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ECON6003/6703 Mathematical Methods for Economics Problem Set 4
发布时间:2022-05-20
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Problem Set 4
ECON6003/6703 Mathematical Methods for Economics
Question 1. (prob. 11.2 and 11.3 pg. 243 from SB Text) In each case, verify whether the set of vectors is linearly independent.
1. 1(2)
and
2(1)
2. 1(2)
and
¡(¡)2(4)
3. 0 1(1) 1
and0
1(0) 1
@ 0 A @ 1 A
4. 0 1(1) 1
0
0(1) 1
and 0
1(0) 1
@ 0 A; @ 1 A @ 1 A
0 1 1 0 1 1 0 0 1
5. B 1(0) C; B 0(0) Cand B 1(0) C
@ A @ A @ A
0 1 1 0 1 1 0 1 1
6. B 1(0) C; B 0¡1 Cand B 0(0) C
@ 0 A @ 1 A @ 0 A
Answer. All answers rely on the various propositions presented in the Lecture class slides. Make sure you understand which one is being used I am not explicitly quoting the propositions
Part 1, Part 2, Part 3, the number of vectors equals the number dimension of the vec- tors, i.e. m = k in terms of the notation in class. We can therefore sort out the linear dependence issue by simply calculating the determinant of the matrix whose columns are these corresponding vectors.
In Part 1 and Part 2, calculate the determinant of the 2x2 matrix formed by the two colum vectors. In Q1, det 1(2) 2(1)
= 4 ¡ 1
0 and in Q2, det
= 0, so the vectors in the former are linearly independent, in the latter they are linearly dependent.
Likewise, in Part (4), we have
A = 0 1
@ 0 1 1 A
and det(A) = ¡2. Hence this is a linearly independent set of vectors.
Part 3, when we take a linear combination of the two vectors,
c1 @(0) A(1)+ c2@(0)
A(1) = @(0) c2
c2 A(1):
The linear combination equals the zero vector, i.e. satisfies the equation
c1@(0) A(1)+ c2@(0)
A(1) = @(0)
A(1)
OR 0 c2
c2 1 00(0) 1
if and only if c1= c2= 0.
Part 5, Part 6, I will leave it to you, follow the procedure above.
Note 1. From here onward, whenever you see a vector expressed as a row, x = (x1 ; : ::; 0 x1 1
xi; :::; xn), treat it as the colum vector: x = B
B
@ x |
Question 2. Do the vectors (1; 2; 3), (4; 5; 12) and (0; 8; 0) span R3? Explain. Hint: (Think in terms the determinants!).
Answer. Again, look at the determinant formed by the three vectors as columns,
A = 0 1
@ 3 12 0 A
Recall the procedure described in class for calculating the determinant, about picking aand working through the rows: In class, when illustrating the procedure, I chose the first column. However in the above, you will find it more conveninent to pick the third column given the zeros. Then the
det(A) = ¡8det
3(1) 12(4)
= ¡8 (12¡12) = 0:
So theese vectors are not linearly independent. Since we need at least three linearly independent vectors to span R3, it then follows these vectors do not span R3 .
Question 3. Do the problems 11.12, 11.13 and 11.14 from the SB Text, (page 249).
Answer. The screenshot of solutions from the publisher:
Question 4. You may want to read up Example 3, pg. 113 of SB Text, but equally, you may wish to give it a try first anyway.
At each date, a typical individual is either employed or not. We may think of these as two states fe ; ug. Let pij denote the probability that someone that is in state i today moves into state j tomorrow. That is, peu for instance is the probability that an employed will lose his job pee is the probability that he keeps his job.
1) Express these probabilities as a 2 2 matrix.
I Solution.
P = pee peu !
pue puu
2) Suppose that a fraction xt of the working population at date t is employed and the remaining fraction yt = 1 ¡ xt is unemployed at date t. Using the above matrix and the vector (xt; yt), write down the relation between (xt+1 ; yt+1).
I Solution. The average fraction of those who are employed tomorrow consist of
those who are employed today and will continue to be employed tomorrow, which is peext and those who are unemployed today and become employed tomorrow, which are pueyt . So
xt+1 = peext + pueyt
Similarly,
yt+1 = peuxt + puuyt:
The above two equations can be written in a matrix form as
( xt+1 yt+1 ) = ( xt yt ) pee peu !
pue puu
or, you can rewrite this in the usual
=
!
y(x)t(t)
(1)
by swapping the off-diagonal terms in the matrix P.
3) Suppose that, in the long run, (xt; yt) converges to some steady-state ratio (x ; y). Calculate (x ; y).
I Solution Since we are dealing with probabilities, we must have pee+ peu = 1 and
pue + puu = 1. Set pee = p and puu = q , and taking limits on both sides of (1), we have
Since x and y are the fractions of employed and unemployed, we also have x + y = 1:
The above equation, together with the two equations determined by (2) deter- mine the entire system. A detailed explanation of these is given Example 3, pg.
113 of SB Text.