Week II: Solutions
. . .
Exercise 1. Consider the proof of the claim that if� is symmetric and transitive then it is reflexive. pf: let x ∈ X and take some y such that x � y. Then y � x by symmetry and hence x � x by transitivity. Why is this proof wrong?
Solution: there may not exist such a y for all x - it would be true if� was also complete.
Exercise 2. Prove the following statement if true, if not, provide a counterexample: if A is a positive definite matrix, then A−1 is a negative definite matrix.
Solution: The statement is false - consider the identity matrix. Exercise 3. Prove theorem 3.1 by using an Indirect Proof.
Proof. Suppose not. Then there exists at least two limits, say L and L�, L �= L�. Let � = 1
2d(L, L�) > 0. Then
there exists some N1 such that for all n ≥ N1, d(xn, L) < � = 12d(L, L�) And, there exists some N2such that
for all n ≥ N2, d(xn, L�) < � = 12d(L, L�) Let, N = max{N1, N2}, then d(L, L�) ≤ d(xN, L) + d(xN, L�) <
2� = d(L, L�) Contradiction. Thus the limit of a sequence, if it exists, is unique.
Exercise 4. Prove theorem 3.2 by using a Direct Proof.
Proof. Let xn(k) be a subsequence of xk. Let � > 0. There exists some positive integer N such that if
k≥ N, then d(xk, L) < �. If k ≥ N, then n(k) ≥ n(N) ≥ N (since n(K) is an increasing function) Thus,
d(xn(k), L) < �.
Exercise 5. Prove that the real sequence {(−1)k
− 1
n} does not converge
Proof. Suppose the sequence converges. Consider the subsequence indexed by even integers x2k→ 1. Next,
consider the subsequence indexed by odd integers x2k+1 → −1. Contradiction ∴ the sequence cannot not
converge.
Exercise 6. Consider the following equation:
f2(x, y) = f1(x, y)h(y) + f(x, y)h�(y)
Solution:
Use the implicit function theorem (or totally differentiate the above wrt x and y and solve for dx/dy): dx
dy = [f12h + f1h
�+ f
2h�+ fh��− f22] / [f21− f11h− f1h�]
Exercise 7. Find the hessians D2f of each of the following functions. Examine the hessian at the given
point and state if it is positive definite, negative definite, positive semidefinite, negative definite or indefinite. (i) f(x) = −x2 1+ √x2 at x = (1, 1) (ii) f(x) = (x1x2)1/2 at x = (4, 4) (iii) f(x) = x1+ x22 Solution: (i) D2f = � −2 0 0 −1 4x− 3 2 � and D2f (1, 1) = � −2 0 0 −1 4 � the hessian is negative definite
(ii) D2f = 1 2 � −1 2x −32 1 x 1 2 2 12x −12 1 x −12 2 1 2x −12 1 x −12 2 −12x 1 2 1x −32 2 � D2f (4, 4) = � −1 16 1 16 1 16 − 1 16 � eigenvalues are 0 and −1
8
The hessian is negative semidefinite (iii) D2f = � 0 0 0 2 � and D2f (1, 0) = � 0 0 0 2 �
eigenvalues of this matrix are λ = 0 and λ = 2, thus this matrix is positive semidefinite. Exercise 8. Use the method of integration by parts to integrate the following
(i) ´ 3x · sin(x)dx (ii) ´1 0 x 2 · exdx Solution:
(i) Let u = 3x and dv
dx = sinx
Then, du
dx = 3 and v = ´ sinx · dx = −cosx and
ˆ
3x · sin(x)dx = u · v −ˆ dudx· vdx = −3x · cosx + 3 ˆ
cosx· dx = −3x · cosx + 3sinx + c (ii) Let u = x2and dv
dx = e x Then, du dx = 2x and v = ´ e x · dx = ex ∴ by integration by parts ˆ 1 0 x2exdx = u· v − ˆ du dx · vdx = x 2ex��1 0− 2 ˆ 1 0 xexdx = e− 2 ˆ 1 0 xexdx
Exercise 9. Use Leibnez’s rule to differentiate the following (i) V (t) = ´∞
t Ae−r(τ−t)dτ w.r.t t
(ii) V (t) = ´t2
0 x + atdF (x)w.r.t t where F is the cdf of x on [0, ∞) and f(x) is its corresponding pdf.
(iii) G(b) = ´b1(d − 1
2b)dF (d)w.r.t. b where F is the cdf of d on [0,1] and f(d) is its corresponding pdf.
Solution: (i) dV dt = r ´ ∞ t Ae−r(τ−t)dτ− A = rV (t) − A ⇒ rV (t) = A + dV dt (ii) ∂V ∂t = a ´ t2 0 dF (x) + 2t(t 2+ at)f(t) = aF (t2) + 2t(t2+ at)f(t2) (iii) ∂G ∂b =− ´ 1 b 1 2dF (d)− 1 2bf (b) =− 1 2[1 − F (b)] − 1 2bf (b) = 1 2[F (b) − 1 − bf(b)]
Exercise 10. Let (a, b) ∈ R2
+. Define the set X = {(x, y) ∈ R2+ : x > a or if x = a and y > b}. Is this set
open, closed or neither? Explain. Solution:
Neither. Not closed: Consider the point (a, b) and the sequence {(a + 1
n, b)} in X that converges to (a, b).
(a, b) is not in X, thus X is not closed. Not open: consider the point (a, b + 1) in X. Let � > 0. Consider the point (a − �/2, b + 1) which is in B�((a, b + 1)). (a − �/2, b + 1) is not in X. Since this holds for an arbitrary
�, X is not open.
Exercise 11. Let f, g be continous real valued functions on R. Prove that f ◦ g is also a continous function. Proof. (by direct proof) let a ∈ R. Let � > 0. Then there exists some δ1 such that if d(x, g(a)) < δ1 then
d(f (x), f (g(a))) < �. And, there then exists some δ such that if d(x, a) < δ then d(g(x), g(a)) < δ1. Thus, if
d(x, a) < δ then d(f(g(x)), f(g(a))) < �. Thus, f ◦ g is continous at a.
Exercise 12. Let (V,+,·) be a vector space,. Prove that ||x|| = √< x.x > is a norm and d(x, y) = √< x
− y, x − y > is a metric given any inner product < ·, · > on V . Proof. Let x, y, z ∈ V .
||x|| ≥ 0 from property (i). ||ax|| =√< ax.ax > = a||x|| from properties (iv) and (v). And ||x + y||2 = < x + y, x + y > = < x, .x > + < y, y > +2 < x, y > And, (||x|| + ||y||)2 = �√ < x, x > +√< y, y >�2 = < x, x > + < y, y > +2√< x, x >< y, y > By cauchy-schwarzt inequality [< x, y >≤√< x, x >· < y, y >] the result follows.
d(x, z) = ||x − z||
= ||x − y + y − z|| ≤ ||x − y|| + ||y − z||
Exercise 13. Prove every cauchy sequence in R is bounded
Proof. Let {xk} be a cauchy sequence in R. Choose some ε > 0. Then ∃ some N such that |xn− xm| < � ∀
n, m > N�
Let k ≥ N,
|xk| ≤ |xk− xN| + |xN|
≤ ε + |xN|
where the first inequality follows from the triangle inequality (property (iii) of metrics) and the second from the property of Cauchy sequences.
∴ |xk| ≤ max{x1, x2, . . . , xN−1,|xN| + �} ∀ k
Exercise 14. Consider the following system of difference equations kt+1 = (1 − δ)kt+ ktα− ct
ct+1 = [1 − δ + αktα−1]ct
Linearize the system by both the Taylor series method and the log-linearization approach. Solution:
First, we must find the steady state values, k∗ and c∗:
k∗ is defined by: 1 = 1–δ + αkα−1 thus, k∗= (α/δ)1/1−α
c∗ is defined by: c∗= k∗α− δk∗
Linearization by Taylor series:
Log-Linearization:
Similar as above but sub xt= elogxt before taking the Taylor series:
ˆkt+1= [(1 − δ) + αk∗α−1]ˆkt–c
∗
k∗ˆct
ˆct+1= α(α − 1)k∗α−1ˆkt+ ˆct
Exercise 15. Draw a phase plane for the following systems of differential equations. (i) k� = f(k) − c − δk c� = [f�(k) − δ − ρ]c where f�(k) > 0, f��(k) < 0 , δ, ρ > 0 (ii) x� = y − δ y� = ay − f�(x) + (y − δ)2 where f�(k) > 0, f��(k) < 0 , a, δ > 0 Solution:
(i)First, consider the k�=0 locus: c = f(k) – δk.
It is a humpshaped line that starts at the origin [f(0)=0 typically if with think of f as a production function – I didn’t indicate this in the problem] as f(k) is positively sloped with diminishing returns. Above this locus, (increasing c) k is falling, and anywhere below this locus (dec c) k is increasing. Mark these dynamics on the phase plane along with the k�=0 locus.
Now, consider the c�=0 locus: f�(k) = δ + ρ
This system exhibits saddlepath dynamics with the saddlepath having positive slope going from the bottom left quadrant to the upper right.
(ii)The x� = 0 locus defines the steady state value of y=δ, thus, the locus is a horizontal line at δ. Above this locus, x is increasing and below, x is decreasing.
The y� = 0 locus is defined implicitly by ay - f�(x) + (y-δ)2 = 0.
We can find the slope of this line in x-y space by using the implicit function theorem: dy/dx = f��(x)/[2(y − δ) + a]
at steady state y = δ. Thus slope of y� = 0 locus is negative at least around the steady state. To the right of this locus (increasing x) then y is increasing and to the left of this locus, y is decreasing.
This system also exhibits saddlepath dynamics.
And, notice that P∗= X∗, So ˆ Pt= 1 2Xˆt+ 1 2Xˆt−1
Exercise 17. Log-linearize around steady state values ¯Y , ¯L, ¯K, ¯A (i) Yt= AtLαtKt1−α
(ii) Yt= Et−1AtLαtKt1−α
Solution:
(i) Take logs of both sides - let xt= logXt
yt= at+ αlt+ (1 − α)kt
Differentiate w.r.t logged variables - let ˆxt= dlogxt− log¯x, where ¯x is steady state value
ˆyt= ˆat+ αˆlt+ (1 − α)ˆkt
(ii) rewrite variables in logs eyt = E
t−1eateαlte(1−α)kt
Take first order aproximation around logged variables eyˆy
t= Et−1[eaeαle(1−α)kˆat+ αeaeαle(1−α)kˆlt+ (1 − α)eaeαle(1−α)kkˆt]
Note that in the steady state: ey= eaeαle(1−α)k
ˆyt= Et−1[ˆat+ αˆlt+ (1 − α)ˆkt]
Exercise 18. Solve this differential equation (i) x� = xt2
solution: lnx = 1 3t
3+ C ⇒ x = κe13t3
(ii) x��= t
solution: use the change of variable v = x�.Then v� = t ⇒ v = 1 2t 2 + C 1. Then x� = 12t2 + C1 ⇒ x = 1 6t 3+ C 1t + C0 (iii) x��= x�
solution : again use change of variables v = x�. So v�= v ⇒ lnv = t + C ⇒ x� = κet
⇒ x = κet+ C
Exercise 19. Find the solution to the difference equation: xt−52xt−1+ xt−2= 0
Solution: Write with lag operators: xt(1 −52L + L2) = xt(1 − 2L)(1 −12L)
xt= A1(
1
2)t+ A22t
Exercise 20. Consider the following system of difference equations xt= Axt−1
Solution:
(i) Eigenvalues λ = 0, 4, 7 this system is unstable but may have a saddlepath (ii) Eigenvalues λ = 3
4 and 1
