Chapter 13

Question 1

(i)

Since
	\(TraditionalForm\`s - s\&_ = \(\(-1\)\/\(u\ v\)\) \((p - p\&_)\)\)
then
	\(TraditionalForm
      \`s = \(s\&_ - 
            \(1\/\(u\ v\)\)[
              p\&_ + \((p\_0 - p\&_)\) e\^\(\(-λ\)\ t\) - p\&_]
	 = 
          s\&_ - \(\((p\_0 - p\&_)\)\/\(u\ v\)\) 
              e\^\(\(-λ\)\ t\)\)\)
But
	\(TraditionalForm
      \`s\_0 - s\&_ = \(\(-1\)\/\(u\ v\)\) \((p\_0 - p\&_)\)\)
Therefore
	\(TraditionalForm
      \`s = s\&_ + \((s\_0 - s\&_)\) e\^\(\(-λ\)\ t\)\)

(ii)

Solve[s == 105 - (1/0.1) (105 - 5  (E^(-0.011 t)) - 105), 
      s]

(iii)

pathp1 = 105 - 5 (E^(-0.011 t))

pathp2 = 105 - 5 (E^(-0.02 t))

Plot[{pathp1, pathp2}, {t, 0, 100},  
        AxesLabel -> {"t", "p(t)"}, 
        PlotStyle -> {{}, {Dashing[{ .02}]}}];

paths1 = 105 + 50  (E^(-0.011 t))

paths2 = 105 + 50  (E^(-0.02 t))

Plot[{paths1, paths2}, {t, 0, 100},  
        AxesLabel -> {"t", "s(t)"}, 
        PlotStyle -> {{}, {Dashing[{ .02}]}}, 
        PlotRange -> {100, 160}];

Question 2

GM line is p = s and AM line is p = 110 - 0.1  s.

Solve[{p == s, p == 110 - 0.1  s}, {s, p}]

am = (m - k y + u rstar + u v sbar - u v s)

am0 = 
      am  /. {m -> 105,  k -> 0.5, y -> 20,  u -> 0.5,  rstar -> 10, 
          v -> 0.2, sbar -> 100}

am1 = 
      am  /. {m -> 105,  k -> 0.5, y -> 20,  u -> 0.5,  rstar -> 12, 
          v -> 0.2, sbar -> 100}

Solve[{p == s, p == 111 - 0.1  s}, {s, p}]

The value on the new AM line at the original price of p=100 is

Solve[100 == 111 - 0.1  s, s]

lines2 = 
        Plot[{s, am0, am1, 100}, {s, 90, 115}, 
          AxesLabel -> {"s", "p"}, 
          PlotStyle -> {{}, {}, {Dashing[{ .02}]}, {Dashing[{ .01}]}}];

\(\(points2 = 
        Graphics[
		{PointSize[ .02], Point[{100, 100}], 
            Point[{101, 101}], Point[{110, 100}]
		}];\)\)

Show[points2, lines2, Axes -> True, 
        PlotLabel -> "Rise in foreign interest rate \n"];

System begins with initial equilibrium (s, p) = (100, 100) then moves horizontally across to new AM curve, with value (s, p) = (110, 100). As prices begin to adjust, the system moves up the new AM curve to the new equilibrium at (s, p) = (101, 101).

Question 3

(i)

Since \(TraditionalForm\`p\&_ = m - k\ y - u\ \(r\^*\)\) and
	\(TraditionalForm
      \`p = \(m - k\ y - u(\(r\^*\) + \(\(s\^e\)\& . \))
	 = 
          \(m\  - k\ y - u\ \(r\^*\) - u\ \(v(\(s\&~\) - s)\)
	 = 
            m - k\ y - u\ \(r\^*\) - u\ \(v(s - \(s\&~\))\)\)\)\)
Subtracting \(TraditionalForm\`\(p\&~\)\) from p we obtain
	\(TraditionalForm\`p - \(p\&~\) = u\ v\ \((s - \(s\&~\))\)\)
Or
	\(TraditionalForm
      \`s - \(s\&~\) = \(1\/\(u\ v\)\) \((p - \(p\&~\))\)\)

(ii)

We have
	\(TraditionalForm
      \`\(p\& . \) = 
        \(a(e - y) = 
          \(a[c\ y - d\ r\  + g + h(s - p) - y]
	 = 
            \(a[\(-\((1 - c)\)\) y + g + h\ s - h\ p - d\ r\)\)\)\)

pdot = a (-(1 - c) y + g + h s - h p  - d r)

But
	m = k y  -  u r + p

Simplify[Solve[m == k y  - u r + p, r]]

newpdot = pdot  /. {r -> ((-m + p + k y)/u)}

\(pdot0 = 
      a \((\(-\((1 - c)\)\) y + g + h\ s - h\ p\  - d\ r)\)\  /. {p -> p\&_, 
          s -> s\&_, r -> \(\(-m\) + p\&_ + k\ y\)\/u}\)

finalpdot = Simplify[newpdot - pdot0]

Which can be written
	\(TraditionalForm
      \`a\ \(h(s - s\&_)\) - a\ \(h(p - p\&_)\) - 
        \(\(a\ d\)\/u\) \((p - p\&_)\)\)
But from part (i) we know
	\(TraditionalForm
      \`s - \(s\&~\) = \(1\/\(u\ v\)\) \((p - \(p\&~\))\)\)
Hence,
	\(TraditionalForm
      \`\(p\& . \) = 
        \(\(\(a\ h\)\/\(u\ v\)\) \((p - p\&_)\) - a\ \(h(p - p\&_)\) - 
            \(\(a\ d\)\/u\) \((p - p\&_)\)
	 = 
          \(-\(a(h + h\/\(u\ v\) + d\/u)\)\) \((p - p\&_)\)\)\)

Let λ = a(h + (h/(u v)) + (d/u)), then

\(DSolve[{\(p'\)[t] == \(-λ\) \((p[t] - p\&_)\), p[0] == p0}, 
      p[t], t]\)

\(Expand[
      E\^\(\(-t\)\ λ\)\ \((p0 - p\&_ + 
            E\^\(t\ λ\)\ p\&_)\)]\)

Which can be expressed \(TraditionalForm\`p\&_ + \((p0 - p\&_)\) E\^\(\(-λ\)\ t\)\). So the adjustment coefficient in this model is,
	λ = a(h + (h/(u v)) + (d/u))

Question 4

(i)

We suppress the t subscript throughout. To derive the GM line let (Subscript[p, t + 1]) - (Subscript[p, t]) = 0.

e = 0.01 (s - p) + 0.8 (20) - 0.1  r + 5

Simplify[Solve[0 == 0.2 (e - 20), p]]

We need to eliminate r using equilibrium in the money market.

Simplify[Solve[105 == p + 0.5 (20) - 0.25  r, r]]

Furthermore, in equilibrium r = r* = 10, so we can solve for equilibrium p from

Solve[105 == p + 0.5 (20) - 0.25 (10), p]

Then the GM line is

Simplify[NSolve[p == 100 - 10 (-380 + 4  p) + s, p]]

Given equilibrium p = 97.5, then equilibrium s is given by

Solve[97.5 == 95.122 + 0.0243902  s, s]

Which is, to one decimal place the same value.

We can now derive the equation for the AM line as follows.

Simplify[
      Solve[105 == p + 0.5 (20) - 0.25 (10 + 0.25 (97.5 - s)), 
        p]]

To summarise:
	GM line:	(Subscript[p, t]) = 95.122 + 0.0243902  (Subscript[s, t])
	AM line:	(Subscript[p, t]) = 103.594 - 0.0625  (Subscript[s, t])
To check these equations

Solve[{p == 95.122 + 0.0243902  s, p == 103.594 - 0.0625  s}, {s, 
        p}]

(ii)

We have already established for this model that
	\(TraditionalForm
      \`p\_\(t + 1\) - p\&_ = 
        \(-\(a(h + h\/\(u\ v\) + d\/u)\)\) \((p\_t - p\&_)\)\)\(TraditionalForm\`	p\_\(t + 1\) = p\&_ - λ(p\_t - p\&_)\)
with \(TraditionalForm\`p\&_ = 97.5\).

λ = a (h + (h/(u v)) + (d/u))

valueλ = λ  /. {a -> 0.2, h -> 0.01, u -> 0.25,  
          v -> 0.25, d -> 0.1}

Solve[s0 - 97.5 == -((1/(0.25 0.25))) (100 - 97.5), 
      s0]

pathp = 97.5 + (100 - 97.5) (E^(-0.114 t))

paths = 
      Simplify[97.5 + (57.5 - 97.5) (E^(-0.114 t))]

Table[{t, pathp, paths}, {t, 0, 50}]  // TableForm

Plot[{pathp, paths, 97.5}, {t, 0, 50}, PlotRange -> {50, 105}, 
        PlotStyle -> {{}, {Dashing[{ .02}]}, {Dashing[{ .01}]}}, 
        AxesLabel -> {"t", "p(t),s(t)"}];

(iii)

The expenditure function in terms of s,p and r remains unaffected. However, the value of r in terms of p now differs.

Simplify[Solve[110 == p + 0.5 (20) - 0.25  r, r]]

Equilibrium price is

Solve[110 == p + 0.5 (20) - 0.25 (10), p]

The GM line is then,

Simplify[NSolve[p == 100 - 10 (-400 + 4  p) + s, p]]

Solve[102.5 == 100 + 0.0243902  s, s]

The AM line is then,

Simplify[
      Solve[110 == p + 0.5 (20) - 0.25 (10 + 0.25 (102.5 - s)), 
        p]]

Solve[{p == 100 + 0.0243902  s, p == 108.906 - 0.0625  s}, {s, 
        p}]

Hence,
	(Subscript[dm, t])=(Subscript[ds, t])=(Subscript[dp, t])=5

Question 5

(i)

The goods market is solved as before.

Clear[e]

e = 0.8 (20) + 4 + 0.01 (s - p)

Simplify[Solve[0 == 0.1 (e - 20), p]]

Which is as it should be, a (45^0) line through the origin. The major difference in this model lies in the asset market, the AM line.

In equilibrium s = \(TraditionalForm\`s\&_\), so \(TraditionalForm\`\(s\& . \)\^e = 0\), bp = 0 and r = r*.

Clear[pbar]

Solve[105 == pbar + 0.5 (20) - 0.5 (10), pbar]

Simplify[
      Solve[0 == 0.01 (s - p) + 0.0045 (r - 10 - 0.2 (100 - s)), 
        r]]

Simplify[
      Solve[105 == p + 0.5 (20) - 0.5 (30 + 2.22222  p - 2.42222  s), 
        p]]

To check

Solve[{p == s, p == -990.01 + 10.9001  s}, {s, p}]

(ii) Rise in ms from 105 to 110

The goods line remains as the (45^0) line through the origin. But,

Solve[110 == pbar + 0.5 (20) - 0.5 (10), pbar]

Simplify[
      Solve[0 == 0.01 (s - p) + 0.0045 (r - 10 - 0.2 (105 - s)), 
        r]]

Simplify[
      Solve[110 == p + 0.5 (20) - 0.5 (31 + 2.22222  p - 2.42222  s), 
        p]]

Solve[{p == s, p == -1039.50 + 10.9001  s}, {s, p}]

Point C arises on the new AM line at a price of 100.

Solve[100 == -1039.51 + 10.90001  s, s]

Hence, coordinates of point C are (s, p) = (104.542, 100).

(iii)

In the new equilibrium p = s = 105 and hence dp=ds=5=dm.

Question 6

(i) Model 13.4

solr = 
      Solve[0 == h (s - p) + b (r - rstar - v (sbar - s)), 
        r]

valuer = solr[[1, 1, 2]]

valuep = m - k y  + u valuer

valuepbar = 
      m - ((u (-h pbar - b rstar + h sbar + b sbar v - b sbar v))/b) - k y

Simplify[valuep - valuepbar]

Which can be written,
	\(TraditionalForm
      \`\(-\((u\ v + \(u\ h\)\/b)\)\) \((s - s\&_)\) + 
        \(\(u\ h\)\/b\) \((p - p\&_)\)\)
Hence,
	\(TraditionalForm
      \`\((p - p\&_)\) = 
        \(-\((u\ v + \(u\ h\)\/b)\)\) \((s - s\&_)\) + 
          \(\(u\ h\)\/b\) \((p - p\&_)\)\)\(TraditionalForm
      \`\((1 - \(u\ h\)\/b)\) \((p - p\&_)\) = 
        \(-\((u\ v + \(u\ h\)\/b)\)\) \((s - s\&_)\)\)\(TraditionalForm
      \`\((s - s\&_)\) = 
        \(-\(\((1 - \(u\ h\)\/b)\)\/\((uv + \(u\ h\)\/b)\)\)\) 
          \((p - p\&_)\)\)

Since in equilibrium \(TraditionalForm
      \`r = \(\(r\^*\) + v(s\&_ - s) = \(r\^*\) - v(s - s\&_)\)\), then

pdot = 
      a (c y - d (rstar - v (s - sbar)) + g + h (s - p) - 
            y)

pdot0 = 
      a (g + h (-pbar + sbar) - 
            d (rstar - (sbar - sbar) v) - y + c y)

Simplify[pdot - pdot0]

Which can be written,
	\(TraditionalForm
      \`a(d\ \(v(s - s\&_)\) + h(s - s\&_) - h(p - p\&_))\)
and substituting the value for \(TraditionalForm\`\((s - s\&_)\)\), we get for \(TraditionalForm\`\(p\& . \)\)
	\(TraditionalForm
      \`\(p\& . \) = 
        \(a {\(-\((h + d\ v)\)[\(1 - \((u\ h/b)\)\)\/\(u\ v + \((u\ h/b)\)\)]
                      \) \((p - p\&_)\) - h(p - p\&_)}
	 = 
          \(-a\) {\((h + d\ v)\)[
                  \(1 - \((u\ h/b)\)\)\/\(u\ v + \((u\ h/b)\)\)] + h} 
            \((p - p\&_)\)\)\)
and for λ we have,
	\(TraditionalForm
      \`λ = 
        a {\((h + d\ v)\)[1 - \((u\ h/b)\)]\/\(u\ v + \((u\ h/b)\)\) + h}
            \)

newλ = 
      a ((((h + d v) (1 - (u h/b)))/(u v + (u h/b))) + h)

(ii) Limit for d → ∞

Limit[newλ, d -> 0]

which is the same as
	\(TraditionalForm
      \`a\ h {\(1 - \((u\ h/b)\)\)\/\(u\ v + \((u\ h/b)\)\) + 1}\)

(iii) Limit for b → ∞

Limit[newλ, b -> ∞]

which is the same as
	a {(h/(u v)) + (d/u) + h}

(iv) Limit for d → 0 and b → ∞

Limit[Limit[newλ, d -> 0], b -> ∞]

which is the same as
	a h((1/(u v)) + 1)

Question 7

(i)

(a) Solve for equilibrium

Solve[105 == p + 0.5 (20) - 0.5 (10), p]

(b) GM line

To derive the GM line we first solve for r and then substitute this into the equation for \(TraditionalForm\`\(p\& . \)\) and then let this in turn have the value of 0 and solve for p.

Simplify[Solve[m == p + k y - u r, r]]

Simplify[
      Solve[0 == -(1 - c) y - d (((-m + p + k y)/u)) + g + 
            h (s - p), p]]

Which can be expressed
	\(TraditionalForm
      \`p = \(\(-\([\((1 - c)\) + \((d\ k/u)\)]\)\) y\)\/\(h + \((d/u)\)\) + 
          \(g + \((d/u)\) m\)\/\(h + \((d/u)\)\) + 
          \((h\/\(h + \((d/u)\)\))\) s\)

intGM4 = ((-((1 - c) + (d k/u)) y + g + (d/u) m)/(h + (d/u))) /. {c -> 0.8, d -> 0.1, k -> 0.5, u -> 0.5, y -> 20, 
          g -> 5, m -> 105, h -> 0.01}

slopeGM4 = 
      (h/(h + (d/u))) /. {h -> 0.01, d -> 0.1, u -> 0.5}

Hence, the equation for the GM line is
	p = 95.2381 + 0.047619  s

(c) AM line

Simplify[
      Solve[0 == h (s - p) + b (r - rstar - v (sbar - s)), 
        r]]

Simplify[
      Solve[m == p + k y - 
            u (rstar + ((h (p - s))/b) + (-s + sbar) v), 
        p]]

Which can be expressed
	\(TraditionalForm
      \`p = \(m - k\ y + u\ \(r\^*\)\)\/\(1 - \((u\ h/b)\)\) + 
          \(u\ v\ s\&_\)\/\(1 - \((u\ h/b)\)\) - 
          \((u\ v + \((u\ h/b)\))\)\/\(1 - \((u\ h/b)\)\)\)

intAM4 = ((m - k y + u rstar + u v sbar)/(1 - (u h/b)))  /. {m -> 105, k -> 0.5, y -> 20, rstar -> 10, u -> 0.5,  v -> 0.2, sbar -> 100, 
          b -> 0.004, h -> 0.01}

slopeAM4 = ((-(u v + (u h/b)))/(1 - (u h/b))) /. {u -> 
            0.5, v -> 0.2, h -> 0.01, b -> 0.004}

Hence, equation for the LM curve is
	p = -440 + 5.4  s

We can check the equilibrium results by solving

Solve[{p == 95.2381 + 0.047619  s, p == -440 + 5.4  s}, {s, 
        p}]

(ii) Rise in money supply from 105 to 110

Solve[110 == p + 0.5 (20) - 0.5 (10), p]

Only the intercept on the GM line and the AM lines are affected.

intGM42 = ((-((1 - c) + (d k/u)) y + g + (d/u) m)/(h + (d/u))) /. {c -> 0.8, d -> 0.1, k -> 0.5, u -> 0.5, y -> 20, 
          g -> 5, m -> 110, h -> 0.01}

intAM42 = ((m - k y + u rstar + u v sbar)/(1 - (u h/b)))  /. {m -> 110, k -> 0.5, y -> 20, rstar -> 10, u -> 0.5,  v -> 0.2, sbar -> 105, 
          b -> 0.004, h -> 0.01}

Hence, the two new equations are
	p = 100 + 0.047619  sp = -462 + 5.4  s

The new equilibrium is

Solve[{p == 100 + 0.047619  s, p == -462 + 5.4  s}, {s, 
        p}]

Point C on the new AM line is at a price of p = 100.

Solve[100 == -462 + 5.4  s, s]

Hence, the coordinates of point C are (s, p) = (104.074, 100).
	Also, we have that dp = ds = 5 = dm.

lines7 = 
        Plot[{s, 95.2381 + 0.047619  s, -440 + 5.4  s, 
            100 + 0.047619  s, -462 + 5.4  s}, {s, 95, 110}, 
          PlotStyle -> {{Dashing[{ .01}]}, {}, {}, {Dashing[{ .02}]}, {Dashing[{ .02}]}}];

points7 = 
        Graphics[{PointSize[ .02], Point[{100, 100}], Point[{105, 105}], 
            Point[{104.074, 100}] }];

Show[points7, lines7,  Axes -> True, 
        AxesLabel -> {"s", "p"}, 
        PlotLabel -> "Model 11.4, Rise in money supply\n"];

Question 8

(i) Equilibrium

Clear[pdot, sdot]

pbar = 105 - 0.5 (20) + 0.5 (18)

Hence, \(TraditionalForm\`p\&_ = \(s\&_ = 104\)\)

(ii) Characteristic roots

Simplify[
      pdot = 0.01 (0.1 (s - p) + 0.8 (20) + 4 - y)]

\(newpdot = \(-0.001\) \((p - p\&_)\) + 0.001 \((s - s\&_)\)\)

\(sdot = \((1/0.5)\) \((s - s\&_)\)\)

matrixA=(({{-0.001, 0.001}, {2, 0}}))

Eigenvalues[matrixA]

(iii) Saddle-path equations

(a)

Eigenvectors[matrixA]

matrixA - 0.0442242 IdentityMatrix[2]

Simplify[
      Solve[-0.0452242 (p - 104) + 0.001 (s - 104) == 0, 
        p]]

(b)

matrixA + 0.0452242 IdentityMatrix[2]

Simplify[
      Solve[0.0442242 (p - 104) + 0.001 (s - 104) == 0, 
        p]]

(iv) Rise in the money supply to 110

The change in the money supply has no affect on the dynamics of the system, i.e., the matrix of the system remains unaffected. However, the equilibrium value changes to

newpbar = 110 - 0.5 (20) + 0.5 (18)

Hence, \(TraditionalForm\`p\&_ = \(s\&_ = 109\)\)

Simplify[
      Solve[-0.0452242 (p - 109) + 0.001 (s - 109) == 0, 
        p]]

Simplify[
      Solve[0.0442242 (p - 109) + 0.001 (s - 109) == 0, 
        p]]

Question 9

(i) Initial situation

Clear[pdot, sdot, pbar]

pbar = 105 - 0.5 (20) + 0.5 (10)

Hence, \(TraditionalForm\`p\&_ = \(s\&_ = 100. \)\)

(ii) Saddlepaths

\(pdot = \(-0.1\) \((0.01)\) \((p - p\&_)\) + 
        0.1 \((0.01)\) \((s - s\&_)\)\)

\(sdot = \((0.8/0.5)\) \((p - p\&_)\) + \((0.2/0.5)\) \((s - 
              s\&_)\)\)

matrixA7=(({{-0.001, 0.001}, {1.6, 0.4}}))

Eigenvalues[matrixA7]

Eigenvectors[matrixA7]

matrixA7 - 0.403951 IdentityMatrix[2]

Simplify[
      Solve[-0.404951 (p - 100) + 0.001 (s - 100) == 0, 
        p]]

matrixA7 + 0.00495109 IdentityMatrix[2]

Simplify[
      Solve[0.00395109 (p - 100) + 0.001 (s - 100) == 0, 
        p]]

The saddlepaths are, then,
	p = 99.7531 + 0.00246943  sp = 125.309 - 0.253095

(iii) Resource discovery

Solve[{p == s + ((4 - (1 - 0.8) (20) + 2 (0.5))/0.01), 
        p == (((105 - 0.5 (20) - 0.3 + 0.5 (10))/0.8)) - ((0.2/0.8)) s}, {s, p}]

The matrix of the system remains unaffected, and so we have the same eigenvalues.

matrixA7 - 0.403951 IdentityMatrix[2]

Simplify[
      Solve[-0.404951 (p - 119.7) + 0.001 (s - 19.7) == 0, 
        p]]

matrixA7 + 0.00495109 IdentityMatrix[2]

Simplify[
      Solve[0.00395109 (p - 119.7) + 0.001 (s - 19.7) == 0, 
        p]]

Hence, the saddle path equations are
	p = 119.651 + 0.00246943  sp = 124.686 - 0.253095  s

Question 11

(i)

Clear[s, sdot]

Simplify[
      Solve[106 - s - 1 == 0.5 (20) - 0.5 (10 + sdot), 
        sdot]]

Hence the differential equation is
	ds/dt = -200 + 2  s

(ii)

Simplify[Solve[0 == -200 + 2  s, s]]

Question 12

Clear[p, s, sdot, pstar, rstar, λ]

m - s - pstar - k y + u (rstar + sdot)

m - s - pstar - k y + u (rstar + sdot)  /.  
      sdot -> λ - pstar

Hence,
	sbar = 
        m - pstar - k y + u(rstar + λ - pstar)