20137 Advanced Statistics for Economic and Social Sciences (ESS-MS) General Exam
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
20137 Advanced Statistics for Economic and Social Sciences (ESS-MS)
General Exam
January 18th , 2022
Notation: Recall that 1(a < x < b) = 1(a,b)(x) and that fx(x; θ) = f9(x)(x).
Question 1 (5 points)
Let (X, Y) be the bivariate vector with joint density
f(x,Y )(x, y) = 1(0 < x < 1) 1(0 < y < 1).
(a) Find the joint density of the vector (U, V) with U = X + Y, V = .
Solution: The inverse transformation is
去 wdet(J)w = 1/v2 ,
therefore
fI,v(u, v) = 1 z0 < < 1、 1 z0 < u … < 1、 = 1(v > 1) 1 z < u < 1 + 、
(b) Find the marginal density of V .
Solution: We have
1+1/u 1 1
fv(v) = 1/u v2 1(v > 1) du = v2 1(v > 1).
Question 2 (11 points)
For i = 1, . . . , n, let Xi be i.i.d. according to the density function
e9 – 1/w
fx(x; θ) = x2 1(0 < x 生 1/θ),
where θ > 0 is an unknown parameter.
(a) Show that Fx(x; θ) = P9(X 生 x) = e9 – 1/w for 0 < x 生 1/θ .
Solution:
Fx(x; θ) = P9(X 生 x) = 0 w du = e9 – 1/t w0(w) = e9 – 1/w
(b) Find T, a minimal sufficient statistic for θ .
Solution: We can write
L(θ; x) = i 1(0 < xi 生 1/θ) = en9 1(x(n) 生 1/θ) = c(x)en9 1(x(n) 生 1/θ)
and
L(θ; x) 1(x(n) 生 1/θ)
L(θ; y) 1(y(n) 生 1/θ) ,
which one can then show to be free of θ if and only if x(n) = y(n) . Thus, T = X(n) is minimal sufficient.
(c) Find the distribution of T, and show that T is complete.
Solution:
FT(t; θ) = FX μ亢) (t; θ) = [FX (t; θ)]n = [e – 1=t]n = en –n=t , t l (0, 1/θ]
thus,
fT (t; θ) = F(t; θ) =
Assuming
1= nen –n=t 1= e –n=t
E [g(T)] = g(t) dt = g(t) dt = 0 Aθ > 0
and applying Leibnitz’s rule, we obtain
… g(1/θ)θ2 e –n = 0 书 g(t) = 0 w.p. 1 Aθ > 0
Thus, T is complete.
(d) Find , the MLE of θ .
L(θ; x) = i 1(0 < xi 生 1/θ) = en 1(x(n) 生 1/θ) = c(x)en 1(θ 生 1/x(n))
an increasing function of θ on the interval (0, 1/x(n)]. Thus, we have = 1/X(n) .
E[] = E[g(T)] = E[1/T] = 1= 1 nen –n=t dt =
(f) Obtain the UMVUE of θ .
Solution: V = … is unbiased and a function of the sufficient and complete statistic, thus it is
Squared Error? [Hint: you do not need to compute the detailed expressions of the MSEs of the two estimators.]
Solution: We have
MSE [ ] = Var [ ] + (Bias [ ]) 2 = Var [ ] + z 、 2 > Var [] = Var [ … 1/n] = Var [V]
Solution: We need to show that n 书(d) θ Aθ > 0 as n 书 k. Since θ is a constant, we will have
Being θ a constant, we have F(u) = 1(u 上 θ). Now, recall that n =T(1) , so that, for u > θ ,
(i) Find the median m = m(θ) of the distribution. Is the MLE for the median consistent for m(θ)?
Solution:
1/2 = Fx(m; θ) = e9 – 1/m 去 m(θ) = 1
for m(θ).
Question 3 (10 points)
Let X be a random variable having density function fx(x; θ) with θ l v1 .
(a) Define
∂
∂θ
Impose suitable assumptions and, under such assumptions, prove that:
(1) E9 [S9(X)] = 0 for any θ l v1 ;
(2) Var9 [S9(X)] = …E9 ┌ log f9 (X)┐ for any θ l v1 .
(b) Carefully define the Fisher Information Ix(θ). Let X = (X1 , . . . , Xn)T be an i.i. d. sample. Suppose
n = 3 and prove from the definition of Ix(θ) that Ix(θ) = 3 尸 Ix(θ).
(c) Consider three statistics T1(X), T2(X) and T3(X) such that T2 = r(T1) and T3 = 5 with probability one, for some function r . Further suppose that T2(X) is sufficient. Without doing calculations, what can be said about the Fisher information of: T1(X), T2(X), T3(X) and (T1(X), T3 (X))? (Motivate your answers)
(d) Considering the same statistics of point (c), prove or disprove the claim I(T1(x),T2 (x))(θ) > Ix(θ).
Solution: Please refer to your notes and to the textbook.
Question 4 (5 points)
State and prove Basu’s theorem. Make sure to include all definitions of statistics that are used in the theorem.
Solution: Please refer to your notes and to the textbook.
2022-08-20