課程名稱︰統計學與計量經濟學暨實習上 課程性質︰必修 課程教師︰張勝凱 開課學院：社會科學院 開課系所︰經濟系 考試日期（年月日）︰2014/10/14 考試時限（分鐘）： 試題 : 1.How do we determine whether a function follows axioms of probability or not? Please write down axioms of probability. Let the function be a function of x i.e., f(x) ;use S to denote sample space, and E to denote empty set. (Hint: at least three axioms) Sol: (i) f(E) = 0, f(S) = 1 (ii) f(A) ≧ 0 for all A ⊆ S (iii) If A, B are mutually exclusive, for A, B ⊆ S if and only if P(A ∪ B) = P(A) + P(B) 2.(a) Please use Venn Diagram and mathematic formular to show Morgan's Law: c c c c c c (A ∪ B) = A ∩ B , (A ∩ B) = A ∪ B , where A, B are events in sample space S. c (b) A, B are events in sample space S. If P(A) = 1/3 , P(B ) = 1/4 , can A and B be disjoint (mutually exclusive)? Why? c c c Sol:(a) Show (A ∪ B) = A ∩ B c Let x be an element in sample space S. Then x ∈ (A ∪ B) => x不屬於 c c c c (A ∪ B) => x不屬於A and x不屬於B => x∈A and x∈B => x ∈ (A ∩ B ) c c c Show (A ∩ B) = A ∪ B c c c x ∈ (A ∩ B) => x不屬於(A ∩ B) => x ∈ A or x ∈ B => c c x ∈ (A ∪ B ) c (b) P(B ) = 1/4 => P(B) = 3/4. If A, B are disjoint, P(A ∪ B) = P(A) + P(B) = 1/3 + 3/4 = 13/12 > 1 there is a contradiction. So it is imposssible. c Another solution is to say that if A, B disjoint, then A ⊆ B => c c P(A) ≦ P(B ). But P(A) = 1/3 > 1/4 = P(B ), so A, B aren't disjoint. 3.(a) What is random variable? (b) We observe the numbers of boys and girls in families in a city. Difine X is the number of boys in a family; Y is the number of girls in a family. The joint probability distribution is as below: ┌────┬────┬────┬────┐ │╲ │ │ │ │ ∣ ╲ Y │ 0 │ 1 │ 2 │ ∣ X ╲ │ │ │ │ │ ╲│ │ │ │ ├────┼────┼────┼────┤ │ 0 │ 0.05 │ 0.15 │ 0.10 │ ├────┼────┼────┼────┤ │ 1 │ 0.20 │ 0.25 │ 0.05 │ ├────┼────┼────┼────┤ │ 2 │ 0.15 │ 0.05 │ 0 │ └────┴────┴────┴────┘ (i) What is the value of f(X = 1)? What is the value of f(X = 1, Y = 1)? What is the value of f(X = 1 | Y = 1)? (ii) Are X and Y moving toward the same direction? What is the value of ρ (Correlation Coefficient of X and Y)? xy (iii) What is the value of Var(X|Y = 1)? Sol: (a) A numerical description of the outcome of an experiment. (b) (i) f(X = 1) = Σ f(X = 1, Y = 1) = f(1,0) + f(1,1) + f(1,2) y = 0.2 + 0.25 + 0.05 = 0.5 f(X = 1, Y = 1) = 0.25 f(X = 1, Y = 1) 0.25 0.25 5 f(X = 1 | Y = 1) = ──────── = ─────── = ── = ─ f(Y = 1) 0.15+0.25+0.05 0.45 9 (ii) To determine whether X and Y move toward the same direction, we can see Covariance or Correlation Coefficient of X and Y. Cov(X,Y) = E[XY] - E[X]E[Y] = 0.45 - (0.9)(0.75) = -0.225 < 0 So X and Y do not move to the same direction. Cov(X,Y) 0.225 Corr(X,Y) = ρ = ───── = - ────── ≒ -0.4605 xy ρ ρ (0.7)(0.698) x y 2 2 (iii) Var(X|Y = 1) = E[X |Y = 1] - (E[X|Y = 1]) 2 2 2 = [(0) (0.15/0.45) + (1) (0.25/0.45) + (2) (0.05/0.45)] 2 32 - [0(0.15/0.45) + 1(0.25/0.45) + 2(0.05/0.45)] = 1-49/81 = ─ 81 4.Prove the law of iterated exceptation: E [E[Y|X]] = E[Y]. You can use x continuous random variable (x,y) or discrete random variable (x,y) to complete the proof. T Sol: Use continuous vector (x,y) E [E[Y|X]] = ∫E[Y|X]f (x)dx = ∫[∫y f (y|x)dy]f (x)dx x x X|Y x = ∫∫y f (x,y) dxdy (∵f (y|x) = f (x,y)/f (x) ) X,Y Y|X X,Y x = ∫y[∫f (x,y) dx]dy = ∫y f (y) dy = E[Y] (∵f (y) = ∫f (x,y) dx) X,Y Y Y X,Y Use discrete random variable X, Y E [E[Y|X]] = ΣE[Y|X = x]p(x) = Σ(ΣE[Y|X = x]p(y|x))p(x) x x x y =ΣΣy p(x,y) = ΣyΣp(x,y) = Σy p(y) = E[Y] (∵Σp(y|x) = p(y)) x y y x y y 5.We use a linear function h(x) = a + bx to predict y. If h(x) is the best linear predictor of y given x, then E[Y|X = x] = α + βx , while σ Cov(X,Y) xy Cov(X,Y) β = ───── = ─── . Why β is equal to ───── ? Var(X) 2 Var(X) σ x --

※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.166.231.126 ※ 文章網址: https://webptt.com/m.aspx?n=bbs/NTU-Exam/M.1422784910.A.51C.html ※ 編輯: Malzahar (118.166.231.126), 02/01/2015 18:03:18 ※ 編輯: Malzahar (118.166.211.193), 02/02/2015 11:54:57 ※ 編輯: Malzahar (118.166.211.193), 02/02/2015 13:09:56 ※ 編輯: Malzahar (118.166.211.193), 02/02/2015 20:18:37