫0yxy⋅xi−1(y−x)j−i−1(1−y)n−jdxdy, And if I We put y y y Regard as y = x + ( y − x ) y=x+(y-x) y=x+(y−x), Then this integral can be reduced to two parts : ∫ 0 1 ∫ 0 y x i + 1 ( y − x ) j − i − 1 ( 1 − y ) n − j d x d y = ( i + 1 ) ! ( j − i − 1 ) ! ( n − j ) ! ( n + 2 ) ! , ∫ 0 1 ∫ 0 y x i ( y − x ) j − i ( 1 − y ) n − j d x d y = i ! ( j − i ) ! ( n − j ) ! ( n + 2 ) ! , \begin{gathered} \int_{0}^{1} \int_{0}^{y} x^{i+1}(y-x)^{j-i-1}(1-y)^{n-j} d x d y=\frac{(i+1) !(j-i-1) !(n-j) !}{(n+2) !}, \\ \int_{0}^{1} \int_{0}^{y} x^{i}(y-x)^{j-i}(1-y)^{n-j} d x d y=\frac{i !(j-i) !(n-j) !}{(n+2) !}, \end{gathered} ∫01∫0yxi+1(y−x)j−i−1(1−y)n−jdxdy=(n+2)!(i+1)!(j−i−1)!(n−j)!,∫01∫0yxi(y−x)j−i(1−y)n−jdxdy=(n+2)!i!(j−i)!(n−j)!,( First of all, look at f i , j ( x , y ) f_{i, j}(x, y) fi,j(x,y) The expression of , Will find
( n + 2 ) ! ( i + 1 ) ! ( j − i − 1 ) ! ( n − j ) ! x i + 1 ( y − x ) j − i − 1 ( 1 − y ) n − j \frac{(n+2) !}{(i+1) !(j-i-1) !(n-j) !} x^{i+1}(y-x)^{j-i-1}(1-y)^{n-j} (i+1)!(j−i−1)!(n−j)!(n+2)!xi+1(y−x)j−i−1(1−y)n−j It is also a density function , Therefore, the above two integral values can be easily obtained through the regularity of probability ) There are E [ Y i Y j ] = i ( j + 1 ) ( n + 1 ) ( n + 2 ) E [Y_{i} Y_{j}]=\frac{i(j+1)}{(n+1)(n+2)} E[YiYj]=(n+1)(n+2)i(j+1), so Cov ( Y i , Y j ) = i ( n + 1 − j ) ( n + 1 ) 2 ( n + 2 ) \operatorname{Cov}\left(Y_{i}, Y_{j}\right)=\frac{i(n+1-j)}{(n+1)^{2}(n+2)} Cov(Yi,Yj)=(n+1)2(n+2)i(n+1−j). So the covariance matrix is Σ = ( a i j ) n × n \Sigma=\left(a_{i j}\right)_{n \times n} Σ=(aij)n×n, among a i j = i ( n + 1 − j ) ( n + 1 ) 2 ( n + 2 ) , i < j a_{i j}=\frac{i(n+1-j)}{(n+1)^{2}(n+2)}, i<j aij=(n+1)2(n+2)i(n+1−j),i<j.
8、 ... and 、(20 branch ) Some come from the general U ( θ , 2 θ ) U(\theta,2\theta) U(θ,2θ) A random sample of X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn, seek θ \theta θ Moment estimation sum of MLE, And verify the unbiasedness and consistency .
Solution:
First find the moment estimation , Expect E X 1 = 3 θ 2 EX_1=\frac{3\theta}{2} EX1=23θ, From the principle of substitution θ ^ M = 2 3 X ˉ \hat{\theta}_M=\frac{2}{3}\bar{X} θ^M=32Xˉ. The expectation is easy to see that it is unbiased , It is known from the strong law of large numbers that it is a strongly consistent estimate .
Ask again MLE, Write the likelihood function as
L ( θ ) = 1 θ n I { X ( n ) < 2 θ } I { X ( 1 ) > θ } = I { X ( n ) 2 < θ < X ( 1 ) } θ n , L\left( \theta \right) =\frac{1}{\theta ^n}I_{\left\{ X_{\left( n \right)}<2\theta \right\}}I_{\left\{ X_{\left( 1 \right)}>\theta \right\}}=\frac{I_{\left\{ \frac{X_{\left( n \right)}}{2}<\theta <X_{\left( 1 \right)} \right\}}}{\theta ^n}, L(θ)=θn1I{ X(n)<2θ}I{ X(1)>θ}=θnI{ 2X(n)<θ<X(1)}, It can be seen that 1 θ n \frac{1}{\theta^n} θn1 About θ \theta θ Monotonic decline , so θ \theta θ The minimum value is MLE, namely θ ^ L = X ( n ) 2 \hat{\theta}_L=\frac{X_{(n)}}{2} θ^L=2X(n). Use transformation Y i = X i − θ θ ∼ U ( 0 , 1 ) Y_i=\frac{X_i-\theta}{\theta}\sim U(0,1) Yi=θXi−θ∼U(0,1), know Y ( n ) = X ( n ) − θ θ ∼ B e t a ( n , 1 ) Y_{(n)}=\frac{X_{(n)}-\theta}{\theta}\sim Beta(n,1) Y(n)=θX(n)−θ∼Beta(n,1), There are
E ( Y ( n ) ) = n n + 1 , E ( θ ^ L ) = 1 2 E ( X ( n ) ) = 1 2 [ θ E ( Y ( n ) ) + θ ] = 2 n + 1 2 n + 2 θ . E\left( Y_{\left( n \right)} \right) =\frac{n}{n+1},\quad E\left( \hat{\theta}_L \right) =\frac{1}{2}E\left( X_{\left( n \right)} \right) =\frac{1}{2}\left[ \theta E\left( Y_{\left( n \right)} \right) +\theta \right] =\frac{2n+1}{2n+2}\theta . E(Y(n))=n+1n,E(θ^L)=21E(X(n))=21[θE(Y(n))+θ]=2n+22n+1θ. therefore θ ^ L \hat{\theta}_L θ^L Not without bias , But asymptotically unbiased , Then look at consistency , Yes
P ( ∣ θ ^ L − θ ∣ > ε 1 ) = P ( ∣ X ( n ) − 2 θ ∣ > ε 2 ) = P ( ∣ Y ( n ) − 1 ∣ > ε 3 ) = P ( Y ( n ) < 1 − ε 3 ) = ( 1 − ε 3 ) n , \begin{aligned} P\left( \left| \hat{\theta}_L-\theta \right|>\varepsilon _1 \right) &=P\left( \left| X_{\left( n \right)}-2\theta \right|>\varepsilon _2 \right)\\ &=P\left( \left| Y_{\left( n \right)}-1 \right|>\varepsilon _3 \right)\\ &=P\left( Y_{\left( n \right)}<1-\varepsilon _3 \right)\\ &=\left( 1-\varepsilon _3 \right) ^n,\\ \end{aligned} P(∣∣∣θ^L−θ∣∣∣>ε1)=P(∣∣X(n)−2θ∣∣>ε2)=P(∣∣Y(n)−1∣∣>ε3)=P(Y(n)<1−ε3)=(1−ε3)n, Convergence of series , So it is a strongly consistent estimate .
Nine 、(20 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn Is from N ( μ , 1 ) N(\mu,1) N(μ,1) A random sample of , Consider the hypothesis test problem H 0 : μ = 1 v s H 1 : μ = 2 H_0:\mu = 1 \quad \mathrm{vs} \quad H_1:\mu=2 H0:μ=1vsH1:μ=2 Given denial domain W = { X ˉ > 1.6 } W=\{\bar{X}>1.6\} W={ Xˉ>1.6}, Answer the following questions :
(1)(10 branch ) n = 10 n=10 n=10, Find the probability of making two kinds of mistakes α \alpha α, β \beta β;
(2)(10 branch ) Request the second kind of error β ≤ 0.01 \beta\le0.01 β≤0.01, Find the value range of the sample size .
Solution:
(1) First count the first kind of errors , When the original assumption comes true X ˉ ∼ N ( 1 , 1 n ) \bar{X}\sim N(1,\frac{1}{n}) Xˉ∼N(1,n1), so
α = P μ = 1 ( X ˉ > 1.6 ) = P μ = 1 ( n ( X ˉ − 1 ) > 0.6 n ) = 1 − Φ ( 0.6 n ) = 1 − Φ ( 0.6 10 ) . \begin{aligned} \alpha &=P_{\mu =1}\left( \bar{X}>1.6 \right)\\ &=P_{\mu =1}\left( \sqrt{n}\left( \bar{X}-1 \right) >0.6\sqrt{n} \right)\\ &=1-\Phi \left( 0.6\sqrt{n} \right)\\ &=1-\Phi \left( 0.6\sqrt{10} \right) .\\ \end{aligned} α=Pμ=1(Xˉ>1.6)=Pμ=1(n(Xˉ−1)>0.6n)=1−Φ(0.6n)=1−Φ(0.610). Calculate the second kind of error , When alternative assumptions come true X ˉ ∼ N ( 2 , 1 n ) \bar{X}\sim N(2,\frac{1}{n}) Xˉ∼N(2,n1), so
β = P μ = 2 ( X ˉ ≤ 1.6 ) = P μ = 1 ( n ( X ˉ − 2 ) ≤ − 0.4 n ) = Φ ( − 0.4 n ) = Φ ( − 0.4 10 ) . \begin{aligned} \beta &=P_{\mu =2}\left( \bar{X}\le 1.6 \right)\\ &=P_{\mu =1}\left( \sqrt{n}\left( \bar{X}-2 \right) \le -0.4\sqrt{n} \right)\\ &=\Phi \left( -0.4\sqrt{n} \right)\\ &=\Phi \left( -0.4\sqrt{10} \right) .\\ \end{aligned} β=Pμ=2(Xˉ≤1.6)=Pμ=1(n(Xˉ−2)≤−0.4n)=Φ(−0.4n)=Φ(−0.410). (2) According to the first (1) Ask calculation , Make Φ ( − 0.4 n ) ≤ 0.01 \Phi \left( -0.4\sqrt{n} \right) \le 0.01 Φ(−0.4n)≤0.01, have to
− 0.4 n ≤ − 2.33 * n ≥ 2.3 3 2 0. 4 2 * n ≥ 34. -0.4\sqrt{n}\le -2.33\Longrightarrow n\ge \frac{2.33^2}{0.4^2}\Longrightarrow n\ge 34. −0.4n≤−2.33*n≥0.422.332*n≥34.
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