Reformat the Exercise and Solution (#271)

* update formats

* temporarily change ci

* set ci back to PR

Author: HumphreyYang

lectures/ar1_processes.md

@@ -359,6 +359,52 @@ Confirm this by simulation at a range of $k$ using the default parameters from t
 ```
 
 
+```{solution-start} ar1p_ex1
+:class: dropdown
+```
+
+Here is one solution:
+
+```{code-cell} python3
+from numba import njit
+from scipy.special import factorial2
+
+@njit
+def sample_moments_ar1(k, m=100_000, mu_0=0.0, sigma_0=1.0, seed=1234):
+    np.random.seed(seed)
+    sample_sum = 0.0
+    x = mu_0 + sigma_0 * np.random.randn()
+    for t in range(m):
+        sample_sum += (x - mu_star)**k
+        x = a * x + b + c * np.random.randn()
+    return sample_sum / m
+
+def true_moments_ar1(k):
+    if k % 2 == 0:
+        return std_star**k * factorial2(k - 1)
+    else:
+        return 0
+
+k_vals = np.arange(6) + 1
+sample_moments = np.empty_like(k_vals)
+true_moments = np.empty_like(k_vals)
+
+for k_idx, k in enumerate(k_vals):
+    sample_moments[k_idx] = sample_moments_ar1(k)
+    true_moments[k_idx] = true_moments_ar1(k)
+
+fig, ax = plt.subplots()
+ax.plot(k_vals, true_moments, label="true moments")
+ax.plot(k_vals, sample_moments, label="sample moments")
+ax.legend()
+
+plt.show()
+```
+
+```{solution-end}
+```
+
+
 ```{exercise}
 :label: ar1p_ex2
 
@@ -405,95 +451,6 @@ of these distributions?)
 ```
 
 
-```{exercise}
-:label: ar1p_ex3
-
-In the lecture we discussed the following fact: for the $AR(1)$ process
-
-$$
-X_{t+1} = a X_t + b + c W_{t+1}
-$$
-
-with $\{ W_t \}$ iid and standard normal,
-
-$$
-\psi_t = N(\mu, s^2) \implies \psi_{t+1}
-= N(a \mu + b, a^2 s^2 + c^2)
-$$
-
-Confirm this, at least approximately, by simulation. Let
-
-- $a = 0.9$
-- $b = 0.0$
-- $c = 0.1$
-- $\mu = -3$
-- $s = 0.2$
-
-First, plot $\psi_t$ and $\psi_{t+1}$ using the true
-distributions described above.
-
-Second, plot $\psi_{t+1}$ on the same figure (in a different
-color) as follows:
-
-1. Generate $n$ draws of $X_t$ from the $N(\mu, s^2)$
-   distribution
-1. Update them all using the rule
-   $X_{t+1} = a X_t + b + c W_{t+1}$
-1. Use the resulting sample of $X_{t+1}$ values to produce a
-   density estimate via kernel density estimation.
-
-Try this for $n=2000$ and confirm that the
-simulation based estimate of $\psi_{t+1}$ does converge to the
-theoretical distribution.
-```
-
-
-## Solutions
-
-```{solution-start} ar1p_ex1
-:class: dropdown
-```
-
-```{code-cell} python3
-from numba import njit
-from scipy.special import factorial2
-
-@njit
-def sample_moments_ar1(k, m=100_000, mu_0=0.0, sigma_0=1.0, seed=1234):
-    np.random.seed(seed)
-    sample_sum = 0.0
-    x = mu_0 + sigma_0 * np.random.randn()
-    for t in range(m):
-        sample_sum += (x - mu_star)**k
-        x = a * x + b + c * np.random.randn()
-    return sample_sum / m
-
-def true_moments_ar1(k):
-    if k % 2 == 0:
-        return std_star**k * factorial2(k - 1)
-    else:
-        return 0
-
-k_vals = np.arange(6) + 1
-sample_moments = np.empty_like(k_vals)
-true_moments = np.empty_like(k_vals)
-
-for k_idx, k in enumerate(k_vals):
-    sample_moments[k_idx] = sample_moments_ar1(k)
-    true_moments[k_idx] = true_moments_ar1(k)
-
-fig, ax = plt.subplots()
-ax.plot(k_vals, true_moments, label="true moments")
-ax.plot(k_vals, sample_moments, label="sample moments")
-ax.legend()
-
-plt.show()
-```
-
-```{solution-end}
-```
-
-
 ```{solution-start} ar1p_ex2
 :class: dropdown
 ```
@@ -553,6 +510,48 @@ distribution is smooth but less so otherwise.
 ```
 
 
+```{exercise}
+:label: ar1p_ex3
+
+In the lecture we discussed the following fact: for the $AR(1)$ process
+
+$$
+X_{t+1} = a X_t + b + c W_{t+1}
+$$
+
+with $\{ W_t \}$ iid and standard normal,
+
+$$
+\psi_t = N(\mu, s^2) \implies \psi_{t+1}
+= N(a \mu + b, a^2 s^2 + c^2)
+$$
+
+Confirm this, at least approximately, by simulation. Let
+
+- $a = 0.9$
+- $b = 0.0$
+- $c = 0.1$
+- $\mu = -3$
+- $s = 0.2$
+
+First, plot $\psi_t$ and $\psi_{t+1}$ using the true
+distributions described above.
+
+Second, plot $\psi_{t+1}$ on the same figure (in a different
+color) as follows:
+
+1. Generate $n$ draws of $X_t$ from the $N(\mu, s^2)$
+   distribution
+1. Update them all using the rule
+   $X_{t+1} = a X_t + b + c W_{t+1}$
+1. Use the resulting sample of $X_{t+1}$ values to produce a
+   density estimate via kernel density estimation.
+
+Try this for $n=2000$ and confirm that the
+simulation based estimate of $\psi_{t+1}$ does converge to the
+theoretical distribution.
+```
+
 ```{solution-start} ar1p_ex3
 :class: dropdown
 ```

lectures/cake_eating_numerical.md

@@ -502,17 +502,6 @@ Make the required changes to value function iteration code and plot the value an
 Try to reuse as much code as possible.
 ```
 
-
-```{exercise}
-:label: cen_ex2
-
-Implement time iteration, returning to the original case (i.e., dropping the
-modification in the exercise above).
-```
-
-
-## Solutions
-
 ```{solution-start} cen_ex1
 :class: dropdown
 ```
@@ -593,6 +582,14 @@ Consumption is higher when $\alpha < 1$ because, at least for large $x$, the ret
 ```
 
 
+```{exercise}
+:label: cen_ex2
+
+Implement time iteration, returning to the original case (i.e., dropping the
+modification in the exercise above).
+```
+
+
 ```{solution-start} cen_ex2
 :class: dropdown
 ```

lectures/cake_eating_problem.md

@@ -526,8 +526,6 @@ In doing so, you will need to use the definition of the value function and the
 Bellman equation.
 ```
 
-## Solutions
-
 ```{solution} cep_ex1
 :class: dropdown
 

lectures/career.md

@@ -383,43 +383,6 @@ To generate the draws from the distributions $F$ and $G$, use `quantecon.random.
 ```
 
 
-```{exercise}
-:label: career_ex2
-
-Let's now consider how long it takes for the worker to settle down to a
-permanent job, given a starting point of $(\theta, \epsilon) = (0, 0)$.
-
-In other words, we want to study the distribution of the random variable
-
-$$
-T^* := \text{the first point in time from which the worker's job no longer changes}
-$$
-
-Evidently, the worker's job becomes permanent if and only if $(\theta_t, \epsilon_t)$ enters the
-"stay put" region of $(\theta, \epsilon)$ space.
-
-Letting $S$ denote this region, $T^*$ can be expressed as the
-first passage time to $S$ under the optimal policy:
-
-$$
-T^* := \inf\{t \geq 0 \,|\, (\theta_t, \epsilon_t) \in S\}
-$$
-
-Collect 25,000 draws of this random variable and compute the median (which should be about 7).
-
-Repeat the exercise with $\beta=0.99$ and interpret the change.
-```
-
-
-```{exercise}
-:label: career_ex3
-
-Set the parameterization to `G_a = G_b = 100` and generate a new optimal policy
-figure -- interpret.
-```
-
-## Solutions
-
 ```{solution-start} career_ex1
 :class: dropdown
 ```
@@ -469,6 +432,32 @@ plt.show()
 ```{solution-end}
 ```
 
+```{exercise}
+:label: career_ex2
+
+Let's now consider how long it takes for the worker to settle down to a
+permanent job, given a starting point of $(\theta, \epsilon) = (0, 0)$.
+
+In other words, we want to study the distribution of the random variable
+
+$$
+T^* := \text{the first point in time from which the worker's job no longer changes}
+$$
+
+Evidently, the worker's job becomes permanent if and only if $(\theta_t, \epsilon_t)$ enters the
+"stay put" region of $(\theta, \epsilon)$ space.
+
+Letting $S$ denote this region, $T^*$ can be expressed as the
+first passage time to $S$ under the optimal policy:
+
+$$
+T^* := \inf\{t \geq 0 \,|\, (\theta_t, \epsilon_t) \in S\}
+$$
+
+Collect 25,000 draws of this random variable and compute the median (which should be about 7).
+
+Repeat the exercise with $\beta=0.99$ and interpret the change.
+```
 
 ```{solution-start} career_ex2
 :class: dropdown
@@ -519,10 +508,19 @@ Not surprisingly, more patient workers will wait longer to settle down to their
 ```
 
 
+```{exercise}
+:label: career_ex3
+
+Set the parameterization to `G_a = G_b = 100` and generate a new optimal policy
+figure -- interpret.
+```
+
 ```{solution-start} career_ex3
 :class: dropdown
 ```
 
+Here is one solution
+
 ```{code-cell} python3
 cw = CareerWorkerProblem(G_a=100, G_b=100)
 T, get_greedy = operator_factory(cw)
@@ -536,7 +534,7 @@ ax.contourf(tg, eg, greedy_star.T, levels=lvls, cmap=cm.winter, alpha=0.5)
 ax.contour(tg, eg, greedy_star.T, colors='k', levels=lvls, linewidths=2)
 ax.set(xlabel='θ', ylabel='ϵ')
 ax.text(1.8, 2.5, 'new life', fontsize=14)
-ax.text(4.5, 2.5, 'new job', fontsize=14, rotation='vertical')
+ax.text(4.5, 1.5, 'new job', fontsize=14, rotation='vertical')
 ax.text(4.0, 4.5, 'stay put', fontsize=14)
 plt.show()
 ```

lectures/cass_koopmans_1.md

@@ -875,8 +875,6 @@ state in which $f'(K)=\rho +\delta$.
 - Why does the saving rate respond as it does?
 ```
 
-### Solution
-
 ```{solution-start} ck1_ex1
 :class: dropdown
 ```

lectures/coleman_policy_iter.md

@@ -438,8 +438,6 @@ Set `γ = 1.5`.
 Compute and plot the optimal policy.
 ```
 
-## Solutions
-
 ```{solution-start} cpi_ex1
 :class: dropdown
 ```

lectures/complex_and_trig.md

@@ -306,13 +306,13 @@ x1 = 2 * r * sqrt(2)
 ## Note: we choose the solution near 0
 eq1 = Eq(x1/x0 - r * cos(ω+θ) / cos(ω), 0)
 ω = nsolve(eq1, ω, 0)
-ω = np.float(ω)
+ω = float(ω)
 print(f'ω = {ω:1.3f}')
 
 # Solve for p
 eq2 = Eq(x0 - 2 * p * cos(ω), 0)
 p = nsolve(eq2, p, 0)
-p = np.float(p)
+p = float(p)
 print(f'p = {p:1.3f}')
 ```
 
@@ -485,6 +485,9 @@ integrate(cos(ω) * sin(ω), (ω, -π, π))
 
 ### Exercises
 
+```{exercise}
+:label: complex_ex1
+
 We invite the reader to verify analytically and with the `sympy` package the following two equalities:
 
 $$
@@ -495,3 +498,4 @@ $$
 \int_{-\pi}^{\pi} \sin (\omega)^2 \, d\omega = \frac{\pi}{2}
 $$
 
+```