2. Is the following question true or false, or are you uncertain?
According to the neoclassical model of consumer behavior, if the rate of time preference is zero, then consumption should NOT be decreasing over time.
Let:
$latex
F_1 = \frac{C_{t+1}}{C_{t}} = (\frac{1+r}{1+\rho})^{\frac{1}{\theta}}
\\\\
F_2 = \frac{C_{t+1}}{C_{t}} = (\frac{1+r}{1+\rho})^{\frac{1}{\theta}-1}
$
(2014 I.2)
7. Is the following true or false, or are you uncertain? Explain.
According to the neoclassical model of consumer behavior, if the interest rate is zero, then consumption should be decreasing over time.
Let:
$latex
F_1 = \frac{C_{t+1}}{C_{t}} = (\frac{1+r}{1+\rho})^{\frac{1}{\theta}}
\\\\
F_2 = \frac{C_{t+1}}{C_{t}} = (\frac{1+r}{1+\rho})^{\frac{1}{\theta}-1}
$
(2014 I.1)
8. Consider the following Mundell-Fleming model:
Suppose we take the total derivative and transform the system of equations into a matrix problem with exogenous variables on the right and endogenous variables on the left.
The result will take the following form:
$latex
\begin{bmatrix}
F_1 & \frac{dI}{dr} & -X_\rho\\
L_Y & \frac{dX}{d\rho} & 0\\
X_Y & \sigma_r & X_\rho
\end{bmatrix}
\begin{bmatrix}
dY\\
di\\
d\rho
\end{bmatrix}
=
\begin{bmatrix}
F_2\\
d\frac{M}{P} – \frac{dL}{dY}dT\\
F_3
\end{bmatrix}
\\\\
$
Identify F1, F2, and F3.
(Based on 2014 #II.2.a)
Hints:
- Endogenous variables include the interest rate, the real exchange rate, and output.
- Suppose that the country of interest is small, so that Y* is not a function of Y.
- π^e is notation for expected π, not π raised to the power of e.
- Let $latex \frac{dI}{d(i – \pi^e)}$ and $latex \frac{dI}{Y_{-1}} = \frac{dI}{dY}$.
- Let $latex \sigma(i – i*)’ = \frac{d\sigma}{dr}i – \frac{d\sigma}{dr}i^*$
Thanks to Bryan, Ennio, and Josh for doing most of the work for this problem!
Given:
$latex
Y = C(Y-T) + I(i-\pi^e, Y_{-1}) + \bar{G} + X(\rho, Y, Y*)
\\\\
\frac{M}{P} = L(i, Y-T)
\\\\
BOP = X(\rho, Y, Y*) + \sigma(i – i*) + K
$
We must write the total derivative in matrix form. The result will have a column of endogenous variables $latex dY, di,$ and $latex d\rho$.
Solving for $latex dY$:
$latex
Y = C(Y-T) + I(i-\pi^e, Y_{-1}) + \bar{G} + X(\rho, Y, Y*)
\\\\
dY = \frac{dC}{dY}(dY – dT) + \frac{dI}{dr}(di – d\pi^e)
+ \frac{dI}{dY}dY_{-1} + d\bar{G} + \frac{dX}{d\rho}d\rho + \frac{dX}{dY}dY + \frac{dX}{dY^*}dY^*
\\\\
dY – dY\frac{dC}{dY} – dY\frac{dX}{dY} – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho}
= -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY*
\\\\
dY(1 – \frac{dC}{dY} – \frac{dX}{dY}) – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho}
= -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY*
\\\\
$
Solving for $latex dL$:
$latex
\frac{M}{P} = L(i, Y-T)
\\\\
d\frac{M}{P} = \frac{dL}{di}di + \frac{dL}{dY}(dY – dT)
\\\\
d\frac{M}{P} = \frac{dL}{di}di + \frac{dL}{dY}dY – \frac{dL}{dY}dT
\\\\
\frac{dL}{dY}dY + \frac{dL}{di}di = d\frac{M}{P} – \frac{dL}{dY}dT
\\\\
$
Solving for $latex d\rho$ from the BOP function:
$latex
BOP = 0 = X(\rho, Y, Y*) + \sigma(i – i*) + K
\\\\
0 = \frac{dX}{d\rho}d\rho + \frac{dX}{dY}dY + \frac{dX}{dY^*}dY^* + \frac{d\sigma}{dr}i – \frac{d\sigma}{dr}i^* + dK
\\\\
\frac{dX}{dY}dY + \frac{d\sigma}{dr}i + \frac{dX}{d\rho}d\rho
= \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK
\\\\
$
Expressing the system of equations for $latex dY, di,$ and $latex d\rho$ in matrix form:
$latex
dY(1 – \frac{dC}{dY} – \frac{dX}{dY}) – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho}
= -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY*
\\\\
\frac{dL}{dY}dY + \frac{dL}{di}di = d\frac{M}{P} – \frac{dL}{dY}dT
\\\\
\frac{dX}{dY}dY + \frac{d\sigma}{dr}i + \frac{dX}{d\rho}d\rho
= \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK
\\\\
\\\\
\implies
\begin{bmatrix}
F_1 & \frac{dI}{dr} & -X_\rho\\
L_Y & \frac{dX}{d\rho} & 0\\
X_Y & \sigma_r & X_\rho
\end{bmatrix}
\begin{bmatrix}
dY\\
di\\
d\rho
\end{bmatrix}
=
\begin{bmatrix}
F_2\\
d\frac{M}{P} – \frac{dL}{dY}dT\\
F_3
\end{bmatrix}
\\\\
\\\\
F_1 = 1 – \frac{dC}{dY} – \frac{dX}{dY}
\\\\
F_2 = -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY*
\\\\
F_3 = \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK
$