fix_way1

Author: shlff

lectures/olg.md

@@ -11,7 +11,6 @@ kernelspec:
   name: python3
 ---
 
-
 # The Overlapping Generations Model
 
 In this lecture we study the overlapping generations (OLG) model.
@@ -57,7 +56,6 @@ import matplotlib.pyplot as plt
 plt.rcParams["figure.figsize"] = (11, 5)  #set default figure size
 ```
 
-
 ## Environment
 
 TODO add timing and basic ideas of OLG
@@ -359,7 +357,6 @@ R_star, K_star = E_star
 plot_ad_as(aggregate_capital_demand, aggregate_capital_supply, m, K_prev=50, E_star=E_star)
 ```
 
-
 Let's observe the dynamics of the equilibrium price $R^*_{t+1}$.
 
 ```{code-cell} ipython3
@@ -375,7 +372,6 @@ ax.set_xlabel("$K_{t}$")
 plt.show()
 ```
 
-
 ## Dynamics and steady state
 
 Let $k_t := K_t / L$.
@@ -476,7 +472,6 @@ def k_star(model):
 plot_45(m, k_update, kstar=k_star(m))
 ```
 
-
 ## Another special case: CRRA preference
 
 
@@ -563,7 +558,6 @@ def aggregate_supply_capital_crra(R, model, K_prev):
 plot_ad_as(aggregate_capital_demand, aggregate_supply_capital_crra, m_crra, K_prev=50, E_star=None)  # John this is to be fixed.
 ```
 
-
 Let's plot the aggregate supply with different values of utility parameter $\gamma$ and observe it's behaviour.
 
 ```{code-cell} ipython3
@@ -585,7 +579,6 @@ ax.legend()
 plt.show()
 ```
 
-
 When $\gamma <1$ the supply curve is downward sloping. When $\gamma > 1$ the supply curve is upward sloping.
 
 TODO: Do we need to add some explanation?
@@ -649,7 +642,6 @@ def f(k_prime, k, model):
     return p - z
 ```
 
-
 Let's define a function `k_next` that finds the value of $k_{t+1}$.
 
 ```{code-cell} ipython3
@@ -661,7 +653,6 @@ def k_next(k_prime, model):
 plot_45(m_crra, k_next, kstar=None)
 ```
 
-
 Unlike the log preference case now a steady state cannot be solved analytically.
 
 To see this recall that, a steady state can be obtained by setting [](law_of_motion_capital_crra) to $k_{t+1} = k_t = k^*$, i.e.,
@@ -699,7 +690,6 @@ print(f"k_star = {k_star}")
 plot_45(m_crra, k_next, k_star)
 ```
 
-
 The next figure shows three time paths for capital, from
 three distinct initial conditions, under the parameterization listed above.
 
@@ -741,7 +731,6 @@ def simulate_ts(m, x0_values, ts_length):
 simulate_ts(m_crra, x0, ts_length)
 ```
 
-
 ## Exercises
 
 
@@ -783,7 +772,6 @@ def find_Kstar(R_star, model):
     return model.L * (R_star / model.α)**(1/(model.α-1))
 ```
 
-
 The following function plots the equilibrium quantity and equilibrium price.
 
 ```{code-cell} ipython3
@@ -810,7 +798,9 @@ def plot_ks_rs(K_t_vals, model):
 ---
 mystnb:
   figure:
-    caption: "Equilibrium price and quantity\n"
+    caption: 'Equilibrium price and quantity
+
+      '
     name: equi_ps_q_crra
   image:
     alt: equi_ps_q_crra
@@ -822,14 +812,13 @@ m_crra = create_olg_model(u=crra, u_params={'γ': 0.5})
 plot_ks_rs(K_t_vals, m_crra)
 ```
 
-
 ```{solution-end}
 ```
 
 ```{exercise}
 :label: olg_ex2
 
-Let's keep the model the same except for replacing the utility function $u(c)$ in {eq}`eq_crra`  with a quasilinear form $u(c)=c + c^{\theta}$.
+Let's keep the model the same except for replacing the utility function $u(c)$ in {eq}`eq_crra`  with a nonlinear form $u(c)=c + c^{\theta}$.
 
 Like what we did in the CRRA case we don't have an analytical solution.
 
@@ -842,7 +831,7 @@ Try to compute the time path capital $\{k_t\}$ in this case.
 
 To get the time path capital $\{k_t\}$ first we need to solve the household's utility maximization problem for the optimal consumption and optimal saving.
 
-With the quasilinear preference the Euler equation becomes
+With the quasilinear preference the Euler equation [](euler_2_olg) becomes
 
 ```{math}
 :label: euler_quasilinear
@@ -855,7 +844,7 @@ Since [](aggregate_supply_capital_log_olg), [](wage_2) and [](interest_rate_2) t
 
 ```{math}
 :label: euler_quasilinear1
-    1 + \theta ((1-\alpha)k^{\alpha}_t - k_{t+1})^{\theta-1} = \beta \alpha k^{\alpha - 1}_t + \beta (\alpha k^{\alpha - 1}_t)^{\theta} \theta k_{t+1}^{\theta - 1}
+    1 + \theta ((1-\alpha)k^{\alpha}_t - k_{t+1})^{\theta-1} = \beta \alpha k^{\alpha - 1}_{t+1} + \beta \theta \alpha^{\theta} k^{\alpha \theta - 1}_{t+1}
 ```
 
 Obviously $k_{t+1}$ cannot be solved by pencil and paper.
@@ -865,11 +854,10 @@ To solve for $k_{t+1}$ we need to turn to the newton's method.
 Let's start by defining the utility function.
 
 ```{code-cell} ipython3
-def u_quasilinear(c, θ=4):
+def u_quasilinear(c, θ=6):
     return c + c**θ
 ```
 
-
 The function `find_k_next` is used to find $k_{t+1}$ by finding
 the root of equation [](euler_quasilinear1) using the helper
 function `solve_for_k_next` for a given value of $k_t$.
@@ -878,8 +866,8 @@ function `solve_for_k_next` for a given value of $k_t$.
 def solve_for_k_next(x, k_t, model):
     α, β, L, θ = model.α, model.β, model.L, model.u_params['θ']
     l = 1 + θ * ((1 - α) * k_t**α - x)**(θ - 1)
-    r = β * α * k_t**(α - 1)
-    r += β * (α * k_t**(α - 1))**θ * θ * x**(θ - 1)
+    r = β * α * x**(α - 1)
+    r += β * θ * α**θ * x**(α * θ - 1)
     return l - r
 ```
 
@@ -893,15 +881,13 @@ def solve_for_k_star_q(x, model):
     α, β, L, θ = model.α, model.β, model.L, model.u_params['θ']
     l = 1 + θ * ((1 - α) * x**α - x)**(θ - 1)
     r = β * α * x**(α - 1)
-    r += β * (α * x**(α - 1))**θ * θ * x**(θ - 1)
+    r += β * θ * α**θ * x**(α * θ - 1)
     return l - r
 
 def find_k_star_q(model):
     return optimize.newton(solve_for_k_star_q, 0.3, args=(model,))
-
 ```
 
-
 Let's simulate and plot the time path capital $\{k_t\}$.
 
 ```{code-cell} ipython3
@@ -932,9 +918,14 @@ def simulate_ts(k0_values, model, ts_length=10):
 
 ```{code-cell} ipython3
 k0_values = [0.2, 10, 50, 100]
-m_quasilinear = create_olg_model(u=u_quasilinear, u_params={'θ': 4})
+m_quasilinear = create_olg_model(u=u_quasilinear, u_params={'θ': 6})
 simulate_ts(k0_values, m_quasilinear)
 ```
 
-```{solution-end}
+```{code-cell} ipython3
+
+```
+
+```{code-cell} ipython3
+
 ```