updates

Author: HumphreyYang

lectures/ar1_processes.md

@@ -356,9 +356,7 @@ In this equation, we can use observed data to evaluate the left hand side of {eq
 
 And we can use a theoretical AR(1) model to calculate the right hand side.
 
-If $\frac{1}{m} \sum_{t = 1}^m X_t$ is not close to $\psi^*(x)$, even for many
-observations, then our theory seems to be incorrect and we will need to revise
-it.
+If $\frac{1}{m} \sum_{t = 1}^m h(X_t)$ is not close to $\int h(x)\psi^*(x) dx$, even for many observations, then our theory seems to be incorrect and we will need to revise it.
 
 
 ## Exercises
@@ -427,8 +425,8 @@ def true_moments_ar1(k):
         return 0
 
 k_vals = np.arange(6) + 1
-sample_moments = np.empty_like(k_vals)
-true_moments = np.empty_like(k_vals)
+sample_moments = np.empty(len(k_vals), dtype=float)
+true_moments = np.empty(len(k_vals), dtype=float)
 
 for k_idx, k in enumerate(k_vals):
     sample_moments[k_idx] = sample_moments_ar1(k)

lectures/cagan_adaptive.md

@@ -505,11 +505,13 @@ for λ in λ_vals:
     print(f'{λ:>6.2f} | {ρ:>10.4f} | {str(abs(ρ) < 1):>8} | {peak:>20.4f}')
 ```
 
-All four values satisfy the stability condition $|\rho| < 1$ for the default $\alpha = 5$.
+All four values satisfy the stability condition $|\rho| < 1$ for the default $\alpha = 5$ and have $\rho > 0$.
 
-The case $\lambda = 0.86$ is closest to the stability boundary and therefore displays the largest oscillatory response.
+The case $\lambda = 0.86$ has the largest initial overshoot among these four values and then converges the fastest.
 
-As $\lambda$ moves closer to one, expectations become more inertial and the post-stabilization response decays more slowly but starts from a smaller jump.
+As $\lambda$ moves closer to one, expectations become more inertial, so the post-stabilization response decays more slowly but starts from a smaller jump.
+
+For $\alpha = 5$, an oscillatory stable response would require $0.8 < \lambda < 5/6$.
 
 ```{solution-end}
 ```

lectures/cagan_ree.md

@@ -580,7 +580,7 @@ The government prints money to finance expenditure with the "velocity dividend"
 
 The next code generates a multi-panel graph that includes outcomes of both experiments 1 and 2.
 
-That allows us to assess how important it is to understand whether the sudden permanent drop in $\mu_t$ at $t=T_1$ is fully unanticipated, as in experiment 1, or completely
+That allows us to assess how important it is to understand whether the sudden permanent drop in $\mu_t$ at $t=T_1$ is fully anticipated, as in experiment 1, or completely
 unanticipated, as in experiment 2.
 
 ```{code-cell} ipython3

lectures/cobweb.md

@@ -530,7 +530,7 @@ ts_plot_supply(m, 5, 15)
 
 Backward looking average expectations refers to the case where producers form
 expectations for the next period price as a linear combination of their last
-guess and the second last guess.
+observed price and the price observed one period before that.
 
 That is,
 
@@ -587,9 +587,9 @@ def ts_plot_price_blae(model, p0, p1, alphas, ts_length=15):
 ```{code-cell} ipython3
 m = Market()
 ts_plot_price_blae(m, 
-                   p0=5, 
-                   p1=6, 
-                   alphas=[0.1, 0.3, 0.5, 0.8], 
+                   p0=1,
+                   p1=2.5,
+                   alphas=[0.1, 0.3, 0.5, 0.8],
                    ts_length=20)
 ```
 

lectures/cons_smooth.md

@@ -522,7 +522,7 @@ $$
 which implies that
 
 $$
-\xi_1 \sum_{t=0}^T \phi_t R^{-t} - \xi_0 \sum_{t=0}^T R^{-t} = 0
+\xi_1 \sum_{t=0}^T \phi^t R^{-t} - \xi_0 \sum_{t=0}^T R^{-t} = 0
 $$
 
 which implies that

lectures/eigen_I.md

@@ -1093,8 +1093,8 @@ Try to compute the trajectory of $v$ after being transformed by $A$ for $n=4$ it
 ```{code-cell} ipython3
 A = np.array([[1, 2],
               [1, 1]])
-v = (0.4, -0.4)
-n = 11
+v = (2, -2)
+n = 4
 
 # Compute eigenvectors and eigenvalues
 eigenvalues, eigenvectors = np.linalg.eig(A)
@@ -1105,7 +1105,7 @@ print(f'eigenvectors:\n {eigenvectors}')
 plot_series(A, v, n)
 ```
 
-The result seems to converge to the eigenvector of $A$ with the largest eigenvalue.
+The trajectory begins to turn toward the eigenvector of $A$ with the largest eigenvalue.
 
 Let's use a [vector field](https://en.wikipedia.org/wiki/Vector_field) to visualize the transformation brought by A.
 

lectures/eigen_II.md

@@ -117,7 +117,7 @@ Left eigenvectors will play important roles in what follows, including that of s
 
 A vector $w$ is called a left eigenvector of $A$ if $w$ is a right eigenvector of $A^\top$.
 
-In other words, if $w$ is a left eigenvector of matrix $A$, then $A^\top w = \lambda w$, where $\lambda$ is the eigenvalue associated with the left eigenvector $v$.
+In other words, if $w$ is a left eigenvector of matrix $A$, then $A^\top w = \lambda w$, where $\lambda$ is the eigenvalue associated with the left eigenvector $w$.
 
 This hints at how to compute left eigenvectors
 
@@ -257,7 +257,7 @@ We can see that if a matrix is primitive, then it implies the matrix is irreduci
 
 Now let's step back to the primitive matrices part of the Perron-Frobenius theorem
 
-```{prf:Theorem} Continous of Perron-Frobenius Theorem
+```{prf:Theorem} Continuity of Perron-Frobenius Theorem
 :label: con-perron-frobenius
 
 If $A$ is primitive then,

lectures/input_output.md

@@ -560,13 +560,13 @@ Another way to rank sectors in input-output networks is via output multipliers.
 The **output multiplier** of sector $j$ denoted by $\mu_j$ is usually defined as the
 total sector-wide impact of a unit change of demand in sector $j$.
 
-Earlier when disussing demand shocks we concluded that for $L = (l_{ij})$ the element
+Earlier when discussing demand shocks we concluded that for $L = (l_{ij})$ the element
 $l_{ij}$ represents the impact on sector $i$ of a unit change in demand in sector $j$.
 
 Thus,
 
 $$
-\mu_j = \sum_{j=1}^n l_{ij}
+\mu_j = \sum_{i=1}^n l_{ij}
 $$
 
 This can be written as $\mu^\top = \mathbb{1}^\top L$ or

lectures/intro_supply_demand.md

@@ -789,8 +789,7 @@ $$
 
 Here $p$ is set to $i_s(q)$.
 
-Social welfare is the sum of consumer and producer surplus under the
-assumption that the price is the same for buyers and sellers:
+Planner welfare is the area between inverse demand and inverse supply:
 
 $$
 W(q)
@@ -801,6 +800,8 @@ Solve the integrals and write a function to compute this quantity numerically
 at given $q$. 
 
 Plot welfare as a function of $q$.
+
+At the competitive equilibrium, the common price lets this same area be interpreted as consumer surplus plus producer surplus.
 ```
 
 
@@ -883,8 +884,7 @@ print(f"{maximizing_q: .5f}")
 ````{exercise}
 :label: isd_ex4
 
-Now compute the equilibrium quantity by finding the price that equates supply
-and demand.
+Now compute the equilibrium quantity by finding the quantity that equates supply and demand.
 
 You can do this numerically by finding the root of the excess demand function
 
@@ -904,8 +904,7 @@ Initialize `newton` with a starting guess somewhere close to 1.0.
 
 (Similar initial conditions will give the same result.)
 
-You should find that the equilibrium price agrees with the welfare maximizing
-price, in line with the first fundamental welfare theorem.
+You should find that the equilibrium quantity agrees with the welfare maximizing quantity, in line with the first fundamental welfare theorem.
 
 ````
 

lectures/lake_model.md

@@ -541,24 +541,37 @@ Eq. {eq}`steady_x` implies that the long-run unemployment rate will increase, an
 if $\alpha$ increases or $\lambda$ decreases.
 
 Suppose first that $\alpha=0.01, \lambda=0.1, d=0.02, b=0.025$.
-Assume that $\alpha$ increases to $0.04$.
 
-The below graph illustrates that the line $D$ shifts clockwise downward, which indicates that
-the fraction of unemployment rises as the separation rate increases.
+Assume that $\alpha$ increases to $0.04$.
 
-```{code-cell} ipython3
-fig, ax = plt.subplots(figsize=(10, 8))
+Then compare this with a decrease in $\lambda$ from $0.1$ to $0.04$.
 
-lm = LakeModel(α=0.01, λ=0.1, d=0.02, b=0.025)
-plot_time_paths(lm, ax=ax)
-s=10
-ax.plot([0, s * lm.u_bar], [0, s * lm.e_bar], "k--", lw=1, label='set $D$, α=0.01')
+The graphs show that both changes rotate the line $D$ clockwise downward, which indicates that the long-run unemployment rate rises.
 
-lm = LakeModel(α=0.04, λ=0.1, d=0.02, b=0.025)
-plot_time_paths(lm, ax=ax)
-ax.plot([0, s * lm.u_bar], [0, s * lm.e_bar], "r--", lw=1, label='set $D$, α=0.04')
+```{code-cell} ipython3
+fig, axes = plt.subplots(1, 2, figsize=(12, 5))
+s = 10
+
+def add_D_line(ax, lm, label, color):
+    ax.plot([0, s * lm.u_bar], [0, s * lm.e_bar],
+            color=color, linestyle='--', lw=2, label=label)
+
+lm_base = LakeModel(α=0.01, λ=0.1, d=0.02, b=0.025)
+lm_high_α = LakeModel(α=0.04, λ=0.1, d=0.02, b=0.025)
+plot_time_paths(lm_base, ax=axes[0])
+add_D_line(axes[0], lm_base, r'set $D$, $\alpha=0.01$', 'black')
+plot_time_paths(lm_high_α, ax=axes[0])
+add_D_line(axes[0], lm_high_α, r'set $D$, $\alpha=0.04$', 'red')
+axes[0].legend(loc='best')
+
+lm_low_λ = LakeModel(α=0.01, λ=0.04, d=0.02, b=0.025)
+plot_time_paths(lm_base, ax=axes[1])
+add_D_line(axes[1], lm_base, r'set $D$, $\lambda=0.1$', 'black')
+plot_time_paths(lm_low_λ, ax=axes[1])
+add_D_line(axes[1], lm_low_λ, r'set $D$, $\lambda=0.04$', 'red')
+axes[1].legend(loc='best')
 
-ax.legend(loc='best')
+plt.tight_layout()
 plt.show()
 ```