minor updates

Author: HumphreyYang

lectures/ar1_processes.md

@@ -464,8 +464,8 @@ $$
 For $K$ use the Gaussian kernel ($K$ is the standard normal
 density).
 
-Write the class so that the bandwidth defaults to Silverman’s rule (see
-the “rule of thumb” discussion on [this
+Write the class so that the bandwidth defaults to Silverman's rule (see
+the "rule of thumb" discussion on [this
 page](https://en.wikipedia.org/wiki/Kernel_density_estimation)). Test
 the class you have written by going through the steps
 

lectures/cagan_adaptive.md

@@ -464,9 +464,9 @@ The sluggish fall in inflation is explained by how anticipated  inflation $\pi_t
 
 For Experiment 1 (sudden stabilization at $T_1 = 60$ from $\mu_0 = 0.5$ to $\mu^* = 0$), solve the model for $\lambda \in \{0.86,\, 0.90,\, 0.95,\, 0.98\}$ and, on a single graph, plot the actual inflation rate $\pi_t$ for each value.
 
-(a) How do the sign and speed of post-stabilization convergence change as $\lambda$ varies within the stable region?
+a. How do the sign and speed of post-stabilization convergence change as $\lambda$ varies within the stable region?
 
-(b) For each $\lambda$, print $\rho$ and the peak absolute value of $\pi_t$ for $t \geq T_1$.
+b. For each $\lambda$, print $\rho$ and the peak absolute value of $\pi_t$ for $t \geq T_1$.
 ```
 
 ```{solution-start} ca_ex1
@@ -494,7 +494,7 @@ ax.set_title('Inflation paths for different λ (sudden stabilization)')
 ax.legend()
 plt.show()
 
-print(f'{"λ":>6} | {"ρ":>10} | {"|ρ|<1":>8} | {"peak |π| after T₁":>20}')
+print(f'{"λ":>6} | {"ρ":>10} | {"|ρ|<1":>8} | {"peak |π| after T1":>20}')
 print('-' * 56)
 for λ in λ_vals:
     m = create_cagan_adaptive_model(λ=λ)
@@ -524,12 +524,12 @@ expectations, in contrast to a rational-expectations equilibrium.
 
 For the default model (`md`) and both experiments:
 
-(a) Compute and plot the forecast error $e_t = \pi_t^* - \pi_t$ for
+a. Compute and plot the forecast error $e_t = \pi_t^* - \pi_t$ for
     $t = 0, 1, \ldots, T$.
 
-(b) For each experiment, is $e_t$ systematically positive or negative during
-    the disinflation?  Explain why this systematic bias could not survive under
-    rational expectations.
+b. For each experiment, determine whether $e_t$ is systematically positive or
+    negative during the disinflation and explain why this systematic bias could
+    not survive under rational expectations.
 
 (Recall that `Eπ_seq` returned by `solve_cagan_adaptive` has $T+2$ elements
 while `π_seq` has $T+1$; use `Eπ_seq[:-1]` to align them.)
@@ -575,13 +575,13 @@ axes[1].set_ylabel(r'$\pi_t^* - \pi_t$')
 plt.tight_layout()
 plt.show()
 
-print(f'Exp 1 — mean forecast error t < T₁:  {e1[:T1].mean():.4f}')
-print(f'Exp 1 — mean forecast error t ≥ T₁:  {e1[T1:].mean():.4f}')
-print(f'Exp 2 — mean forecast error overall:  {e2.mean():.4f}')
+print(f'Exp 1: mean forecast error t < T1:  {e1[:T1].mean():.4f}')
+print(f'Exp 1: mean forecast error t >= T1: {e1[T1:].mean():.4f}')
+print(f'Exp 2: mean forecast error overall: {e2.mean():.4f}')
 ```
 
 During disinflation, actual inflation falls *below* expected inflation, so
-$e_t = \pi_t^* - \pi_t > 0$ throughout the transition — the public
+$e_t = \pi_t^* - \pi_t > 0$ throughout the transition, so the public
 systematically **over-predicts** inflation.
 
 Under rational expectations this persistent one-sided bias would be immediately arbitraged away as agents adjust their forecasting rule until $e_t$ has mean zero.
@@ -604,14 +604,14 @@ $$
 
 Using Experiment 1 and the default model `md`:
 
-(a) Compute $\rho$ analytically from the model parameters and verify that
+a. Compute $\rho$ analytically from the model parameters and verify that
     $|\rho| < 1$ (the stability condition {eq}`eq:suffcond`).
 
-(b) From the solved path `π_seq`, compute the empirical ratios
+b. From the solved path `π_seq`, compute the empirical ratios
     $\pi_{t+1}/\pi_t$ for $t = T_1 + 1, \ldots, T_1 + 10$ and compare them
     to $\rho$.
 
-(c) Plot $\log|\pi_t|$ against $t$ for $t \geq T_1$ and verify that it is
+c. Plot $\log|\pi_t|$ against $t$ for $t \geq T_1$ and verify that it is
     linear with slope $\log|\rho|$.
 ```
 
@@ -632,7 +632,7 @@ print(f'ρ = {ρ:.6f}   (|ρ| < 1: {abs(ρ) < 1})')
 μ_seq = np.append(μ0 * np.ones(T1), μ_star * np.ones(md.T + 1 - T1))
 π_seq, _, _, _ = solve_cagan_adaptive(md, μ_seq)
 
-# (b) empirical successive ratios
+# Part b: empirical successive ratios
 print(f'\n{"t":>5} | {"π_t":>12} | {"π_{t+1}/π_t":>14} | {"ρ":>8}')
 print('-' * 46)
 for t in range(T1, T1 + 10):
@@ -641,7 +641,7 @@ for t in range(T1, T1 + 10):
 ```
 
 ```{code-cell} ipython3
-# (c) log|π_t| is linear after T1
+# Part c: log|π_t| is linear after T1
 t_post = np.arange(T1, md.T + 1)
 log_π  = np.log(np.abs(π_seq[T1:]))
 
@@ -676,16 +676,16 @@ $\pi_t = \rho^{t-T_1} \pi_{T_1}$ for $t \geq T_1$.
 Experiment 2 uses a gradual decline in money growth
 $\mu_t = \phi^t \mu_0 + (1-\phi^t)\mu^*$ with $\phi = 0.9$.
 
-(a) For the same gradual $\mu$ path, compare the inflation $\pi_t$ and expected inflation $\pi_t^*$ paths for two stable cases:
+a. For the same gradual $\mu$ path, compare the inflation $\pi_t$ and expected inflation $\pi_t^*$ paths for two stable cases:
 
     * **Faster adjustment**: $\lambda = 0.86$
     * **Slower adjustment**: $\lambda = 0.95$
 
     Plot $\pi_t$, $\pi_t^*$, and $\mu_t$ for each case on side-by-side graphs.
 
-(b) For each case, compute the mean absolute forecast error $\bar{e} = \frac{1}{T+1}\sum_{t=0}^T |\pi_t^* - \pi_t|$.
+b. For each case, compute the mean absolute forecast error $\bar{e} = \frac{1}{T+1}\sum_{t=0}^T |\pi_t^* - \pi_t|$.
 
-(c) Explain why the faster-adjustment case can move below the money-growth path while the slower-adjustment case displays more persistent forecast errors.
+c. Explain why the faster-adjustment case can move below the money-growth path while the slower-adjustment case displays more persistent forecast errors.
 ```
 
 ```{solution-start} ca_ex4

lectures/cagan_ree.md

@@ -461,8 +461,9 @@ Such a completely unanticipated shock is popularly known as an "MIT shock".
 
 The mental experiment involves switching at time $T_1$ from an initial "continuation path" for $\{\mu_t, \pi_t\} $ to another path that involves a permanently lower inflation rate.
 
-**Initial Path:** $\mu_t = \mu_0$ for all $t \geq 0$. So this path is for $\{\mu_t\}_{t=0}^\infty$; the associated
-path for $\pi_t$ has $\pi_t = \mu_0$.
+**Initial Path:** $\mu_t = \mu_0$ for all $t \geq 0$.
+
+This path is for $\{\mu_t\}_{t=0}^\infty$; the associated path for $\pi_t$ has $\pi_t = \mu_0$.
 
 **Revised Continuation Path** Where $ \mu_0 > \mu^*$, we construct a continuation path $\{\mu_s\}_{s=T_1}^\infty$
 by setting $\mu_s = \mu^*$ for all $s \geq T_1$. The perfect foresight continuation path for
@@ -661,8 +662,8 @@ at $T_1 = 60$, with $T = 80$), solve the model for
 $\alpha \in \{1,\, 3,\, 5,\, 10,\, 25\}$ and plot the inflation path $\pi_t$ for
 each value on a single graph.
 
-Describe how the **anticipation effect** — the pre-stabilization fall in inflation
-— changes with $\alpha$.
+Describe how the **anticipation effect**, the pre-stabilization fall in inflation,
+changes with $\alpha$.
 ```
 
 ```{solution-start} cagan_ex1
@@ -716,9 +717,9 @@ $$
 \pi_t = (1-\delta) \sum_{s=t}^T \delta^{s-t} \mu_s + \delta^{T+1-t} \pi_{T+1}^*
 $$
 
-Compute $\pi_t$ directly from this formula for each $t = 0, 1, \ldots, T$ and
-compare it to the matrix solution returned by `solve`.  Plot both on the same
-graph and print the maximum absolute difference.
+Compute $\pi_t$ directly from this formula for each $t = 0, 1, \ldots, T$,
+compare it to the matrix solution returned by `solve`, plot both on the same
+graph, and print the maximum absolute difference.
 ```
 
 ```{solution-start} cagan_ex2
@@ -770,14 +771,15 @@ The two methods agree to machine precision, confirming that the matrix system
 **Foreseen gradual vs sudden stabilization.**
 
 Experiment 1 features a *sudden* foreseen drop in money growth at $T_1 = 60$.
+
 Experiment 3 features a *gradual* foreseen path $\mu_t = \phi^t \mu_0 + (1-\phi^t)\mu^*$.
 
 On a single graph, plot the inflation paths for:
 - Experiment 1 (sudden), and
 - Experiment 3 with $\phi \in \{0.95, 0.85, 0.70\}$ (increasingly fast gradualism).
 
-Use $\alpha = 5$, $\mu_0 = 0.5$, $\mu^* = 0$, $T = 80$.  Which approach generates
-the smoothest pre-stabilization decline in inflation?
+Use $\alpha = 5$, $\mu_0 = 0.5$, $\mu^* = 0$, and $T = 80$ to determine which
+approach generates the smoothest pre-stabilization decline in inflation.
 ```
 
 ```{solution-start} cagan_ex3

lectures/cobweb.md

@@ -82,7 +82,6 @@ ax.plot(years, hog_prices, '-o', ms=4, label='hog price')
 ax.set_xlabel('year')
 ax.set_ylabel('dollars')
 ax.legend()
-ax.grid()
 plt.show()
 ```
 
@@ -595,4 +594,4 @@ ts_plot_price_blae(m,
 ```
 
 ```{solution-end}
-```
+```

lectures/complex_and_trig.md

@@ -153,7 +153,6 @@ ax.text(0-0.2, 0.5, '$x = 1$')                            # Label x
 ax.text(0.5, 1.2, r'$y = \sqrt{3}$')                      # Label y
 ax.text(0.25, 0.15, r'$\theta = 60^o$')                   # Label θ
 
-ax.grid(True)
 plt.show()
 ```
 
@@ -351,7 +350,6 @@ ticklab = ax.yaxis.get_ticklabels()[0] # Set y-label position
 trans = ticklab.get_transform()
 ax.yaxis.set_label_coords(0, 5, transform=trans)
 
-ax.grid()
 plt.show()
 ```
 
@@ -582,16 +580,16 @@ from sympy import Symbol, cos, sin, simplify
 
 θ = Symbol('θ', real=True)
 
-print("cos(2θ) = cos²θ − sin²θ:",
+print("cos(2θ) = cos(θ)**2 - sin(θ)**2:",
       simplify(cos(2*θ) - (cos(θ)**2 - sin(θ)**2)))
 
 print("sin(2θ) = 2sinθcosθ:",
       simplify(sin(2*θ) - 2*sin(θ)*cos(θ)))
 
-print("cos(2θ) = 2cos²θ − 1:",
+print("cos(2θ) = 2*cos(θ)**2 - 1:",
       simplify(cos(2*θ) - (2*cos(θ)**2 - 1)))
 
-print("cos(2θ) = 1 − 2sin²θ:",
+print("cos(2θ) = 1 - 2*sin(θ)**2:",
       simplify(cos(2*θ) - (1 - 2*sin(θ)**2)))
 ```
 
@@ -657,13 +655,13 @@ from sympy import symbols, cos, sin, simplify
 
 θ, w = symbols('θ w', real=True)
 
-print("cos(θ+w) + cos(θ-w) − 2cos(θ)cos(w) =",
+print("cos(θ+w) + cos(θ-w) - 2cos(θ)cos(w) =",
       simplify(cos(θ+w) + cos(θ-w) - 2*cos(θ)*cos(w)))
 
-print("cos(θ-w) − cos(θ+w) − 2sin(θ)sin(w) =",
+print("cos(θ-w) - cos(θ+w) - 2sin(θ)sin(w) =",
       simplify(cos(θ-w) - cos(θ+w) - 2*sin(θ)*sin(w)))
 
-print("sin(θ+w) + sin(θ-w) − 2sin(θ)cos(w) =",
+print("sin(θ+w) + sin(θ-w) - 2sin(θ)cos(w) =",
       simplify(sin(θ+w) + sin(θ-w) - 2*sin(θ)*cos(w)))
 ```
 

lectures/cons_smooth.md

@@ -551,7 +551,14 @@ Now let's compute and plot consumption path variations
 def compute_variation(model, ξ1, ϕ, a0, y_seq, verbose=1):
     R, T, β_seq = model.R, model.T, model.β_seq
 
-    ξ0 = ξ1*((1 - 1/R) / (1 - (1/R)**(T+1))) * ((1 - (ϕ/R)**(T+1)) / (1 - ϕ/R))
+    growth = ϕ / R
+    if np.isclose(growth, 1):
+        pv_sum = T + 1
+    else:
+        pv_sum = (1 - growth**(T+1)) / (1 - growth)
+
+    annuity = (1 - 1/R) / (1 - (1/R)**(T+1))
+    ξ0 = ξ1 * annuity * pv_sum
     v_seq = np.array([(ξ1*ϕ**t - ξ0) for t in range(T+1)])
 
     if verbose == 1:
@@ -901,11 +908,11 @@ $$
 
 and the default model parameters.
 
-(a) Plot all four consumption paths on a single graph.  What do you notice about
-    their shapes relative to one another?
+a. Plot all four consumption paths on a single graph and describe their shapes
+    relative to one another.
 
-(b) Plot $c_0$ against $a_0$ and compute the slope of the resulting line.
-    Verify that the slope equals the annuity factor
+b. Plot $c_0$ against $a_0$, compute the slope of the resulting line, and verify
+    that the slope equals the annuity factor
     $\left(\frac{1-R^{-1}}{1-R^{-(T+1)}}\right)$ from equation {eq}`eq:conssmoothing`.
 ```
 
@@ -946,7 +953,7 @@ R = cs_model.R
 slope = (c0_vals[-1] - c0_vals[0]) / (a0_vals[-1] - a0_vals[0])
 annuity = (1 - 1/R) / (1 - (1/R)**(T+1))
 print(f'Numerical slope of c0 w.r.t. a0: {slope:.8f}')
-print(f'Annuity factor (1 - R⁻¹)/(1 - R⁻⁽ᵀ⁺¹⁾):  {annuity:.8f}')
+print(f'Annuity factor (1 - R**(-1))/(1 - R**(-(T+1))): {annuity:.8f}')
 print(f'Match: {np.isclose(slope, annuity)}')
 ```
 
@@ -968,13 +975,13 @@ maximizes welfare {eq}`welfare` among all budget-feasible paths.
 Using `compute_variation` with $\xi_1 = 0.1$ and the Experiment 1 income sequence
 ($W_0 = 2.5$ windfall at $t=21$, with $a_0 = -2$):
 
-(a) Compute welfare for the optimal flat path and for variations with $\phi \in \{0.7,\, 0.9,\, 0.98,\, 1.02,\, 1.1\}$.
+a. Compute welfare for the optimal flat path and for variations with $\phi \in \{0.7,\, 0.9,\, 0.98,\, 1.02,\, 1.1\}$.
 
     Print the results in a table.
 
-(b) Plot welfare as a function of $\phi$ on a fine grid in $(0.5, 1.5)$.  Mark the
-    welfare of the optimal flat path as a dashed horizontal line and confirm it is
-    the global maximum.
+b. Plot welfare as a function of $\phi$ on a fine grid in $[0.7, 1.1]$ and mark
+    the welfare of the optimal flat path as a dashed horizontal line to confirm
+    that it lies above these budget-feasible variations.
 ```
 
 ```{solution-start} consmooth_ex4
@@ -1001,7 +1008,7 @@ for ϕ in ϕ_vals:
     print(f'{ϕ:>6.2f} | {w:>12.6f} | {w - w_opt:>+14.6f}')
 
 # Fine grid
-ϕ_grid = np.linspace(0.5, 1.5, 200)
+ϕ_grid = np.linspace(0.7, 1.1, 200)
 w_grid = np.array([
     welfare(cs_model,
             compute_variation(cs_model, ξ1=0.1, ϕ=ϕ, a0=a0,
@@ -1021,6 +1028,8 @@ plt.show()
 
 Every non-zero variation in the plotted family delivers strictly lower welfare than the flat path marked by the horizontal dashed line.
 
+The wider interval $(0.5, 1.5)$ is not informative here because values of $\phi$ well above one make the late-life variation $\xi_1\phi^t$ very large.
+
 This numerically confirms the variational principle that the constant consumption path is the global welfare maximizer when $\beta R = 1$.
 
 ```{solution-end}

lectures/eigen_I.md

@@ -222,13 +222,11 @@ def grid_transform(A=np.array([[1, -1], [1, 1]])):
 
     # Plot x-y grid points
     ax[0].scatter(xygrid[0], xygrid[1], s=36, c=colors, edgecolor="none")
-    # ax[0].grid(True)
     # ax[0].axis("equal")
     ax[0].set_title("points $x_1, x_2, \cdots, x_k$")
 
     # Plot transformed grid points
     ax[1].scatter(uvgrid[0], uvgrid[1], s=36, c=colors, edgecolor="none")
-    # ax[1].grid(True)
     # ax[1].axis("equal")
     ax[1].set_title("points $Ax_1, Ax_2, \cdots, Ax_k$")
 
@@ -746,7 +744,6 @@ for spine in ['left', 'bottom']:
     ax.spines[spine].set_position('zero')
 for spine in ['right', 'top']:
     ax.spines[spine].set_color('none')
-# ax.grid(alpha=0.4)
 
 xmin, xmax = -3, 3
 ymin, ymax = -3, 3
@@ -1148,7 +1145,6 @@ plt.legend(lines, labels, loc='center left',
 plt.xlabel("x")
 plt.ylabel("y")
 plt.title("Convergence towards eigenvectors")
-plt.grid()
 plt.gca().set_aspect('equal', adjustable='box')
 plt.show()
 ```
@@ -1237,7 +1233,6 @@ for i, example in enumerate(examples):
 
     ax[i].set_xlabel("x-axis")
     ax[i].set_ylabel("y-axis")
-    ax[i].grid()
     ax[i].set_aspect('equal', adjustable='box')
 
 fig.suptitle("Vector fields of the three matrices")