Question 2.1

a) Minimize costs: \(wAL + rK\)
Subject to the production function: \(Y = ALf(k)\)
Setting up the Lagrangian gives: \(\mathcal{L} = wAL + rK + \lambda (Y - ALf(k)) = 0\)
\(\frac{\partial \mathcal{L}}{\partial K} = r - \lambda [ALf'(\frac{K}{AL}) \frac{1}{AL}] = 0\)
This leads to: \(r = \lambda f'(k)\)
\(\frac{\partial \mathcal{L}}{\partial AL} = w - \lambda [f(\frac{K}{AL} + ALf'(\frac{K}{AL})\frac{-K}{AL^2}] = 0\)
This gives us: \(w = \lambda [f(k) = kf'(k)]\)
Since the cost-minimizing condition requires that the ratio of the marginal cost of K and AL are equal to their marginal products, we can say that: 
\(\frac{r}{w} = \frac{f'(k)}{f(k) - kf'(k)}\)
Thus, we see that this choice does not depend on the level of output, Y. 
b) Since we know that each firm chooses the same value of k and has the same value of A, the total amount produced by N cost-minimizing firms will be:
\(\sum_{i = 1}^N Y_i = \sum^N _{i = 1} AL_i f(k) = Af(k) \sum ^N _{i = 1} L_i = A \bar L f(k)\)
Here, \(\bar L\) signifies the total amount of labor employed. Therefore, we can see that if using all the labor employed by all the firms, the single firm would produce at the same level: \(Y = A \bar L f(k)\)

Question 2.3

a) Using the equation given: \(U = \frac{C_1^{1 - \theta}}{1 - \theta} + \frac{1}{1 + \rho} \frac{C_2 ^{1 - \theta}}{1 - \theta}\)
We make this subject to the budget constraint: \(P_1C_1 + P_2C_2 = W\).
To rearrange and solve for the second period consumption: \(C_2 = \frac{W - P_1C_1}{P_2}\)
Now substituting this into the utility function: \(U = \frac{C_1^{1 - \theta}}{1 - \theta} + \frac{1}{1 + \rho} \frac{\frac{W - P_1C_1^{1 - \theta}}{P_2}}{1 - \theta}\)
Taking the first order condition with respect to the first period consumption to solve an unconstrained problem of maximizing utility: \(\frac{\partial U}{\partial C_1} = C_1^{- \theta} + \frac{1}{1 + \rho} {C_2}^{- \theta} (\frac{P_1}{P_2}) = 0\)
Solving for the first period consumption yields:
\(C_1=(1+\rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1}{\theta}}C_2\)
Then substituting this into the budget constraint to solve for second period consumption: \(C_2=\frac{W-P_1}{P_2}[(1+\rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1}{\theta}}C_2]\)
This will simplify down to:  \(C_2 = \frac{\frac{W}{P_2}}{[1 + (1 + \rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1 - \theta}{\theta}}]}\)
To find the optimal choice of \(C_1\) we can substitute this into the equation: 
\(C_1 =(1+\rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1}{\theta}}\frac{\frac{W}{P_2}}{[1 + (1 + \rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1 - \theta}{\theta}}]}\)
b) The optimal ratio of first period to second period consumption equals: \(\frac{C_1}{C_2} = (1+\rho)^{\frac{1}{\theta}}(\frac{P_2}{P_1})^{\frac{1}{\theta}}\)
Taking logs gives us: \(\ln \frac{C_1}{C_2} = (\frac{1}{\theta}) \ln (1+\rho) + (\frac{1}{\theta}) \ln(\frac{P_2}{P_1})\)
Since the elasticity of substitution between \(C_1, C_2\) depends on price, it will be:
\(\frac{\partial \frac{C_1}{C_2}}{\partial \frac{P_2}{P_1}} = \frac{\frac{P_2}{P_1}}{\frac{C_1}{C_2}} = \frac{1}{\theta}\)

Question 2.4

The equation given in 2.1 for household lifetime utility is:
\(U = \int ^\infty _{t = 0} e^{-\rho t} \ln c \frac{L(t)}{H}dt\)
This is subject to the constraint:
\(\int ^\infty _{t = 0} e^{-R(t)} \ln c \frac{L(t)}{H} dt \leq \frac{K(0)}{H} + \int ^\infty _{t = 0} e^{-R(t)} W(t) \frac{L(t)}{H} dt \)
And let the function of initial wealth be: 
\(W \equiv \frac{K(0)}{H} + \int ^\infty _{t = 0} e^{-R(t)} A(t)w(t) \frac{L(t)}{H} dt \)
Using the Lagrangian to maximize: 
\(\mathcal{L} = \int ^\infty _{t = 0} e^{- \rho t} \ln c \frac{L(t)}{H} dt + \lambda [W - \int ^\infty _{t = 0} e^{- \rho t} \ln c \frac{L(t)}{H} dt]\)
\(\frac{\partial \mathcal{L}} {\partial C(t)} = e ^{-\rho t} c^{-1} \frac{L(t)}{H} - \lambda e^{R(t)} \frac{L(t)}{H} = 0 \)
Rearranging and simplifying gives: 
\(e ^{-\rho t} c^{-1} = \lambda e^{R(t)} \Longrightarrow c(t) = e^{-\rho t} \lambda^{-1} e^{R(t)}\)
Substituting this into the budget constraint:
\(\int ^\infty _{t = 0} e^{-R(t)} [e^{-\rho t} \lambda^{-1} e^{R(t)}] \frac{L(t)}{H} dt = W\)
Since \(L(t) = e^{nt} L(0)\) we have: 
\(\lambda ^{-1} \frac{L(0)}{H} \int ^\infty _{t = 0} e^{-(\rho - n)t} dt = W\)
Since \(\rho - n > 0\) its integral is equal to \(\frac{1}{\rho - n}\):
\(\lambda ^{-1} = \frac{W}{\frac{L(0)}{H}} (\rho - n)\)
Finally, substituting this into the equation for consumption we have: 
\(c(t) = e^{-\rho t} [\frac{W}{\frac{L(0)}{H}} (\rho - n)] e^{R(t)}\)
Initial consumption is thus: 
\(C(0) = \frac{W}{\frac{L(0)}{H}} (\rho - n)\)
This equation states that the initial consumption per person is a constant fraction of initial wealth per person. \(\rho - n\) is the marginal propensity to consume from that wealth, thus independent of the real interest rate. 

Question 2.6

a) The dynamic equation for capital stock per unit of effective labor is
\(\dot k (t) = f(k(t)) - c(t) - (n + g)k(t)\)
When \(\dot k = 0\), we can see that \(c = f(k) - (n _ g)k \). Thus, a fall in g (growth rate of A) would impact the curve by shifting it up. This is due to the fact that consumption would be allowed to increase if the original level of k was to be maintained. Since the term is multiplied, the higher the level of k, then the more the line would shift in response. 
b) The equation for dynamics of consumption per unit of effective labor is
\(\frac{\dot c(t)}{c(t)} = \frac{f'(k(t)) - \rho - \theta g}{\theta}\)
When \(\dot c = 0 \),  we can see that \(f'(k(t)) = \rho + \theta g\). Thus, a fall in g would impact the curve by shifting it right. This is because the a lower g corresponds to a higher amount of k needed for consumption to equal zero. Because \(f''(k) < 0\) then the amount of capital needed to bring  \(f'(k(t)) \) down is greater. 
c) At the time of the change, consumption adjusts immediately to the new balanced growth path. Though we cannot tell whether the new saddle path passes above or below the original k*, c* level, we do know that c will gradually rise to the new k* and c*. See below:
d) On the balanced growth path, the fraction saved is: \(\frac{f(k*) - c*}{c(k*)}\). Since k is constant on the balanced growth path, \(f(k*) - c* = (n + g)k*\), we can rewrite the fraction of output saved as: \(s = \frac{[(n + g)k*]}{f(k*)}\)
Differentiating with respect to g yields: \(\frac{\partial s}{\partial g} = \frac{(n + g)(\frac{\partial k*}{\partial g} + k*] - (n + g)k* f'(k*) \frac{\partial k*}{\partial g}}{[f(k*)]^2}\)
This simplifies to \(\frac{\partial s}{\partial g} = \frac{(n + g)[f(k*) - k*f'(k*)] (\frac{\partial k*}{\partial g} + f(k*) k*} {[f'(k*)]^2}\)
Since \(k* \rightarrow f'(k*) = \rho + \theta g\), differentiating both sides gives \(f''(k*)(\frac{\partial k*}{\partial g}) = \theta\). Solving this gives: \(\frac{\theta}{f''(k*)} < 0\). Now we have:
\(\frac{\partial s}{\partial g} =\frac{(n + g)[f(k*) - k*f'(k*)] \theta + f(k*) k* f''(k*)} {[f'(k*)]^2}\). Since the first part in the numerator is positive and the second is negative, we cannot tell whether the fall in g will raise or lower the saving rate on the new balanced growth path.
e) Using the Cobb-Douglas production function: \(f(k) = k^\alpha ; f'(k) = \alpha k^{\alpha - 1} ; f''(k) = \alpha (\alpha - 1) k^{\alpha - 2}\)
We can substitute this into the equation for d to yield: 
\(\frac{\partial s}{\partial g} = \frac{(n + g) [k* ^\alpha -k* \alpha k*^{\alpha - 1}] \theta + k* ^\alpha k* \alpha(\alpha -1)k*^{\alpha -2}}{k*^\alpha k*^\alpha \alpha (\alpha -1) k*^{\alpha -2}}\)
This simplifies to:
 \(\frac{\partial s}{\partial g} = -\alpha \frac{[(n + g) \theta - (\rho + \theta g)]}{(\rho + \theta g)^2}\)
Finally yielding: 
\(\frac{\partial s}{\partial g} = -\alpha \frac{(n \theta - \rho)}{(\rho + \theta g)^2} \Longrightarrow \alpha \frac{(\rho - n \theta)}{(\rho + \theta g)^2}\)

Question 2.7

a) Using the equations of motion, we can show:
\(\frac{\dot c(t)}{c(t)} = \frac{f'(k(t)) - \rho - \theta g}{\theta}\)
\(\dot k (t) = f(k(t)) - c(t) - (n + g)k(t)\)
A rise in \(\theta\), meaning a falling elasticity of substitution, reveals that households are less willing to substitute consumption between periods. Thus, the marginal utility of consumption falls down rapidly as consumption rises. The capital-accumulation equation is unaffected. 
b) A downward shift of the production function means that for any given k, both f(k) and f'(k) are lower than before. The condition required for \(\dot k = 0\) is given by \(c = f(k) - (n + g)k\). From the graphs, we can see that \(\dot k = 0\) will shift down more at higher levels of k. Whether c initially jumps up or down depends on whether the new saddle path passes above or below the k*, c*.  
c) Given positive depreciation, the new capital-accumulation is \(\dot k(t) = f(k(t)) - c(t) - (n + g + \delta)k(t)\). The extra amount of saving and investment needed to keep the k constant is now higher, and now consumption is lower. Additionally, the real return on capital is now \(f'(k(t)) - \delta\), meaning the household's maximization yields:
\(\frac{\dot c(t)}{c(t)} = \frac{f'(k(t)) - \rho - \theta g}{\theta}\)
The condition required for \(\dot c = 0 \) is now \(f'(k) = \delta + \rho + \theta g\). Thus the \(\dot c = 0\) locus shifts to the left. Again, the initial jump of c depends on the saddle path.

Question 2.11

a, b, and c) Before a tax, the dynamics are