当前位置:网站首页>[the Nine Yang Manual] 2020 Fudan University Applied Statistics real problem + analysis
[the Nine Yang Manual] 2020 Fudan University Applied Statistics real problem + analysis
2022-07-06 13:31:00 【Elder martial brother statistics】
Catalog
The real part
One 、(20 branch ) A family has two children , Find the probability of the following events :
(1)(10 branch ) The first one is known to be a girl , Find the probability that the second is a girl ;
(2)(10 branch ) One is known to be a girl , Find the probability that the other is a girl .
Two 、(15 branch ) Jiayou 21 A coin , B has 20 A coin , Both of them toss all the coins at the same time , Find the probability that the number of coins with a facing up is more than that of B .
3、 ... and 、(15 branch ) There are countless parallel lines on the plane , Every two parallel lines are spaced 2 rice , Use side length 1 An equilateral triangle of meters is thrown at the plane , Find the probability of triangle pressing to a straight line .
Four 、(15 branch ) 8 A boy 、7 Two girls sit in a row , set up X i = 1 X_{i}=1 Xi=1 It means the first one i i i The first position is the same as i + 1 i+1 i+1 The opposite sex sits in this position , X i = 0 X_{i}=0 Xi=0 It means the first one i i i The first position is the same as i + 1 i+1 i+1 Sitting in the same seat , ξ = ∑ i = 1 14 X i , \xi=\sum_{i=1}^{14} X_{i}, ξ=∑i=114Xi, seek E ξ . E \xi . Eξ.
5、 ... and 、(15 branch ) Cite an expectation that tends to be positive and infinite , But it converges to 0 A sequence of random variables { X n } \left\{X_{n}\right\} { Xn}.
6、 ... and 、(20 branch ) Some come from the general X ∼ f ( x ) = θ x θ − 1 I { 0 < x < 1 } X \sim f(x)=\theta x^{\theta-1} I\{0<x<1\} X∼f(x)=θxθ−1I{ 0<x<1} Of n n n Random sample , seek
(1)(5 branch ) θ \theta θ Of M L E , \mathrm{MLE}, MLE, And verify the unbiased ;
(2)(5 branch ) verification MLE The consistency of ;
(3)(5 branch ) θ \theta θ The moment estimate of ;
(4)(5 branch ) Use the sample median pair θ \theta θ Estimate .
7、 ... and 、(20 branch ) X 1 , … , X n , X_{1}, \ldots, X_{n}, X1,…,Xn, i.i.d ∼ N ( μ , σ 2 ) , \sim N\left(\mu, \sigma^{2}\right), ∼N(μ,σ2), prove [ X ( 1 ) , X ( n ) ] \left[X_{(1)}, X_{(n)}\right] [X(1),X(n)] yes μ \mu μ The confidence level of is 1 − 2 1 − n 1-2^{1-n} 1−21−n The confidence interval of .
8、 ... and 、(20 branch ) Some come from the general X ∼ f ( x ) = 1 2 e − ∣ x − θ ∣ X \sim f(x)=\frac{1}{2} e^{-|x-\theta|} X∼f(x)=21e−∣x−θ∣ Of 7 Random sample , seek θ \theta θ Of MLE.
Nine 、(10 branch ) ( X 1 , X 2 ) ∼ N ( 0 , 0 ; 1 , 1 ; 0 ) , (X_1, X_2) \sim N(0,0 ; 1,1 ; 0), (X1,X2)∼N(0,0;1,1;0), seek X 1 X 2 \frac{X_1}{X_2} X2X1 Probability distribution of .
The analysis part
One 、(20 branch ) A family has two children , Find the probability of the following events :
(1)(10 branch ) The first one is known to be a girl , Find the probability that the second is a girl ;
(2)(10 branch ) One is known to be a girl , Find the probability that the other is a girl .
Solution:
(1) First of all, suppose : The probability of a child being a girl without any information is 0.5 0.5 0.5.
In the event A i A_{i} Ai It means the first one i i i This is a girl ( i = 1 , 2 ) (i=1,2) (i=1,2), be P ( A 2 ∣ A 1 ) = P ( A 1 A 2 ) P ( A 1 ) = 0.25 0.5 = 0.5 P\left(A_{2} \mid A_{1}\right)=\frac{P\left(A_{1} A_{2}\right)}{P\left(A_{1}\right)}=\frac{0.25}{0.5}=0.5 P(A2∣A1)=P(A1)P(A1A2)=0.50.25=0.5.
(2) P ( A 1 A 2 ∣ A 1 ∪ A 2 ) = P ( A 1 A 2 ) P ( A 1 ∪ A 2 ) = 0.25 0.75 = 1 3 P\left(A_{1} A_{2} \mid A_{1} \cup A_{2}\right)=\frac{P\left(A_{1} A_{2}\right)}{P\left(A_{1} \cup A_{2}\right)}=\frac{0.25}{0.75}=\frac{1}{3} P(A1A2∣A1∪A2)=P(A1∪A2)P(A1A2)=0.750.25=31.
Two 、(15 branch ) Jiayou 21 A coin , B has 20 A coin , Both of them toss all the coins at the same time , Find the probability that the number of coins with a facing up is more than that of B .
Solution:
According to the symmetry, we can know , P { P\{ P{ There are more coins with a facing up than with B } = P { \}=P\{ }=P{ There are more coins with a side down than with B } \} }. use A random variable X X X Indicates the number of coins with a facing up , A random variable Y Y Y Indicates the number of coins with B facing up . be :
P { X > Y } = P { 21 − X > 20 − Y } = P { 1 − X > − Y } = P { X < Y + 1 } = P { X ⩽ Y } \begin{aligned} P\{X>Y\} &=P\{21-X>20-Y\}=P\{1-X>-Y\} \\ &=P\{X<Y+1\}=P\{X \leqslant Y\} \end{aligned} P{ X>Y}=P{ 21−X>20−Y}=P{ 1−X>−Y}=P{ X<Y+1}=P{ X⩽Y} also P { X > Y } + P { X ⩽ Y } = 1 P\{X>Y\}+P\{X \leqslant Y\}=1 P{ X>Y}+P{ X⩽Y}=1, therefore P { X > Y } = P { X ⩽ Y } = 0.5 P\{X>Y\}=P\{X \leqslant Y\}=0.5 P{ X>Y}=P{ X⩽Y}=0.5.
3、 ... and 、(15 branch ) There are countless parallel lines on the plane , Every two parallel lines are spaced 2 rice , Use side length 1 An equilateral triangle of meters is thrown at the plane , Find the probability of triangle pressing to a straight line .
Solution:
remember △ A B C \triangle A B C △ABC The three sides of the are a , b , c a, b, c a,b,c. Then there are the following situations when a triangle intersects a parallel line :
(1) One vertex of the triangle is on the parallel line ;
(2) One side of the triangle coincides with the straight line ;
(3) Two lines of triangle Edges intersect parallel lines .
According to the geometric probability P ( 1 ) = P ( 2 ) = 0 P(1)=P(2)=0 P(1)=P(2)=0, So just consider the situation (3). and P ( 3 ) = P a b + P a c + P b c P(3)=P_{a b}+P_{a c}+P_{b c} P(3)=Pab+Pac+Pbc, among P a b P_{a b} Pab edge a 、 b a 、 b a、b Intersect with parallel lines . So , remember P a P_{a} Pa edge a a a Intersect with parallel lines , be P a = P a c + P a b P_{a}=P_{a c}+P_{a b} Pa=Pac+Pab. so P ( 3 ) = 1 2 ( P a + P b + P c ) , P(3)=\frac{1}{2}\left(P_{a}+P_{b}+P_{c}\right), P(3)=21(Pa+Pb+Pc), Now we only need to find P a 、 P b 、 P c P_{a} 、 P_{b} 、 P_{c} Pa、Pb、Pc. This is a Buffon Injection model , The probability is P a = 2 a d π P_{a}=\frac{2 a}{d \pi} Pa=dπ2a, among a a a Is the edge a a a The length of , d d d It's parallel Spacing between lines , Substituting data can be calculated P a = 2 2 π = 1 π P_{a}=\frac{2}{2 \pi}=\frac{1}{\pi} Pa=2π2=π1. Empathy P b = P c = 1 π P_{b}=P_{c}=\frac{1}{\pi} Pb=Pc=π1. so
P { Triangle pressed to a straight line } = P ( 3 ) = 1 2 ( P a + P b + P c ) = 3 2 π . P\{\text { Triangle pressed to a straight line }\}=P(3)=\frac{1}{2}\left(P_{a}+P_{b}+P_{c}\right)=\frac{3}{2 \pi}. P{ Triangle pressed to a straight line }=P(3)=21(Pa+Pb+Pc)=2π3.
Four 、(15 branch ) 8 A boy 、7 Two girls sit in a row , set up X i = 1 X_{i}=1 Xi=1 It means the first one i i i The first position is the same as i + 1 i+1 i+1 The opposite sex sits in this position , X i = 0 X_{i}=0 Xi=0 It means the first one i i i The first position is the same as i + 1 i+1 i+1 Sitting in the same seat , ξ = ∑ i = 1 14 X i , \xi=\sum_{i=1}^{14} X_{i}, ξ=∑i=114Xi, seek E ξ . E \xi . Eξ.
Solution:
E ξ = E ( ∑ i = 1 14 X i ) = ∑ i = 1 14 E X i E \xi=E\left(\sum_{i=1}^{14} X_{i}\right)=\sum_{i=1}^{14} E X_{i} Eξ=E(∑i=114Xi)=∑i=114EXi, Considering all X i X_{i} Xi It's identically distributed , Now E X 1 E X_{1} EX1.
E X 1 = P ( X 1 = 1 ) = C 8 1 C 7 1 C 15 2 = 8 × 7 15 × 14 2 × 1 = 8 15 E X_{1}=P\left(X_{1}=1\right)=\frac{C_{8}^{1} C_{7}^{1}}{C_{15}^{2}}=\frac{8 \times 7}{\frac{15 \times 14}{2 \times 1}}=\frac{8}{15} EX1=P(X1=1)=C152C81C71=2×115×148×7=158 therefore E ξ = ∑ i = 1 14 E X i = 14 E X 1 = 112 15 E \xi=\sum_{i=1}^{14} E X_{i}=14 E X_{1}=\frac{112}{15} Eξ=∑i=114EXi=14EX1=15112.
5、 ... and 、(15 branch ) Cite an expectation that tends to be positive and infinite , But it converges to 0 A sequence of random variables { X n } \left\{X_{n}\right\} { Xn}.
Solution:
Give such a sequence of random variables : P ( X n = 0 ) = 1 − 1 n , P ( X n = n 2 ) = 1 n P\left(X_{n}=0\right)=1-\frac{1}{n}, P\left(X_{n}=n^{2}\right)=\frac{1}{n} P(Xn=0)=1−n1,P(Xn=n2)=n1. E X n = n → + ∞ E X_{n}=n \rightarrow+\infty EXn=n→+∞. and P ( X n ≠ 0 ) = 1 n P\left(X_{n} \neq 0\right)=\frac{1}{n} P(Xn=0)=n1, be X n → P 0 X_{n} \stackrel{P}{\rightarrow} 0 Xn→P0.
6、 ... and 、(20 branch ) Some come from the general X ∼ f ( x ) = θ x θ − 1 I { 0 < x < 1 } X \sim f(x)=\theta x^{\theta-1} I\{0<x<1\} X∼f(x)=θxθ−1I{ 0<x<1} Of n n n Random sample , seek
(1)(5 branch ) θ \theta θ Of M L E , \mathrm{MLE}, MLE, And verify the unbiased ;
(2)(5 branch ) verification MLE The consistency of ;
(3)(5 branch ) θ \theta θ The moment estimate of ;
(4)(5 branch ) Use the sample median pair θ \theta θ Estimate .
Solution:
(1) Likelihood function L ( X ; θ ) = θ n ( ∏ i = 1 n x i ) θ − 1 L(\mathbf{X} ; \theta)=\theta^{n}\left(\prod_{i=1}^{n} x_{i}\right)^{\theta-1} L(X;θ)=θn(∏i=1nxi)θ−1, Log likelihood function ln L = n ln θ + ( θ − 1 ) ∑ i = 1 n ln x i \ln L=n \ln \theta+(\theta-1) \sum_{i=1}^{n} \ln x_{i} lnL=nlnθ+(θ−1)∑i=1nlnxi. Make ∂ ln L ∂ θ = n θ + ∑ i = 1 n ln x i = 0 \frac{\partial \ln L}{\partial \theta}=\frac{n}{\theta}+\sum_{i=1}^{n} \ln x_{i}=0 ∂θ∂lnL=θn+∑i=1nlnxi=0, Solution θ ^ L = n ∑ i = 1 n ( − ln x i ) \hat{\theta}_{L}=\frac{n}{\sum_{i=1}^{n}\left(-\ln x_{i}\right)} θ^L=∑i=1n(−lnxi)n. And because the whole obeys beta distribution , Belong to Exponential family distribution , The stationary point of its log likelihood function must be maximum likelihood estimation . therefore θ ^ L = n ∑ i = 1 n ( − ln x i ) \hat{\theta}_{L}=\frac{n}{\sum_{i=1}^{n}\left(-\ln x_{i}\right)} θ^L=∑i=1n(−lnxi)n yes θ \theta θ Maximum likelihood estimation of . If order Y i = − ln X i ∼ Exp ( θ ) Y_{i}=-\ln X_{i} \sim \operatorname{Exp}(\theta) Yi=−lnXi∼Exp(θ), And from the additivity of gamma distribution T = ∑ i = 1 n y i ∼ G a ( n , θ ) T=\sum_{i=1}^{n} y_{i} \sim G a(n, \theta) T=∑i=1nyi∼Ga(n,θ), Zeji Large likelihood estimation can be written as θ ^ L = n T \hat{\theta}_{L}=\frac{n}{T} θ^L=Tn.
E θ ^ L = E n T = ∫ 0 + ∞ n t θ n Γ ( n ) t n − 1 e − θ t d t = n θ Γ ( n ) ∫ 0 + ∞ ( θ t ) n − 2 e − θ t d ( θ t ) = n θ Γ ( n ) Γ ( n − 1 ) = n n − 1 θ E \hat{\theta}_{L}=E \frac{n}{T}=\int_{0}^{+\infty} \frac{n}{t} \frac{\theta^{n}}{\Gamma(n)} t^{n-1} e^{-\theta t} d t=\frac{n \theta}{\Gamma(n)} \int_{0}^{+\infty}(\theta t)^{n-2} e^{-\theta t} d(\theta t)=\frac{n \theta}{\Gamma(n)} \Gamma(n-1)=\frac{n}{n-1} \theta Eθ^L=ETn=∫0+∞tnΓ(n)θntn−1e−θtdt=Γ(n)nθ∫0+∞(θt)n−2e−θtd(θt)=Γ(n)nθΓ(n−1)=n−1nθ
therefore θ ^ L \hat{\theta}_{L} θ^L No θ \theta θ Unbiased estimation of , But it is gradual and unbiased .
(2) In the last question, we have calculated E θ ^ L = n n − 1 θ → θ E \hat{\theta}_{L}=\frac{n}{n-1} \theta \rightarrow \theta Eθ^L=n−1nθ→θ, Now consider its consistency .
E θ ^ L 2 = E n 2 T 2 = n 2 θ 2 Γ ( n ) ∫ 0 + ∞ ( θ t ) n − 3 e − θ t d ( θ t ) = n 2 θ 2 Γ ( n ) Γ ( n − 2 ) = n 2 ( n − 1 ) ( n − 2 ) θ 2 , E \hat{\theta}_{L}^{2}=E \frac{n^{2}}{T^{2}}=\frac{n^{2} \theta^{2}}{\Gamma(n)} \int_{0}^{+\infty}(\theta t)^{n-3} e^{-\theta t} d(\theta t)=\frac{n^{2} \theta^{2}}{\Gamma(n)} \Gamma(n-2)=\frac{n^{2}}{(n-1)(n-2)} \theta^{2}, Eθ^L2=ET2n2=Γ(n)n2θ2∫0+∞(θt)n−3e−θtd(θt)=Γ(n)n2θ2Γ(n−2)=(n−1)(n−2)n2θ2,
be Var ( θ ^ L ) = E θ ^ L 2 − ( E θ ^ L ) 2 = n 2 θ 2 ( n − 1 ) ( n − 2 ) − n 2 θ 2 ( n − 1 ) 2 = n 2 ( n − 1 ) 2 ( n − 2 ) θ 2 \operatorname{Var}\left(\hat{\theta}_{L}\right)=E \hat{\theta}_{L}^{2}-\left(E \hat{\theta}_{L}\right)^{2}=\frac{n^{2} \theta^{2}}{(n-1)(n-2)}-\frac{n^{2} \theta^{2}}{(n-1)^{2}}=\frac{n^{2}}{(n-1)^{2}(n-2)} \theta^{2} Var(θ^L)=Eθ^L2−(Eθ^L)2=(n−1)(n−2)n2θ2−(n−1)2n2θ2=(n−1)2(n−2)n2θ2.
P ( ∣ θ ^ L − θ ∣ ≥ ε ) = P ( ∣ θ ^ L − n n − 1 θ + n n − 1 θ − θ ∣ ≥ ε ) P\left(\left|\hat{\theta}_{L}-\theta\right| \geq \varepsilon\right)=P\left(\left|\hat{\theta}_{L}-\frac{n}{n-1} \theta+\frac{n}{n-1} \theta-\theta\right| \geq \varepsilon\right) P(∣∣∣θ^L−θ∣∣∣≥ε)=P(∣∣∣∣θ^L−n−1nθ+n−1nθ−θ∣∣∣∣≥ε) ≤ P ( ∣ θ ^ L − n n − 1 θ ∣ ≥ ε 2 ) + P ( ∣ n n − 1 θ − θ ∣ ≥ ε 2 ) \leq P\left(\left|\hat{\theta}_{L}-\frac{n}{n-1} \theta\right| \geq \frac{\varepsilon}{2}\right)+P\left(\left|\frac{n}{n-1} \theta-\theta\right| \geq \frac{\varepsilon}{2}\right) ≤P(∣∣∣∣θ^L−n−1nθ∣∣∣∣≥2ε)+P(∣∣∣∣n−1nθ−θ∣∣∣∣≥2ε) According to Chebyshev inequality P ( ∣ θ ^ L − n n − 1 θ ∣ ≥ ε 2 ) ≤ 4 Var ( θ ^ L ) ε 2 = 4 ε 2 n 2 ( n − 1 ) 2 ( n − 2 ) θ 2 → 0 P\left(\left|\hat{\theta}_{L}-\frac{n}{n-1} \theta\right| \geq \frac{\varepsilon}{2}\right) \leq \frac{4 \operatorname{Var}\left(\hat{\theta}_{L}\right)}{\varepsilon^{2}}=\frac{4}{\varepsilon^{2}} \frac{n^{2}}{(n-1)^{2}(n-2)} \theta^{2} \rightarrow 0 P(∣∣∣θ^L−n−1nθ∣∣∣≥2ε)≤ε24Var(θ^L)=ε24(n−1)2(n−2)n2θ2→0 And for the larger n , P ( ∣ n n − 1 θ − θ ∣ ≥ ε 2 ) = 0 n, P\left(\left|\frac{n}{n-1} \theta-\theta\right| \geq \frac{\varepsilon}{2}\right)=0 n,P(∣∣n−1nθ−θ∣∣≥2ε)=0. therefore P ( ∣ θ ^ L − θ ∣ ≥ ε ) → 0 P\left(\left|\hat{\theta}_{L}-\theta\right| \geq \varepsilon\right) \rightarrow 0 P(∣∣∣θ^L−θ∣∣∣≥ε)→0, in other words θ ^ L \hat{\theta}_{L} θ^L yes θ \theta θ A consistent estimate of .
(3) Overall obedience Beta ( θ , 1 ) \operatorname{Beta}(\theta, 1) Beta(θ,1), Digital features distributed by beta , We know E X = θ θ + 1 E X=\frac{\theta}{\theta+1} EX=θ+1θ. According to the inverse solution θ \theta θ The moment estimate of θ ^ M = x ˉ 1 − x ˉ \hat{\theta}_{M}=\frac{\bar{x}}{1-\bar{x}} θ^M=1−xˉxˉ.
(4) The overall distribution function is F ( x ) = { 0 , x < 0 x θ , 0 ≤ x < 1 1 , x ≥ 1 F(x)=\left\{\begin{array}{cc}0, & x<0 \\ x^{\theta}, & 0 \leq x<1 \\ 1, & x \geq 1\end{array}\right. F(x)=⎩⎨⎧0,xθ,1,x<00≤x<1x≥1 Make F ( x ) = 1 2 F(x)=\frac{1}{2} F(x)=21, Solution x 0.5 = ( 1 2 ) 1 θ x_{0.5}=\left(\frac{1}{2}\right)^{\frac{1}{\theta}} x0.5=(21)θ1, Use the median of the sample m 0.5 m_{0.5} m0.5 Replace the overall median x 0.5 x_{0.5} x0.5, And inverse solution Out θ ^ = 1 log 1 2 m 0.5 = log 1 2 1 2 log 1 2 m 0.5 = log m 0.5 1 2 \hat{\theta}=\frac{1}{\log _{\frac{1}{2}} m_{0.5}}=\frac{\log _{\frac{1}{2}} \frac{1}{2}}{\log _{\frac{1}{2}} m_{0.5}}=\log _{m_{0.5}} \frac{1}{2} θ^=log21m0.51=log21m0.5log2121=logm0.521 It is based on the median pair of samples θ \theta θ Estimation .
7、 ... and 、(20 branch ) X 1 , … , X n , X_{1}, \ldots, X_{n}, X1,…,Xn, i.i.d ∼ N ( μ , σ 2 ) , \sim N\left(\mu, \sigma^{2}\right), ∼N(μ,σ2), prove [ X ( 1 ) , X ( n ) ] \left[X_{(1)}, X_{(n)}\right] [X(1),X(n)] yes μ \mu μ The confidence level of is 1 − 2 1 − n 1-2^{1-n} 1−21−n The confidence interval of .
Solution:
First consider asking U = x ( 1 ) U=x_{(1)} U=x(1) The distribution of , According to the calculation formula of the minimum value distribution
F U ( u ) = 1 − [ 1 − Φ ( u − μ σ ) ] n F_{U}(u)=1-\left[1-\Phi\left(\frac{u-\mu}{\sigma}\right)\right]^{n} FU(u)=1−[1−Φ(σu−μ)]n be P ( μ < x ( 1 ) ) = 1 − F U ( μ ) = [ 1 − Φ ( 0 ) ] n = ( 1 2 ) n = 2 − n P\left(\mu<x_{(1)}\right)=1-F_{U}(\mu)=[1-\Phi(0)]^{n}=\left(\frac{1}{2}\right)^{n}=2^{-n} P(μ<x(1))=1−FU(μ)=[1−Φ(0)]n=(21)n=2−n.
According to the symmetry P ( μ > x ( n ) ) = 2 − n P\left(\mu>x_{(n)}\right)=2^{-n} P(μ>x(n))=2−n. therefore P ( x ( 1 ) ⩽ μ ⩽ x ( n ) ) = 1 − 2 ⋅ 2 − n = 1 − 2 1 − n P\left(x_{(1)} \leqslant \mu \leqslant x_{(n)}\right)=1-2 \cdot 2^{-n}=1-2^{1-n} P(x(1)⩽μ⩽x(n))=1−2⋅2−n=1−21−n
8、 ... and 、(20 branch ) Some come from the general X ∼ f ( x ) = 1 2 e − ∣ x − θ ∣ X \sim f(x)=\frac{1}{2} e^{-|x-\theta|} X∼f(x)=21e−∣x−θ∣ Of 7 Random sample , seek θ \theta θ Of MLE.
Solution:
Likelihood function L ( X ; θ ) = ( 1 2 ) 7 e − ∑ i = 1 7 ∣ x i − θ ∣ = ( 1 2 ) 7 e − ∑ i = 1 7 ∣ x ( i ) − θ ∣ L(\mathbf{X} ; \theta)=\left(\frac{1}{2}\right)^{7} e^{-\sum_{i=1}^{7}\left|x_{i}-\theta\right|}=\left(\frac{1}{2}\right)^{7} e^{-\sum_{i=1}^{7}\left|x_{(i)}-\theta\right|} L(X;θ)=(21)7e−∑i=17∣xi−θ∣=(21)7e−∑i=17∣x(i)−θ∣. here x ( i ) x_{(i)} x(i) It means No i i i Order statistics . In order to make the likelihood function as large as possible , Should make e − ∑ i = 1 7 ∣ x ( 0 ) − θ ∣ e^{-\sum_{i=1}^{7}\left|x_{(0)}-\theta\right|} e−∑i=17∣x(0)−θ∣ As big as possible , That is to make ∑ i = 1 7 ∣ x ( i ) − θ ∣ \sum_{i=1}^{7}\left|x_{(i)}-\theta\right| ∑i=17∣∣x(i)−θ∣∣ As small as possible . The following research ∑ i = 1 7 ∣ x ( i ) − θ ∣ \sum_{i=1}^{7}\left|x_{(i)}-\theta\right| ∑i=17∣∣x(i)−θ∣∣ The state of being :
∑ i = 1 7 ∣ x ( i ) − θ ∣ = ∑ i = 1 3 ( ∣ x ( i ) − θ ∣ + ∣ x ( 7 − i + 1 ) − θ ∣ ) + ∣ x ( 4 ) − θ ∣ \sum_{i=1}^{7}\left|x_{(i)}-\theta\right|=\sum_{i=1}^{3}\left(\left|x_{(i)}-\theta\right|+\left|x_{(7-i+1)}-\theta\right|\right)+\left|x_{(4)}-\theta\right| i=1∑7∣∣x(i)−θ∣∣=i=1∑3(∣∣x(i)−θ∣∣+∣∣x(7−i+1)−θ∣∣)+∣∣x(4)−θ∣∣ ( The above formula will x ( 1 ) , x ( 7 ) x_{(1)}, x_{(7)} x(1),x(7) Divided into one group , x ( 2 ) , x ( 6 ) x_{(2)}, x_{(6)} x(2),x(6) Divided into one group , x ( 3 ) , x ( 5 ) x_{(3)}, x_{(5)} x(3),x(5) Divided into one group , x ( 4 ) x_{(4)} x(4) A single group )
among ∣ x ( i ) − θ ∣ + ∣ x ( 7 − i + 1 ) − θ ∣ \left|x_{(i)}-\theta\right|+\left|x_{(7-i+1)}-\theta\right| ∣∣x(i)−θ∣∣+∣∣x(7−i+1)−θ∣∣ stay θ ∈ [ x ( i ) , x ( 7 − i + 1 ) ] \theta \in\left[x_{(i)}, x_{(7-i+1)}\right] θ∈[x(i),x(7−i+1)] Take the minimum value when ; ∣ x ( 4 ) − θ ∣ \left|x_{(4)}-\theta\right| ∣∣x(4)−θ∣∣ stay θ = x ( 4 ) \theta=x_{(4)} θ=x(4) Take the minimum value when . and ( ⋂ i = 1 3 [ x ( i ) , x ( 7 − i + 1 ) ] ) ∩ { x ( 4 ) } = { x ( 4 ) } \left(\bigcap_{i=1}^{3}\left[x_{(i)}, x_{(7-i+1)}\right]\right) \cap\left\{x_{(4)}\right\}=\left\{x_{(4)}\right\} (⋂i=13[x(i),x(7−i+1)])∩{ x(4)}={ x(4)}, therefore θ ^ = x ( 4 ) \hat{\theta}=x_{(4)} θ^=x(4) yes θ \theta θ Of MLE.
Nine 、(10 branch ) ( X 1 , X 2 ) ∼ N ( 0 , 0 ; 1 , 1 ; 0 ) , (X_1, X_2) \sim N(0,0 ; 1,1 ; 0), (X1,X2)∼N(0,0;1,1;0), seek X 1 X 2 \frac{X_1}{X_2} X2X1 Probability 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 .
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