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【九阳神功】2018复旦大学应用统计真题+解析
2022-07-06 09:20:00 【大师兄统计】
真题部分
一、(20分) 从1-10中不放回选3个数字, 求以下概率
(1)(5分) 最小数字是5;
(2)(5分) 最大数字是5;
(3)(5分) 至少一个小于6;
(4)(5分) 一个小于5, 一个等于5, 一个大于5.
二、(15分) 你在重复尝试一个成功概率为 p p p的事件, 直至连续出现两次成功或两次失败才停止, 求你以两次成功停止的概率.
三、(15分) 求二项分布, ( a , b ) (a,b) (a,b)上均匀分布, 伽马分布的期望和方差.
四、(20分) 证明 E ( X 2 ) < ∞ E\left(X^{2}\right)<\infty E(X2)<∞的充要条件是级数 ∑ n P ( ∣ X ∣ > n ) \sum n P(|X|>n) ∑nP(∣X∣>n)收敛.
五、(20分) X 1 , X 2 , X 3 X_{1}, X_{2}, X_{3} X1,X2,X3是取自期望为 α \alpha α的指数分布的随机样本, 求概率 P ( X 1 < X 2 < X 3 ) P\left(X_{1}<X_{2}<X_{3}\right) P(X1<X2<X3) 以及 X ( 1 ) X_{(1)} X(1) 的概率密度.
六、(20分) P ( X i = − 0.3 ) = P ( X i = 0.4 ) = 1 2 , i = 1 , 2 , … , n , P\left(X_{i}=-0.3\right)=P\left(X_{i}=0.4\right)=\frac{1}{2}, i=1,2, \ldots, n, P(Xi=−0.3)=P(Xi=0.4)=21,i=1,2,…,n, 相互独立, 构造随机变量序列 Y n = ∏ i = 1 n ( X i + 1 ) , Y_{n}=\prod_{i=1}^{n}\left(X_{i}+1\right), Yn=∏i=1n(Xi+1), 求 Y n Y_{n} Yn的极限并证明 Y n Y_{n} Yn的期望趋于无穷.
七、(20分) 有一堆球: 2红, 3黑, 4白. 从中随机摸一个球, 如果是黑色则记你赢, 如果是其他颜色, 则有放回的继续摸球, 直至重复出现该颜色或黑色为止, 如果出现你第一次摸到的颜色, 则你赢, 否则你输. 求你赢的概率.
八、(20分)
(1)(10分) 解释相合估计;
(2)(10分) X 1 , … , X n X_{1}, \ldots, X_{n} X1,…,Xn是来自一个同一个总体的样本, 写出一个中位数的相合估计, 并说明理由.
解析部分
一、(20分) 从1-10中不放回选3个数字, 求以下概率
(1)(5分) 最小数字是5;
(2)(5分) 最大数字是5;
(3)(5分) 至少一个小于6;
(4)(5分) 一个小于5, 一个等于5, 一个大于5.
Solution:
(1) # Ω = C 10 3 = 120 \# \Omega=C_{10}^{3}=120 #Ω=C103=120, #A A 1 = 1 ⋅ C 5 2 = 10 , P ( A 1 ) = # A 1 # Ω = 1 12 A_{1}=1 \cdot C_{5}^{2}=10, P\left(A_{1}\right)=\frac{\# A_{1}}{\# \Omega}=\frac{1}{12} A1=1⋅C52=10,P(A1)=#Ω#A1=121.
(2) # A 2 = 1 ⋅ C 4 2 = 6 , P ( A 2 ) = # A 2 # Ω = 1 20 \# A_{2}=1 \cdot C_{4}^{2}=6, P\left(A_{2}\right)=\frac{\# A_{2}}{\# \Omega}=\frac{1}{20} #A2=1⋅C42=6,P(A2)=#Ω#A2=201.
(3) # A 3 ‾ = C 5 3 = 10 , P ( A 3 ) = 1 − # A 3 ‾ # Ω = 11 12 \# \overline{A_{3}}=C_{5}^{3}=10, P\left(A_{3}\right)=1-\frac{\# \overline{A_{3}}}{\# \Omega}=\frac{11}{12} #A3=C53=10,P(A3)=1−#Ω#A3=1211.
(4) # A 4 = 4 ⋅ 1 ⋅ 5 = 20 , P ( A 4 ) = # A 4 # Ω = 1 6 \# A_{4}=4 \cdot 1 \cdot 5=20, P\left(A_{4}\right)=\frac{\# A_{4}}{\# \Omega}=\frac{1}{6} #A4=4⋅1⋅5=20,P(A4)=#Ω#A4=61.
二、(15分) 你在重复尝试一个成功概率为 p p p 的事件, 直至连续出现两次成功或两次失败才停止, 求你 以两次成功停止的概率.
Solution:
设 A A A “以两次成功停止”", p 0 = P ( A ) , p 1 = P ( A ∣ p_{0}=P(A), p_{1}=P(A \mid p0=P(A),p1=P(A∣ 第一次成功 ) , p − 1 = P ), p_{-1}=P ),p−1=P ( A ∣ A \mid A∣ 第一次失败 ) ) ), 根据全概率公式: { p 0 = p 1 ⋅ p + p − 1 ⋅ ( 1 − p ) p 1 = p + p − 1 ⋅ ( 1 − p ) p − 1 = p 1 ⋅ p \left\{\begin{array}{l} p_{0}=p_{1} \cdot p+p_{-1} \cdot(1-p) \\ p_{1}=p+p_{-1} \cdot(1-p) \\ p_{-1}=p_{1} \cdot p \end{array}\right. ⎩⎨⎧p0=p1⋅p+p−1⋅(1−p)p1=p+p−1⋅(1−p)p−1=p1⋅p 解得 p 0 = p 2 ( 2 − p ) 1 − p ( 1 − p ) p_{0}=\frac{p^{2}(2-p)}{1-p(1-p)} p0=1−p(1−p)p2(2−p).
三、(15分) 求二项分布, ( a , b ) (a,b) (a,b)上均匀分布, 伽马分布的期望和方差.
Solution:
(1) 二项分布: X ∼ B ( n , p ) , P ( X = k ) = C n k p k ( 1 − p ) n − k , k = 0 , 1 , … , n X \sim B(n, p), P(X=k)=C_{n}^{k} p^{k}(1-p)^{n-k}, k=0,1, \ldots, n X∼B(n,p),P(X=k)=Cnkpk(1−p)n−k,k=0,1,…,n,
E X = ∑ k = 0 n k n ! k ! ( n − k ) ! p k ( 1 − p ) n − k = n p ∑ k = 1 n ( n − 1 ) ! ( k − 1 ) ! ( n − k ) ! p k − 1 ( 1 − p ) n − k = n p , E X=\sum_{k=0}^{n} k \frac{n !}{k !(n-k) !} p^{k}(1-p)^{n-k}=n p \sum_{k=1}^{n} \frac{(n-1) !}{(k-1) !(n-k) !} p^{k-1}(1-p)^{n-k}=n p, EX=k=0∑nkk!(n−k)!n!pk(1−p)n−k=npk=1∑n(k−1)!(n−k)!(n−1)!pk−1(1−p)n−k=np,
E X ( X − 1 ) = ∑ k = 0 n k ( k − 1 ) C n k p k ( 1 − p ) n − k = n ( n − 1 ) p 2 ∑ k = 2 n C n − 2 k − 2 p k − 2 ( 1 − p ) n − k = n ( n − 1 ) p 2 , E X(X-1)=\sum_{k=0}^{n} k(k-1) C_{n}^{k} p^{k}(1-p)^{n-k}=n(n-1) p^{2} \sum_{k=2}^{n} C_{n-2}^{k-2} p^{k-2}(1-p)^{n-k}=n(n-1) p^{2}, EX(X−1)=k=0∑nk(k−1)Cnkpk(1−p)n−k=n(n−1)p2k=2∑nCn−2k−2pk−2(1−p)n−k=n(n−1)p2,
因此 E X 2 = n ( n − 1 ) p 2 + n p , D X = E X 2 − ( E X ) 2 = n ( n − 1 ) p 2 + n p − n 2 p 2 = n p ( 1 − p ) E X^{2}=n(n-1) p^{2}+n p, D X=E X^{2}-(E X)^{2}=n(n-1) p^{2}+n p-n^{2} p^{2}=n p(1-p) EX2=n(n−1)p2+np,DX=EX2−(EX)2=n(n−1)p2+np−n2p2=np(1−p).(2)均匀分布: X ∼ U ( a , b ) , f ( x ) = 1 b − a , a < x < b X \sim U(a, b), f(x)=\frac{1}{b-a}, a<x<b X∼U(a,b),f(x)=b−a1,a<x<b,
E X = ∫ a b x b − a d x = a + b 2 , D X = ∫ a b ( x − a + b 2 ) 2 1 b − a d x = ( b − a ) 2 12 . E X=\int_{a}^{b} \frac{x}{b-a} d x=\frac{a+b}{2}, D X=\int_{a}^{b}\left(x-\frac{a+b}{2}\right)^{2} \frac{1}{b-a} d x=\frac{(b-a)^{2}}{12} . EX=∫abb−axdx=2a+b,DX=∫ab(x−2a+b)2b−a1dx=12(b−a)2.
(3)伽马分布: X ∼ G a ( α , λ ) , f ( x ) = λ α Γ ( α ) x α − 1 e − λ x , x > 0 X \sim G a(\alpha, \lambda), f(x)=\frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x}, x>0 X∼Ga(α,λ),f(x)=Γ(α)λαxα−1e−λx,x>0,
E X = ∫ 0 + ∞ λ α Γ ( α ) x α e − λ x d x = 1 λ Γ ( α ) ∫ 0 + ∞ ( λ x ) α e − λ x d ( λ x ) = Γ ( α + 1 ) λ Γ ( α ) = α λ , E X=\int_{0}^{+\infty} \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha} e^{-\lambda x} d x=\frac{1}{\lambda \Gamma(\alpha)} \int_{0}^{+\infty}(\lambda x)^{\alpha} e^{-\lambda x} d(\lambda x)=\frac{\Gamma(\alpha+1)}{\lambda \Gamma(\alpha)}=\frac{\alpha}{\lambda}, EX=∫0+∞Γ(α)λαxαe−λxdx=λΓ(α)1∫0+∞(λx)αe−λxd(λx)=λΓ(α)Γ(α+1)=λα,
E X 2 = ∫ 0 + ∞ λ α Γ ( α ) x α + 1 e − λ x d x = 1 λ 2 Γ ( α ) ∫ 0 + ∞ ( λ x ) α + 1 e − λ x d ( λ x ) = Γ ( α + 2 ) λ 2 Γ ( α ) = α ( α + 1 ) λ E X^{2}=\int_{0}^{+\infty} \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha+1} e^{-\lambda x} d x=\frac{1}{\lambda^{2} \Gamma(\alpha)} \int_{0}^{+\infty}(\lambda x)^{\alpha+1} e^{-\lambda x} d(\lambda x)=\frac{\Gamma(\alpha+2)}{\lambda^{2} \Gamma(\alpha)}=\frac{\alpha(\alpha+1)}{\lambda} EX2=∫0+∞Γ(α)λαxα+1e−λxdx=λ2Γ(α)1∫0+∞(λx)α+1e−λxd(λx)=λ2Γ(α)Γ(α+2)=λα(α+1),
故 D X = E X 2 − ( E X ) 2 = α ( α + 1 ) λ 2 − α 2 λ 2 = α λ 2 . D X=E X^{2}-(E X)^{2}=\frac{\alpha(\alpha+1)}{\lambda^{2}}-\frac{\alpha^{2}}{\lambda^{2}}=\frac{\alpha}{\lambda^{2}} . DX=EX2−(EX)2=λ2α(α+1)−λ2α2=λ2α.
四、(20分) 证明 E ( X 2 ) < ∞ E\left(X^{2}\right)<\infty E(X2)<∞的充要条件是级数 ∑ n P ( ∣ X ∣ > n ) \sum n P(|X|>n) ∑nP(∣X∣>n)收敛.
Solution:
(1) 先说明 E ∣ X ∣ < + ∞ E|X|<+\infty E∣X∣<+∞ 的充要条件是级数 ∑ n = 1 ∞ P ( ∣ X ∣ > n ) \sum_{n=1}^{\infty} P(|X|>n) ∑n=1∞P(∣X∣>n) 收敛:因为 ∑ n = 1 ∞ P ( ∣ X ∣ > n ) = ∑ n = 1 ∞ ∑ k = n ∞ P ( k < ∣ X ∣ ≤ k + 1 ) = ∑ k = 1 ∞ ∑ n = 1 k P ( k < ∣ X ∣ ≤ k + 1 ) = ∑ k = 1 ∞ k P ( k < ∣ X ∣ ≤ k + 1 ) \begin{aligned} \sum_{n=1}^{\infty} P(|X|>n) &=\sum_{n=1}^{\infty} \sum_{k=n}^{\infty} P(k<|X| \leq k+1) \\ &=\sum_{k=1}^{\infty} \sum_{n=1}^{k} P(k<|X| \leq k+1) \\ &=\sum_{k=1}^{\infty} k P(k<|X| \leq k+1) \end{aligned} n=1∑∞P(∣X∣>n)=n=1∑∞k=n∑∞P(k<∣X∣≤k+1)=k=1∑∞n=1∑kP(k<∣X∣≤k+1)=k=1∑∞kP(k<∣X∣≤k+1)同时由正项级数的比较判别法, 上级数与 ∑ k = 1 ∞ ( k + 1 ) P ( k < ∣ X ∣ ≤ k + 1 ) \sum_{k=1}^{\infty}(k+1) P(k<|X| \leq k+1) ∑k=1∞(k+1)P(k<∣X∣≤k+1)也是同敛散的, 考虑到 E ∣ X ∣ = ∫ 0 + ∞ x d F ∣ X ∣ ( x ) E|X|=\int_{0}^{+\infty} x d F_{|X|}(x) E∣X∣=∫0+∞xdF∣X∣(x),一方面有 ∫ 0 + ∞ x d F ∣ X ∣ ( x ) = ∑ k = 0 ∞ ∫ k k + 1 x d F ∣ X ∣ ( x ) ≤ ∑ k = 0 ∞ ∫ k k + 1 ( k + 1 ) d F ∣ X ∣ ( x ) = ∑ k = 0 ∞ ( k + 1 ) P ( k < ∣ X ∣ ≤ k + 1 ) , \begin{aligned} \int_{0}^{+\infty} x d F_{|X|}(x) &=\sum_{k=0}^{\infty} \int_{k}^{k+1} x d F_{|X|}(x) \\ & \leq \sum_{k=0}^{\infty} \int_{k}^{k+1}(k+1) d F_{|X|}(x) \\ &=\sum_{k=0}^{\infty}(k+1) P(k<|X| \leq k+1), \end{aligned} ∫0+∞xdF∣X∣(x)=k=0∑∞∫kk+1xdF∣X∣(x)≤k=0∑∞∫kk+1(k+1)dF∣X∣(x)=k=0∑∞(k+1)P(k<∣X∣≤k+1), 另一方面有
∫ 0 + ∞ x d F ∣ X ∣ ( x ) = ∑ k = 0 ∞ ∫ k k + 1 x d F ∣ X ∣ ( x ) ≥ ∑ k = 0 ∞ ∫ k k + 1 k d F ∣ X ∣ ( x ) = ∑ k = 0 ∞ k P ( k < ∣ X ∣ ≤ k + 1 ) , \begin{aligned} \int_{0}^{+\infty} x d F_{|X|}(x) &=\sum_{k=0}^{\infty} \int_{k}^{k+1} x d F_{|X|}(x) \\ & \geq \sum_{k=0}^{\infty} \int_{k}^{k+1} k d F_{|X|}(x) \\ &=\sum_{k=0}^{\infty} k P(k<|X| \leq k+1), \end{aligned} ∫0+∞xdF∣X∣(x)=k=0∑∞∫kk+1xdF∣X∣(x)≥k=0∑∞∫kk+1kdF∣X∣(x)=k=0∑∞kP(k<∣X∣≤k+1), 综上所述, E ∣ X ∣ < + ∞ E|X|<+\infty E∣X∣<+∞ 的充要条件是级数 ∑ n = 1 ∞ P ( ∣ X ∣ > n ) \sum_{n=1}^{\infty} P(|X|>n) ∑n=1∞P(∣X∣>n) 收敛.(2) 再说明 E X 2 < + ∞ E X^{2}<+\infty EX2<+∞ 的充要条件是级数 ∑ n = 1 + ∞ n P ( ∣ X ∣ > n ) \sum_{n=1}^{+\infty} n P(|X|>n) ∑n=1+∞nP(∣X∣>n)收敛: 因为 ∑ n = 1 ∞ n P ( ∣ X ∣ > n ) = ∑ n = 1 ∞ ∑ k = n ∞ n P ( k < ∣ X ∣ ≤ k + 1 ) = ∑ k = 1 ∞ ∑ n = 1 ∞ n P ( k < ∣ X ∣ ≤ k + 1 ) = ∑ k = 1 ∞ k ( k + 1 ) 2 P ( k < ∣ X ∣ ≤ k + 1 ) \begin{aligned} \sum_{n=1}^{\infty} n P(|X|>n) &=\sum_{n=1}^{\infty} \sum_{k=n}^{\infty} n P(k<|X| \leq k+1) \\ &=\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} n P(k<|X| \leq k+1) \\ &=\sum_{k=1}^{\infty} \frac{k(k+1)}{2} P(k<|X| \leq k+1) \end{aligned} n=1∑∞nP(∣X∣>n)=n=1∑∞k=n∑∞nP(k<∣X∣≤k+1)=k=1∑∞n=1∑∞nP(k<∣X∣≤k+1)=k=1∑∞2k(k+1)P(k<∣X∣≤k+1) 同时由正项级数的比较判别法, 上式的敛散性显然等同于 ∑ n = 1 ∞ n 2 P ( n < ∣ X ∣ ≤ n + 1 ) \sum_{n=1}^{\infty} n^{2} P(n<|X| \leq n+1) ∑n=1∞n2P(n<∣X∣≤n+1) 的敛散性,也等价于 ∑ n = 1 ∞ ( n + 1 ) 2 P ( n < ∣ X ∣ ≤ n + 1 ) \sum_{n=1}^{\infty}(n+1)^{2} P(n<|X| \leq n+1) ∑n=1∞(n+1)2P(n<∣X∣≤n+1) 的敛散性, 同样借助二阶矩的定 义式 E X 2 = ∫ 0 + ∞ x 2 d F ∣ X ∣ ( x ) E X^{2}=\int_{0}^{+\infty} x^{2} d F_{|X|}(x) EX2=∫0+∞x2dF∣X∣(x), 一方面有 ∫ 0 + ∞ x 2 d F ∣ X ∣ ( x ) = ∑ n = 0 ∞ ∫ n n + 1 x 2 d F ∣ X ∣ ( x ) ≤ ∑ n = 0 ∞ ∫ n n + 1 ( n + 1 ) 2 d F ∣ X ∣ ( x ) = ∑ n = 0 ∞ ( n + 1 ) 2 P ( n < ∣ X ∣ ≤ n + 1 ) \begin{aligned} \int_{0}^{+\infty} x^{2} d F_{|X|}(x) &=\sum_{n=0}^{\infty} \int_{n}^{n+1} x^{2} d F_{|X|}(x) \\ & \leq \sum_{n=0}^{\infty} \int_{n}^{n+1}(n+1)^{2} d F_{|X|}(x) \\ &=\sum_{n=0}^{\infty}(n+1)^{2} P(n<|X| \leq n+1) \end{aligned} ∫0+∞x2dF∣X∣(x)=n=0∑∞∫nn+1x2dF∣X∣(x)≤n=0∑∞∫nn+1(n+1)2dF∣X∣(x)=n=0∑∞(n+1)2P(n<∣X∣≤n+1)
另一方面有 ∫ 0 + ∞ x 2 d F ∣ X ∣ ( x ) = ∑ n = 0 ∞ ∫ n n + 1 x 2 d F ∣ X ∣ ( x ) ≤ ∑ n = 0 ∞ ∫ n n + 1 n 2 d F ∣ X ∣ ( x ) = ∑ n = 0 ∞ n 2 P ( n < ∣ X ∣ ≤ n + 1 ) , \begin{aligned} \int_{0}^{+\infty} x^{2} d F_{|X|}(x) &=\sum_{n=0}^{\infty} \int_{n}^{n+1} x^{2} d F_{|X|}(x) \\ & \leq \sum_{n=0}^{\infty} \int_{n}^{n+1} n^{2} d F_{|X|}(x) \\ &=\sum_{n=0}^{\infty} n^{2} P(n<|X| \leq n+1), \end{aligned} ∫0+∞x2dF∣X∣(x)=n=0∑∞∫nn+1x2dF∣X∣(x)≤n=0∑∞∫nn+1n2dF∣X∣(x)=n=0∑∞n2P(n<∣X∣≤n+1), 综上所述, E X 2 < + ∞ E X^{2}<+\infty EX2<+∞ 的充要条件是级数 ∑ n = 1 + ∞ n P ( ∣ X ∣ > n ) \sum_{n=1}^{+\infty} n P(|X|>n) ∑n=1+∞nP(∣X∣>n) 收敛.
五、(20分) X 1 , X 2 , X 3 X_{1}, X_{2}, X_{3} X1,X2,X3是取自期望为 α \alpha α的指数分布的随机样本, 求概率 P ( X 1 < X 2 < X 3 ) P\left(X_{1}<X_{2}<X_{3}\right) P(X1<X2<X3) 以及 X ( 1 ) X_{(1)} X(1) 的概率密度.
Solution:
根据轮换对称性, P ( X 1 < X 2 < X 3 ) = 1 6 P\left(X_{1}<X_{2}<X_{3}\right)=\frac{1}{6} P(X1<X2<X3)=61. 令 Y = X ( 1 ) Y=X_{(1)} Y=X(1), 则当 y > 0 y>0 y>0 时, 1 − F ( y ) = P { Y > y } = P 3 { X 1 > y } = e − 3 α y 1-F(y)=P\{Y>y\}=P^{3}\left\{X_{1}>y\right\}=e^{-\frac{3}{\alpha} y} 1−F(y)=P{ Y>y}=P3{ X1>y}=e−α3y, 故 f ( y ) = 3 α e − 3 α y , y > 0 f(y)=\frac{3}{\alpha} e^{-\frac{3}{\alpha} y}, y>0 f(y)=α3e−α3y,y>0. 这恰好是 Exp ( 3 α ) \operatorname{Exp}\left(\frac{3}{\alpha}\right) Exp(α3).
六、(20分) P ( X i = − 0.3 ) = P ( X i = 0.4 ) = 1 2 , i = 1 , 2 , … , n , P\left(X_{i}=-0.3\right)=P\left(X_{i}=0.4\right)=\frac{1}{2}, i=1,2, \ldots, n, P(Xi=−0.3)=P(Xi=0.4)=21,i=1,2,…,n, 相互独立, 构造随机变量序列 Y n = ∏ i = 1 n ( X i + 1 ) , Y_{n}=\prod_{i=1}^{n}\left(X_{i}+1\right), Yn=∏i=1n(Xi+1), 求 Y n Y_{n} Yn的极限并证明 Y n Y_{n} Yn的期望趋于无穷.
Solution:
由强大数律, 1 n ln Y n = 1 n ∑ i = 1 n ln ( X i + 1 ) * a.s. E ln ( X 1 + 1 ) = 1 2 ln 0.98 < 0 \frac{1}{n} \ln Y_{n}=\frac{1}{n} \sum_{i=1}^{n} \ln \left(X_{i}+1\right) \stackrel{\text { a.s. }}{\longrightarrow} E \ln \left(X_{1}+1\right)=\frac{1}{2} \ln 0.98<0 n1lnYn=n1∑i=1nln(Xi+1)* a.s. Eln(X1+1)=21ln0.98<0, 故 ln Y n * a.s. − ∞ , Y n * a.s. 0 \ln Y_{n} \stackrel{\text { a.s. }}{\longrightarrow}-\infty, Y_{n} \stackrel{\text { a.s. }}{\longrightarrow} 0 lnYn* a.s. −∞,Yn* a.s. 0, 因此 Y n Y_{n} Yn 的极限是单点分布, 以概率 1 取 0 . 而 E Y n = ∏ i = 1 n E ( X i + 1 ) = ∏ i = 1 n ( 0.7 + 1.4 2 ) = 1.0 5 n → + ∞ . E Y_{n}=\prod_{i=1}^{n} E\left(X_{i}+1\right)=\prod_{i=1}^{n}\left(\frac{0.7+1.4}{2}\right)=1.05^{n} \rightarrow+\infty. EYn=i=1∏nE(Xi+1)=i=1∏n(20.7+1.4)=1.05n→+∞.
七、(20分) 有一堆球: 2红, 3黑, 4白. 从中随机摸一个球, 如果是黑色则记你赢, 如果是其他颜色, 则有放回的继续摸球, 直至重复出现该颜色或黑色为止, 如果出现你第一次摸到的颜色, 则你赢, 否则你输. 求你赢的概率.
Solution:
设 A k A_k Ak为第 k k k次赢的概率, 待求结果为 P ( A ) = P ( ⋃ k = 1 ∞ A k ) = ∑ k = 1 ∞ P ( A k ) P(A)=P(\bigcup_{k=1}^{\infty}A_k)=\sum_{k=1}^{\infty}P(A_k) P(A)=P(⋃k=1∞Ak)=∑k=1∞P(Ak).
(i) 第一次赢的概率 P ( A 1 ) = 1 3 P(A_1)=\frac{1}{3} P(A1)=31.
(ii) 第 k k k次赢( k > 1 k>1 k>1)表明: 第一次未摸到黑球, 后续的 2 , 3 , . . . , k − 1 2,3,...,k-1 2,3,...,k−1次没有摸到黑也没有摸到你第一次摸中的颜色, 最后第 k k k次摸中了最初的颜色, 这一概率肯定与你最初摸到的颜色有关, 考虑全概率公式 P ( A k ) = P ( R ) P ( A k ∣ R ) + P ( W ) P ( A k ∣ W ) P(A_k)=P(R)P(A_k|R)+P(W)P(A_k|W) P(Ak)=P(R)P(Ak∣R)+P(W)P(Ak∣W), 其中 R R R表示摸到红, W W W表示摸到白, 有
P ( A k ∣ R ) = ( 4 9 ) k − 2 2 9 , P ( A k ∣ W ) = ( 2 9 ) k − 2 4 9 . P\left( A_k|R \right) =\left( \frac{4}{9} \right) ^{k-2}\frac{2}{9},\quad P\left( A_k\mid W \right) =\left( \frac{2}{9} \right) ^{k-2}\frac{4}{9}. P(Ak∣R)=(94)k−292,P(Ak∣W)=(92)k−294. 因此 P ( A k ) = P ( R ) 2 9 ( 4 9 ) k − 2 + P ( W ) 4 9 ( 2 9 ) k − 2 = 4 81 ( 4 9 ) k − 2 + 16 81 ( 2 9 ) k − 2 . P\left( A_k \right) =P\left( R \right) \frac{2}{9}\left( \frac{4}{9} \right) ^{k-2}+P\left( W \right) \frac{4}{9}\left( \frac{2}{9} \right) ^{k-2}=\frac{4}{81}\left( \frac{4}{9} \right) ^{k-2}+\frac{16}{81}\left( \frac{2}{9} \right) ^{k-2}. P(Ak)=P(R)92(94)k−2+P(W)94(92)k−2=814(94)k−2+8116(92)k−2. 进行求和, 有
P ( A ) = P ( A 1 ) + 4 81 ( 1 − 4 9 ) + 16 81 ( 1 − 2 9 ) = 1 3 + 4 45 + 16 63 = 71 105 . P\left( A \right) =P\left( A_1 \right) +\frac{4}{81\left( 1-\frac{4}{9} \right)}+\frac{16}{81\left( 1-\frac{2}{9} \right)}=\frac{1}{3}+\frac{4}{45}+\frac{16}{63}=\frac{71}{105}. P(A)=P(A1)+81(1−94)4+81(1−92)16=31+454+6316=10571.
八、(20分)
(1)(10分) 解释相合估计;
(2)(10分) X 1 , … , X n X_{1}, \ldots, X_{n} X1,…,Xn是来自一个同一个总体的样本, 写出一个中位数的相合估计, 并说明理由.
Solution:
(1) g ^ \hat{g} g^ 是 g g g 的相合估计意味着 g ^ → p g \hat{g} \stackrel{p}{\rightarrow} g g^→pg, 这说明了 g ^ \hat{g} g^ 是一个很好的估计, 至少在样本量 n n n 较大时, 它偏离 g g g 的概率极小. 更进一步, 如果说 g ^ \hat{g} g^ 是 g g g 的强相合估计意味着 g ^ * a.s. g \hat{g} \stackrel{\text { a.s. }}{\longrightarrow} g g^* a.s. g.(2) 样本中位数 X [ n 2 ] X_{\left[\frac{n}{2}\right]} X[2n] 就是总体中位数 x 0.5 x_{0.5} x0.5 的相合估计量, 设总体密度函数为 f ( x ) f(x) f(x), 有样本中位数的渐近正态分布 X [ n 2 ] ∼ N ( x 0.5 , 1 4 n f 2 ( x 0.5 ) ) , X_{\left[\frac{n}{2}\right]} \sim N\left(x_{0.5}, \frac{1}{4 n f^{2}\left(x_{0.5}\right)}\right), X[2n]∼N(x0.5,4nf2(x0.5)1), 由其渐近正态分布可以得到 P ( ∣ X [ n 2 ] − x 0.5 ∣ < ε ) = P ( 2 n f ( x 0.5 ) ∣ X [ n 2 ] − x 0.5 ∣ < 2 n f ( x 0.5 ) ε ) ∼ Φ ( 2 n f ( x 0.5 ) ε ) − Φ ( − 2 n f ( x 0.5 ) ε ) → 1. P\left( \left| X_{\left[ \frac{n}{2} \right]}-x_{0.5} \right|<\varepsilon \right) =P\left( 2\sqrt{n}f\left( x_{0.5} \right) \left| X_{\left[ \frac{n}{2} \right]}-x_{0.5} \right|<2\sqrt{n}f\left( x_{0.5} \right) \varepsilon \right) \sim \Phi \left( 2\sqrt{n}f\left( x_{0.5} \right) \varepsilon \right) -\Phi \left( -2\sqrt{n}f\left( x_{0.5} \right) \varepsilon \right) \rightarrow 1. P(∣∣∣X[2n]−x0.5∣∣∣<ε)=P(2nf(x0.5)∣∣∣X[2n]−x0.5∣∣∣<2nf(x0.5)ε)∼Φ(2nf(x0.5)ε)−Φ(−2nf(x0.5)ε)→1.上式说明了样本中位数 X [ n 2 ] X_{\left[\frac{n}{2}\right]} X[2n] 就是总体中位数 x 0.5 x_{0.5} x0.5 的相合估计量.
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