UBC, ECONOMICS 527
2014 MIDTERM EXAMINATION
Suggested solutions
1. Let Y and X be two random variables with second moments.
(a) First, note that
E (Y E(Y |X) = EE ( Y E(Y |X)| X)
= E (E(Y |X) E(Y |X)
= 0.
Hence, we can write
Cov(Y E(Y |X), E(Y |X) = E [(Y
UBC, ECONOMICS 527
Solution to 2014 Final Examination
1. (a) In this model
N (, 25/100) = N (, 0.25).
Hence, under H0 :
(1) H0
N (0, 1).
0.5
One should reject H0 agains the two-sided alternative when |S| > z0.975 = 1.96.
We obtain:
1.8 + 1
= 1.6.
S=
0.
Econ 527
2010 Midterm: Suggested Answer
Q1
(a)
= (W X)1 W Y
= (W X)1 W (X + U )
= + (W X)1 W U
E(|X, W ) = E[ + (W X)1 W U |X, W ]
= + (W X)1 W E[U |X, W ]
Use the condition E[U |X, W ] = 0
=
Use LIE,
E[] = E[E(|X, W )] =
(b)
V ar(|X, W ) = V ar( + (W
UBC, ECONOMICS 527
2013 FINAL EXAMINATION
Suggested solution
1. Using the denition of Fq,m distribution, we can write:
Y /q
Xq,m =
m1
m
2,
j=1 Zj
where Z1 , . . . , Zm are iid N (0, 1), and
Y 2
q
2
and independent from Zs. Since EZj = 1, by WLLN,
m
m1
2
Z
UBC, ECONOMICS 527
2012 FINAL EXAMINATION
Suggested Solution
1. By the denition of a t-distribution, Xm = Z0 /
are iid N (0, 1). As m ,
2
2
(Z1 + . . . + Zm )/m , where Z0 , Z1 , . . . , Zm
1 2
2
2
(Z + . . . + Zm ) p EZ1 = 1,
m 1
and therefore,
Xm p Z0 N
UBC, ECONOMICS 527
2011 FINAL EXAMINATION
Suggested Solution
1. (a) 2 = 1 + . Equality of 1 and 2 can be tested by testing = 0 in Model 2.
(b) This part is the same as Question 2 in Assignment 9.
2. (a) exp n1/2 n c exp n1/2 c = exp n1/2 (n ) c . Function