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