Tom's edits and renaming of two prospective lectures on present values

Author: thomassargent30

sandpit/cons_smooth.md renamed to sandpit/cons_smooth_tom_v2.md

@@ -11,13 +11,20 @@ kernelspec:
   name: python3
 ---
 
++++ {"user_expressions": []}
+
 ## Some dynamic models with matrices
 
 In this notebook, we'll present  some useful models of economic dynamics using only linear algebra -- matrix multiplication and matrix inversion.
 
 **Present value formulas** are at the core of the models.
 
-+++
+```{code-cell} ipython3
+import numpy as np
+import matplotlib.pyplot as plt
+```
+
++++ {"user_expressions": []}
 
 ## Consumption smoothing
 
@@ -71,7 +78,7 @@ where $g_1 > 0, g_2 > 0$.
 
 We shall see that when $\beta R = 1$ (a condition assumed by Milton Friedman and Robert Hall), this criterion assigns higher welfare to **smoother** consumption paths.
 
-+++
++++ {"user_expressions": []}
 
 ## Difference equations with linear algebra ##
 
@@ -94,14 +101,20 @@ where  $y_0$ is a given  initial condition.
 We can cast this set of $T$ equations as a single  matrix equation
 
 $$
-\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 \cr
-              -\lambda & 1 & 0 & \cdots & 0 & 0 \cr
-                0 & -\lambda & 1 & \cdots & 0 & 0 \cr
-                 \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
-                0 & 0 & 0 & \cdots & -\lambda & 1 
-
-\end{bmatrix} \begin{bmatrix} y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T \end{bmatrix}
-= \begin{bmatrix} \lambda y_0 \cr 0 \cr 0 \cr \vdots \cr 0 \end{bmatrix}
+\begin{bmatrix} 
+1 & 0 & 0 & \cdots & 0 & 0 \cr
+-\lambda & 1 & 0 & \cdots & 0 & 0 \cr
+0 & -\lambda & 1 & \cdots & 0 & 0 \cr
+ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
+0 & 0 & 0 & \cdots & -\lambda & 1 
+\end{bmatrix} 
+\begin{bmatrix}
+y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T 
+\end{bmatrix}
+= 
+\begin{bmatrix} 
+\lambda y_0 \cr 0 \cr 0 \cr \vdots \cr 0 
+\end{bmatrix}
 $$
 
 
@@ -110,30 +123,41 @@ $$
 Multiplying both sides by  inverse of the matrix on the left provides the solution
 
 $$
-\begin{bmatrix} y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T \end{bmatrix} = 
-\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 \cr
-              \lambda & 1 & 0 & \cdots & 0 & 0 \cr
-                \lambda^2 & \lambda & 1 & \cdots & 0 & 0 \cr
-                 \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
-                \lambda^{T-1} & \lambda^{T-2} & \lambda^{T-3} & \cdots & -\lambda & 1 
+\begin{bmatrix} 
+y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T 
+\end{bmatrix} 
+= 
+\begin{bmatrix} 
+1 & 0 & 0 & \cdots & 0 & 0 \cr
+\lambda & 1 & 0 & \cdots & 0 & 0 \cr
+\lambda^2 & \lambda & 1 & \cdots & 0 & 0 \cr
+ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
+\lambda^{T-1} & \lambda^{T-2} & \lambda^{T-3} & \cdots & -\lambda & 1 
+\end{bmatrix}
+\begin{bmatrix} 
+\lambda y_0 \cr 0 \cr 0 \cr \vdots \cr 0 
 \end{bmatrix}
-\begin{bmatrix} \lambda y_0 \cr 0 \cr 0 \cr \vdots \cr 0 \end{bmatrix}
-
 $$
 
 
 #### Second order difference equation
 
 
 $$
-\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0 & 0 \cr
-              -\lambda_1 & 1 & 0 & \cdots & 0 & 0 & 0 \cr
-                -\lambda_2 & -\lambda_2 & 1 & \cdots & 0 & 0 & 0 \cr
-                 \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
-                0 & 0 & 0 & \cdots & \lambda_2 & -\lambda_1 & 1 
-
-\end{bmatrix} \begin{bmatrix} y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T \end{bmatrix}
-= \begin{bmatrix} \lambda_1 y_0 + \lambda_2 y_{-1} \cr \lambda_2 y_0 \cr 0 \cr \vdots \cr 0 \end{bmatrix}
+\begin{bmatrix} 
+1 & 0 & 0 & \cdots & 0 & 0 & 0 \cr
+-\lambda_1 & 1 & 0 & \cdots & 0 & 0 & 0 \cr
+-\lambda_2 & -\lambda_2 & 1 & \cdots & 0 & 0 & 0 \cr
+ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \cr
+0 & 0 & 0 & \cdots & \lambda_2 & -\lambda_1 & 1 
+\end{bmatrix} 
+\begin{bmatrix} 
+y_1 \cr y_2 \cr y_3 \cr \vdots \cr y_T 
+\end{bmatrix}
+= 
+\begin{bmatrix} 
+\lambda_1 y_0 + \lambda_2 y_{-1} \cr \lambda_2 y_0 \cr 0 \cr \vdots \cr 0 
+\end{bmatrix}
 $$
 
 Multiplying both sides by  inverse of the matrix on the left again provides the solution.
@@ -143,7 +167,7 @@ Multiplying both sides by  inverse of the matrix on the left again provides the
 As an exercise, we ask you to represent and solve a **third order linear difference equation**.
 How many initial conditions must you specify?
 
-+++
++++ {"user_expressions": []}
 
 ## Friedman-Hall consumption-smoothing model
 
@@ -152,7 +176,7 @@ A key object is what Milton Friedman called "non-human" or "non-financial" wealt
 
 
 $$
-h_0 \equiv \sum_{t=0}^T R^t y_t = \begin{bmatrix} 1 & R & \cdots & R^T \end{bmatrix}
+h_0 \equiv \sum_{t=0}^T R^{-t} y_t = \begin{bmatrix} 1 & R^{-1} & \cdots & R^{-T} \end{bmatrix}
 \begin{bmatrix} y_0 \cr y_1  \cr \vdots \cr y_T \end{bmatrix}
 $$
 
@@ -165,7 +189,7 @@ $$
 it is possible to convert a sequence of budget constraints into the single intertemporal constraint
 
 $$
-\sum_{t=0}^T R^t c_t = a_0 + h_0,
+\sum_{t=0}^T R^{-t} c_t = a_0 + h_0,
 $$
 
 which says that the present value of the consumption stream equals the sum of finanical and non-financial wealth.
@@ -180,14 +204,14 @@ $$
 In this case, we can use the intertemporal budget constraint to write 
 
 $$
-c_0 = \left(\sum_{t=0}^T R^t\right)^{-1} (a_0 + h_0)
+c_0 = \left(\sum_{t=0}^T R^{-t}\right)^{-1} (a_0 + h_0)
 $$
 
 This is the consumption-smoothing model in a nutshell.
 
 We'll put the model through some paces with Python code below.
 
-+++
++++ {"user_expressions": []}
 
 ## Permanent income model of consumption 
 
@@ -203,7 +227,7 @@ In the calculations below, please we'll  set default values of  $R > 1$, e.g., $
 For some $T+1 \times 1$ $y$ vector, use matrix algebra to compute 
 
 $$
-\sum_{t=0}^T R^t y_t = \begin{bmatrix} 1 & R & \cdots & R^T \end{bmatrix}
+\sum_{t=0}^T R^{-t} y_t = \begin{bmatrix} 1 & R^{-1} & \cdots & R^{-T} \end{bmatrix}
 \begin{bmatrix} y_0 \cr y_1  \cr \vdots \cr y_T \end{bmatrix}
 $$
 
@@ -215,6 +239,7 @@ $$
 c_0 = \left( \frac{1 - R^{-1}}{1 - R^{-(T+1)}} \right) (a_0 + \sum_{t=0}^T R^t y_t )
 $$
 
+**Jiacheng:** The same for $R^t$ here.
 
 #### Step 3 ####
 
@@ -228,10 +253,12 @@ $$
 \vdots  &\vdots & \vdots & \cdots & \vdots & \vdots & \vdots \cr
 0 & 0 & 0 & \cdots & -R & 1 & 0 \cr
 0 & 0 & 0 & \cdots & 0 & -R & 1
-\end{bmatrix} \begin{bmatrix} a_1 \cr a_2 \cr a_3 \cr \vdots \cr a_T \cr a_{T+1} \end{bmatrix}
-= R \begin{bmatrix} y_0 + a_0 - c_0 \cr y_1 - c_0 \cr y_2 - c_0 \cr \vdots\cr y_T - y_0 \cr 0
+\end{bmatrix} 
+\begin{bmatrix} a_1 \cr a_2 \cr a_3 \cr \vdots \cr a_T \cr a_{T+1} 
+\end{bmatrix}
+= R 
+\begin{bmatrix} y_0 + a_0 - c_0 \cr y_1 - c_0 \cr y_2 - c_0 \cr \vdots\cr y_{T-1} - c_0 \cr y_T - c_0
 \end{bmatrix}
-
 $$
 
 Multiply both sides by the inverse of the matrix on the left side to compute
@@ -250,13 +277,14 @@ Let's verify this with our Python code.
 
 
 
+
 ### Feasible consumption variations ###
 
 To explore what types of consumption paths are welfare-improving, we shall create an **admissible consumption path variation sequence** $\{v_t\}_{t=0}^T$
 that satisfies
 
 $$
-\sum_{t=0}^T v_t = 0
+\sum_{t=0}^T R^{-t} v_t = 0
 $$
 
 We'll compute a two-parameter class of admissible variations
@@ -305,7 +333,155 @@ to compute alternative consumption paths, then evaluate their welfare.
 **Note to John:** We can do some fun simple experiments with these variations -- we can use
 graphs to show that, when $\beta R =1$ and  starting from the smooth path, all nontrivial budget-feasible variations lower welfare according to the criterion above.  
 
-We can even use the Python numpy grad command to compute derivatives of welfare with respect to our two parameters.  Notice that we are teaching the key idea beneath the calculus of variations.
+We can even use the Python numpy grad command to compute derivatives of welfare with respect to our two parameters.  
+
+We are teaching the key idea beneath the **calculus of variations**.
+
+```{code-cell} ipython3
+class Consumption_smoothing:
+    "A class of the Permanent Income model of consumption"
+    
+    def __init__(self, R, y_seq, a0, g1, g2, T):
+        self.a0, self.y_seq, self.R, self.β = a0, y_seq, R, 1/R    # set β = 1/R
+        self.g1, self.g2 = g1, g2       # welfare parameter
+        self.T = T
+        
+        self.β_seq = np.array([self.β**i for i in range(T+1)])
+        
+    def compute_optimal(self, verbose=1):
+        R, y_seq, a0, T = self.R, self.y_seq, self.a0, self.T
+        
+        # non-financial wealth
+        h0 = self.β_seq @ y_seq     # since β = 1/R
+        
+        # c0
+        c0 = (1 - 1/R) / (1 - (1/R)**(T+1)) * (a0 + h0)
+        c_seq = c0*np.ones(T+1)
+        
+        # verify
+        A = np.diag(-R*np.ones(T), k=-1) + np.eye(T+1)
+        b = y_seq - c_seq
+        b[0] = b[0] + a0
+        
+        a_seq = np.linalg.inv(A) @ b
+        a_seq = np.concatenate([[a0], a_seq])
+        
+        # check that a_T+1 = 0
+        if verbose==1:
+            print('check a_T+1=0:', np.abs(a_seq[-1] - 0) <= 1e-8)
+        
+        return c_seq, a_seq
+    
+    def welfare(self, c_seq):
+        β_seq, g1, g2 = self.β_seq, self.g1, self.g2
+        
+        u_seq = g1 * c_seq - g2/2 * c_seq**2
+        return β_seq @ u_seq
+        
+    
+    def compute_variation(self, ξ1, ϕ, verbose=1):
+        R, T, β_seq = self.R, self.T, self.β_seq
+        
+        ξ0 = ξ1*((1 - 1/R) / (1 - (1/R)**(T+1))) * ((1 - (ϕ/R)**(T+1)) / (1 - ϕ/R))
+        v_seq = np.array([(ξ1*ϕ**t - ξ0) for t in range(T+1)])
+        
+        # check if it is feasible
+        if verbose==1:
+            print('check feasible:', np.round(β_seq @ v_seq, 7)==0)     # since β = 1/R
+        
+        c_opt, _ = self.compute_optimal(verbose=verbose)
+        cvar_seq = c_opt + v_seq
+        
+        return cvar_seq
+```
+
++++ {"user_expressions": []}
+
+Below is an example where the consumer inherits $a_0<0$ (which can be interpreted as a student debt).
+
+The income process $\{y_t\}_{t=0}^{T}$ is constant and positive up to $t=45$ and then becomes zero afterward.
+
+```{code-cell} ipython3
+# parameters
+T=65
+R = 1.05
+g1 = 1
+g2 = 1/2
+
+# financial wealth
+a0 = -2     # such as "student debt"
+
+# income process
+y_seq = np.concatenate([np.ones(46), np.zeros(20)])
+
+# create an instance
+mc = Consumption_smoothing(R=R, y_seq=y_seq, a0=a0, g1=g1, g2=g2, T=T)
+c_seq, a_seq = mc.compute_optimal()
+
+# compute welfare 
+print('Welfare:', mc.welfare(c_seq))
+```
+
+```{code-cell} ipython3
+plt.plot(range(T+1), y_seq, label='income')
+plt.plot(range(T+1), c_seq, label='consumption')
+plt.plot(range(T+2), a_seq, label='asset')
+plt.plot(range(T+2), np.zeros(T+2), '--')
+
+plt.legend()
+plt.xlabel(r'$t$')
+plt.ylabel(r'$c_t,y_t,a_t$')
+plt.show()
+```
+
++++ {"user_expressions": []}
+
+We can visualize how $\xi_1$ and $\phi$ controls **budget-feasible variations**.
+
+```{code-cell} ipython3
+# visualize variational paths
+cvar_seq1 = mc.compute_variation(ξ1=.01, ϕ=.95)
+cvar_seq2 = mc.compute_variation(ξ1=.05, ϕ=.95)
+cvar_seq3 = mc.compute_variation(ξ1=.01, ϕ=1.02)
+cvar_seq4 = mc.compute_variation(ξ1=.05, ϕ=1.02)
+```
+
+```{code-cell} ipython3
+print('welfare of optimal c: ', mc.welfare(c_seq))
+print('variation 1: ', mc.welfare(cvar_seq1))
+print('variation 2:', mc.welfare(cvar_seq2))
+print('variation 3: ', mc.welfare(cvar_seq3))
+print('variation 4:', mc.welfare(cvar_seq4))
+```
+
+```{code-cell} ipython3
+plt.plot(range(T+1), c_seq, color='orange', label=r'Optimal $\vec{c}$ ')
+plt.plot(range(T+1), cvar_seq1, color='tab:blue', label=r'$\xi_1 = 0.01, \phi = 0.95$')
+plt.plot(range(T+1), cvar_seq2, color='tab:blue', ls='-.', label=r'$\xi_1 = 0.05, \phi = 0.95$')
+plt.plot(range(T+1), cvar_seq3, color='tab:green', label=r'$\xi_1 = 0.01, \phi = 1.02$')
+plt.plot(range(T+1), cvar_seq4, color='tab:green', ls='-.', label=r'$\xi_1 = 0.05, \phi = 1.02$')
+
+
+plt.legend()
+plt.xlabel(r'$t$')
+plt.ylabel(r'$c_t$')
+plt.show()
+```
+
+```{code-cell} ipython3
+def welfare_ϕ(mc, ξ1, ϕ):
+    "Compute welfare of variation sequence for given ϕ, ξ1 with an instance of our model mc"
+    cvar_seq = mc.compute_variation(ξ1=ξ1, ϕ=ϕ, verbose=0)
+    return mc.welfare(cvar_seq)
+
+welfare_φ = np.vectorize(welfare_φ)
+ξ1_arr = np.linspace(-0.5, 0.5, 20)
+
+plt.plot(ξ1_arr, welfare_φ(mc, ξ1=ξ1_arr , ϕ=1.02))
+plt.ylabel('welfare')
+plt.xlabel(r'$\xi_1$')
+plt.show()
+```
 
 ```{code-cell} ipython3