课程名称︰统计学与计量经济学暨实习上
课程性质︰必修
课程教师︰张胜凯
开课学院：社会科学院
开课系所︰经济系
考试日期（年月日）︰2013/10/16
考试时限（分钟）：180分钟
试题 :
Econ 2013 Statistics and Econometrics I First Midterm Exam (10/16/2013)
Problem 1. (15 points (7,8))
A candy maker produces mints that have a label weight of 20.4 grams. Assume
that the distribution of the weight of these mints is N(21.37,0.16).
(a) Let denote the weight of a single mint selected at random from the
production line. Find P(X > 22.07).
(b) Suppose that 5 mints are selected independently and weighted. Let Y equal
the numberr of these mints that weigh less than 20.857 grams. Find P(Y ≦ 2).
Sol:
(a) P(X > 22.07) = P([x-μ]/σ > [22.07-21.37]/√0.16) = P(Z > 1.75)
= 0.5 - P(0 < Z < 1.75) = 0.5 - 0.46 = 0.04
(b) P(Y ≦ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2)
╭5╮
P(Y = 0) =│ │((X < 20.857)^0 [P(X≧20.857)]^5) = [(0.1)^0][(0.9)^5]≒0.59
╰0╯
╭5╮ 5!
P(Y = 1) =│ │P(X<20.857)[P(X≧20.857]^4 = ──P(X<20.857)[P(X≧20.857)]^4
╰1╯ 1!4!
P((x<20.857)) = P(Z < [20.857-21.37]/0.4) = P(Z<-1.2825) = P(Z>1.2825)
≒ 0.5 - P(0<Z<1.2825) ≒ 0.5 - 0.4 = 0.1
P(Y = 1) ≒ 5˙0.1˙(0.9)^4 = 0.32805
╭5╮ 5!
P(Y = 2) = │ │[P(X<20.857)]^2 [P(X≧20.857]^3 = ──(0.1)^2˙(0.9)^3
╰2╯ 2!3!
= 10˙0.01˙0.729 = 0.0729
P(Y ≦ 2) = 0.59 + 0.32805 + 0.0729 ≒ 0.99
Problem 2. (15 points (5,5,5))
Suppose P(A) = 0.3, P(B) = 0.5, P(C) = 0.4, and A is mutually exclusive with C;
A and B are independent; P(B ∩ C) = 0.1. Find the following probabilities.
c
(a) P(A ∪ C).
c
(b) P(C ∩ B).
c
(c) P(B |A).
Sol: c c
(a) P(A ∪ C) = P(A ) = 1 - P(A) = 1 - 0.3 = 0.7.
c
(b) P(C ∩ B) = P(B) - P(B ∩ C) = 0.5 - 0.1 = 0.4.
c
c P(B ∩ A) (P(A) - P(A ∩ B))
(c) P(B |A) = ───── =──────────= 1-P(B|A) = 1-P(B) = 1-0.5 = 0.5
P(A) P(A)
Problem 3. (10 points (5,5))
Given the following probability distribution,
────────────────
Y = 0 Y = 1 Y = 2
────────────────
X = 0 0.1 0.1 0.15
X = 1 0.1 r 0.05
X = 2 q 0.1 0.1
────────────────
(a) If P(Y > 1|X ≦ 1) = 1/3, find q and r.
(b) Find P(X ≦ 1|Y < 1).
Sol: P(Y > 1, X ≦ 1) P(Y=2,X=1)+P(Y=2,X=0) 0.05+0.15
(a) P(Y > 1|X ≦ 1) =─────────=───────────= ───── =1/3
P(X ≦ 1) P(X=1) + P(X=0) 0.5+r
r = 0.1;
and since P(S) = 1,
P(X = 0) + P(X = 1) + P(X = 2) = 0.35 + 0.25 + (0.2 + q) = 0.8 + q,
q = 0.2
P(X≦1, Y<1) 0.2
(b) P(X ≦ 1|Y < 1) = ─────── = ─── = 0.5.
P(Y<1) 0.4
Problem 4. (10 points)
X and Y are two random variables which only take two values, 1 and -1. From the
following joint probability table, calculate E[X], Var(X), Cov(X,Y),and E[X^3].
───┬────────
│ Y = 1 Y = -1
───┼────────
X = 1│ 0.3 0.2
│
X =-1│ 0.2 0.3
───┴────────
Sol:
E[X] = 1˙P(X = 1) + (-1)˙P(X = -1) = 0
Var(X) = E[X^2] - (E[X] )^2 = 1^2˙P(X = 1) + (-1)^2˙P(X = -1) + 0^2 = 1
Cor(X,Y) = E[XY] - E[X]E[Y]
= 1˙[P(X=1,Y=1)+P(X=-1,Y=-1)]+(-1)˙[P(X=-1,Y=1)+P(X=1,Y=-1)] = 0.2
E[X^3] = 1^3˙P(X = 1) + (-1)^3˙P(X = -1) = 0.
Problem 5. (20 points (5,5,5,5))
The salaries of the employees of a corporation are normally distributed with a
mean of 25,000 and a standard deviation of 5,000.
(a) What is the probability that a randomly selected employee will have a
starting salary of at least 31,000?
(b) What percentage of employees has salaries of less than 12,200?
(c) What are the minimum and the maximum salaries of the middle 95% of the
employees?
(d) If sixty-eight of the employees have incomes of at least 35,600, how many
individuals are employed in the corporation?
Sol:
Let X be amount of salaty. X_i~N(25000,500^2).
(a) P(X_i ≧ 31000) = P([X_i-μ]/σ ≧ [31000-25000]/5000) = P(Z ≧ 1.2)
≒ 0.5 - 0.3849 = 0.1151
(b) P(X < 12200) = P(Z < -2.56) = 0.5 - 0.4948 = 0.0052
95%
(c) ── = 0.475, P(0 ≦ Z ≦ z) = 0.475, z ≒ 1.96, -z = -1.96;
2
P(0 ≦ Z ≦ 1.96) = P(0≦Z≦[X*-25000]/5000), [X*-25000]/5000 = 1.96,
X* = 34800;
P(-1.96 ≦ Z ≦ 0) = P([X*-2500]/5000≦Z≦0), [X*-25000]/5000 = -1.96,
X* = 15200
minimum: 15000; maximum: 348000
(d) If there are N employees in the corporation.
P(X ≧ 35600) = P(Z≧[35600-25000]/5000) = P(Z ≧2.12) = 0.5-0.483 = 0.017
Problem 6. (20 points (5,5,5,5))
i.i.d.
Let X_i ~ N(μ,σ^2), we are interested in estimating σ^2. Suppose we know
i.i.d.
μ. Moreover, if Z_i ~ N(0,1), then
n
C_n = Σ(Z_i)^2 ~ Χ^2(n).
i=1
We know that if C_n ~ Χ^2(n), then E[C_n] = n and Var(C_n) = 2n. Let
n
V_n = (1/n)Σ (X_i - μ)^2
i=1
and
n
W_n = (1/[n-1])Σ (X_i - μ)^2
i=1
(Χ^2(n) denotes Chi-square distribution)
(a) Is V_n an unbiased estimator of σ^2? If not, what is the bias of V_n?
(b) Is V_n a consistent estimator of σ^2?
(c) Is W_n an unbiased estimator of σ^2? If not, what is the bias of W_n?
(d) Is W_n a consistent estimator of σ^2?
Sol: n
(a) E[V_n] = E[(1/n)Σ (X_i-μ)^2] = (1/n)E[(X_1-μ)^2 + … + (X_n-μ)^2]
i=1
= (1/n)nσ^2 = σ^2.
V_n is an unbiased estimator of σ^2.
(b) We see if MSE(V_n) will converge to zero or not when n goes to infinity.
n
(X_i-μ)/σ = Z_i~N(0,1), Σ [(X_i-μ)^2/σ^2] ~ Χ^2(n),
i=1
Var({Σ[(X_i-μ)^2]/σ^2}) = 2n, → Var(Σ(X_i-μ)^2) = 2nσ^2
→ Var([1/n]Σ(X_i-μ)^2) = (2σ^2)/n
MSE(V_n) = (2σ^4)/n + 0^2,
lim MSE(V_n) = 0, V_n is a consistent estimator of σ^2.
n→∞
n
(c) E[W_n] = E[(1/[n-1])Σ (X_i-μ)^2] = [1/(n-1)]nσ^2 = nσ^2/[(n-1)^2]
i=1
W_n is not an unbiased estimator of σ^2, bias = nσ^2/[(n-1)^2] - σ^2
= (-n^2+3n-1)σ^2/[(n-1)^2].
(d) Var(Σ(X_i-μ)^2) = 2nσ^4 → Var([1/(n-1)]Σ(X_i-μ)^2) = 2nσ^4/[(n-1)^2]
MSE(W_n) = 2nσ^4/[(n-1)^2] + {[nσ^2/(n-1)]-σ^2}^2
lim MSE(W_n) = 0, V_n is a consistent estimator of σ^2.
n→∞
Problem 7. (20 points (5,5,5,5))
i.i.d.
X_i ~ N(μ,σ^2),
_
μ_1 = (1/4)(X_1 + 2X_2 + X_3)
_
μ_2 = X_5
_
μ_3 = (1/n)(X_1 + … +X_n)
(a) Which estimator(s) is an unbiased estimator(s) of μ? Explain.
(b) Which estimator(s) is relatively efficient? Explain.
(c) Which estimator(s) has the smallest mean squared errors (MSE)? Explain.
(d) Which estimator(s) is a consistent estimator(s) of μ? Explain.
Sol:
(a) Three of them.
_ _ _
(b) Var(μ_1) = (3/8)σ^2, Var(μ_2) = σ^2, Var(μ_3) = σ^2/n,
_ _
so μ_3 has the smallest variance (as long as n ≧ 3), μ_3 is relative
efficient.
_ _
(c) MSE(μ) = Var(μ) + (Bias)^2, since three of them has Bias = 0, so we only
_
compare variance. From (b), we know μ_3 has the smallest mean squared
errors (MSE).
_ _ _
(d) lim MSE(μ_1) = (3/8)σ^2, lim MSE(μ_2) = σ^2, lim MSE(μ_3) = 0,
n→∞ n→∞ n→∞
_
so μ_3 is a consistent estimator of μ.