ENH: Migrate exercise and solutions: Part 2 (#118)

* lqramsey exercise

* arellano exercise

* update matsuyama

* update coase

* lucas_model update

* lu_tricks update

* amss update

* asl update

* classical_filtering update

* estspec

* fix dollar signs

* adding dollar sign back

* aligned update

* adding a sentence

* update smoothing_tax

Author: HumphreyYang

lectures/amss.md

@@ -616,10 +616,16 @@ $$
 \Pi(\tilde s| s_-) u_c(\tilde s)}}
 $$
 
-**Exercise**: Please verify that $\check \Pi(s|s_-)$ is a valid Markov
+```{exercise-start}
+:label: amss_ex1
+```
+Please verify that $\check \Pi(s|s_-)$ is a valid Markov
 transition density, i.e., that its elements are all non-negative and
 that for each $s_-$, the sum over $s$ equals unity.
 
+```{exercise-end}
+```
+
 ### Absence of State Variable Degeneracy
 
 Along a Ramsey plan, the state variable $x_t = x_t(s^t, b_0)$

lectures/arellano.md

@@ -689,14 +689,21 @@ Periods of relative stability are followed by sharp spikes in the discount rate
 ## Exercises
 
 (arellano_ex1)=
-### Exercise 1
+```{exercise-start}
+:label: arella_ex1
+```
 
 To the extent that you can, replicate the figures shown above
 
 * Use the parameter values listed as defaults in `Arellano_Economy`.
 * The time series will of course vary depending on the shock draws.
 
-## Solutions
+```{exercise-end}
+```
+
+```{solution-start} arella_ex1
+:class: dropdown
+```
 
 Compute the value function, policy and equilibrium prices
 
@@ -818,3 +825,5 @@ for ax, series, title in zip(axes, plot_series, titles):
 
 plt.show()
 ```
+```{solution-end}
+```

lectures/asset_pricing_lph.md

@@ -464,7 +464,9 @@ def simple_ols(X, Y, constant=False):
     return β_hat, σ_hat
 ```
 
-### Exercise 1
+```{exercise-start}
+:label: apl_ex1
+```
 
 Look at the equation, 
 
@@ -474,11 +476,60 @@ $$
 
 Verify that this equation is a regression equation.
 
-### Exercise 2
+```{exercise-end}
+```
+
+```{solution-start} apl_ex1
+:class: dropdown
+```
+
+To verify that it is a **regression equation** we must show that the residual is orthogonal to the regressor.
+
+Our assumptions about mutual orthogonality imply that
+
+$$
+E\left[\epsilon_{i,t}\right]=0,\quad E\left[\epsilon_{i,t}u_{t}\right]=0
+$$
+
+It follows that
+
+$$
+\begin{aligned}
+E\left[\sigma_{i}\epsilon_{i,t}\left(R_{t}^{m}-R^{f}\right)\right]&=E\left[\sigma_{i}\epsilon_{i,t}\left(\xi+\lambda u_{t}\right)\right] \\
+	&=\sigma_{i}\xi E\left[\epsilon_{i,t}\right]+\sigma_{i}\lambda E\left[\epsilon_{i,t}u_{t}\right] \\
+	&=0
+\end{aligned}
+$$
+
+```{solution-end}
+```
+
+
+```{exercise-start}
+:label: apl_ex2
+```
 
 Give a formula for the regression coefficient $\beta_{i, R^m}$.
 
-### Exercise 3
+```{exercise-end}
+```
+
+```{solution-start} apl_ex2
+:class: dropdown
+```
+
+The regression coefficient $\beta_{i, R^m}$ is
+
+$$
+\beta_{i,R^{m}}=\frac{Cov\left(R_{t}^{i}-R^{f},R_{t}^{m}-R^{f}\right)}{Var\left(R_{t}^{m}-R^{f}\right)}
+$$
+
+```{solution-end}
+```
+
+```{exercise-start}
+:label: apl_ex3
+```
 
 As in many sciences, it is useful to distinguish a  **direct problem** from  an **inverse problem**.
 
@@ -522,51 +573,12 @@ E[(a + b R_t^m) R^m_t)] &= 1 \\
 E[(a + b R_t^m) R^f_t)] &= 1
 \end{align*}
 
-### Exercise 4
-
-Using the equations above, find a system of two **linear** equations that you can solve for $a$ and $b$ as functions of the parameters $(\lambda, \xi, E[R_f])$.
-
-Write a function that can solve these equations.
-
-Please check the **condition number** of a key matrix that must be inverted to determine a, b
-
-### Exercise 5
-
-Using the estimates of the parameters that you generated above, compute the implied stochastic discount factor.
-
-
-
-## Solutions 
-
-### Solution to Exercise 1
-
-To verify that it is a **regression equation** we must show that the residual is orthogonal to the regressor.
-
-Our assumptions about mutual orthogonality imply that
-
-$$
-E\left[\epsilon_{i,t}\right]=0,\quad E\left[\epsilon_{i,t}u_{t}\right]=0
-$$
-
-It follows that
-
-$$
-\begin{aligned}
-E\left[\sigma_{i}\epsilon_{i,t}\left(R_{t}^{m}-R^{f}\right)\right]&=E\left[\sigma_{i}\epsilon_{i,t}\left(\xi+\lambda u_{t}\right)\right] \\
-	&=\sigma_{i}\xi E\left[\epsilon_{i,t}\right]+\sigma_{i}\lambda E\left[\epsilon_{i,t}u_{t}\right] \\
-	&=0
-\end{aligned}
-$$
-
-### Solution to Exercise 2
-
-The regression coefficient $\beta_{i, R^m}$ is
-
-$$
-\beta_{i,R^{m}}=\frac{Cov\left(R_{t}^{i}-R^{f},R_{t}^{m}-R^{f}\right)}{Var\left(R_{t}^{m}-R^{f}\right)}
-$$
+```{exercise-end}
+```
 
-### Solution to Exercise 3
+```{solution-start} apl_ex3
+:class: dropdown
+```
 
 **Direct Problem:**
 
@@ -670,14 +682,36 @@ for i in range(N):
 
 Q: How close did your estimates come to the parameters we specified?
 
+```{solution-end}
+```
+
+```{exercise-start}
+:label: apl_ex4
+```
+
+Using the equations above, find a system of two **linear** equations that you can solve for $a$ and $b$ as functions of the parameters $(\lambda, \xi, E[R_f])$.
+
+Write a function that can solve these equations.
+
+Please check the **condition number** of a key matrix that must be inverted to determine a, b
 
+```{exercise-end}
+```
+
+```{solution-start} apl_ex4
+:class: dropdown
+```
+
+The system of two linear equations is shown below:
 
-### Solution to Exercise 4
 
-\begin{align}
+$$
+\begin{aligned}
 a ((E(R^f) + \xi) + b ((E(R^f) + \xi)^2 + \lambda^2 + \sigma_f^2) & =1 \cr
 a E(R^f) + b (E(R^f)^2 + \xi E(R^f) + \sigma_f ^ 2) & = 1 
-\end{align}
+\end{aligned}
+$$
+
 
 ```{code-cell} ipython3
 # Code here
@@ -705,7 +739,21 @@ a, b, condM = solve_ab(ERf, σf, λ, ξ)
 a, b, condM
 ```
 
-### Solution to Exercise 5
+```{solution-end}
+```
+
+```{exercise-start}
+:label: apl_ex5
+```
+
+Using the estimates of the parameters that you generated above, compute the implied stochastic discount factor.
+
+```{exercise-end}
+```
+
+```{solution-start} apl_ex5
+:class: dropdown
+```
 
 Now let's pass $\hat{E}(R^f), \hat{\sigma}^f, \hat{\lambda}, \hat{\xi}$ to the function `solve_ab`.
 
@@ -715,4 +763,7 @@ a_hat, b_hat, M_hat = solve_ab(ERf_hat, σf_hat, λ_hat, ξ_hat)
 
 ```{code-cell} ipython3
 a_hat, b_hat, M_hat
+```
+
+```{solution-end}
 ```

lectures/classical_filtering.md

@@ -935,7 +935,9 @@ and that the zeros of $\theta (z)$ are not inside the unit circle.
 
 ## Exercises
 
-### Exercise 1
+```{exercise-start}
+:label: cf_ex1
+```
 
 Let $Y_t = (1 - 2 L ) u_t$ where $u_t$ is a mean zero
 white noise with $\mathbb{E} u^2_t = 1$. Let
@@ -964,7 +966,13 @@ $$
 X_{t-j}
 $$
 
-### Exercise 2
+```{exercise-end}
+```
+
+```{exercise-start}
+:label: cf_ex2
+```
+
 
 **Multivariable Prediction:** Let $Y_t$ be an $(n\times 1)$
 vector stochastic process with moving average representation
@@ -1038,3 +1046,6 @@ $$
     = \left[{C(L) \over L^J} \right]_+ C(L)^{-1}\, X_t
 $$
 
+```{exercise-end}
+```
+

lectures/coase.md

@@ -575,27 +575,22 @@ Note that downstream firms choose to be larger, a point we return to below.
 ## Exercises
 
 (lucas_asset_exer1)=
-### Exercise 1
-
+```{exercise-start}
+:label: coa_ex1
+```
 The number of firms is endogenously determined by the primitives.
 
 What do you think will happen in terms of the number of firms as $\delta$ increases?  Why?
 
 Check your intuition by computing the number of firms at delta in (1.01, 1.05, 1.1).
 
-### Exercise 2
-
-The **value added** of firm $i$ is $v_i := p^*(t_{i-1}) - p^*(t_i)$.
-
-One of the interesting predictions of the model is that value added is increasing with downstreamness, as are several other measures of firm size.
-
-Can you give any intution?
-
-Try to verify this phenomenon (value added increasing with downstreamness) using the code above.
-
-## Solutions
+```{exercise-end}
+```
 
-### Exercise 1
+```{solution-start}  coa_ex1
+:class: dropdown
+```
+Here is one solution 
 
 ```{code-cell} python3
 for delta in (1.01, 1.05, 1.1):
@@ -606,8 +601,26 @@ for delta in (1.01, 1.05, 1.1):
    num_firms = len(transaction_stages)
    print(f"When delta={delta} there are {num_firms} firms")
 ```
+```{solution-end}
+```
+
+```{exercise-start}
+:label: coa_ex2
+```
+The **value added** of firm $i$ is $v_i := p^*(t_{i-1}) - p^*(t_i)$.
 
-### Exercise 2
+One of the interesting predictions of the model is that value added is increasing with downstreamness, as are several other measures of firm size.
+
+Can you give any intution?
+
+Try to verify this phenomenon (value added increasing with downstreamness) using the code above.
+
+```{exercise-end}
+```
+
+```{solution-start}  coa_ex2
+:class: dropdown
+```
 
 Firm size increases with downstreamness because $p^*$, the equilibrium price function, is increasing and strictly convex.
 
@@ -640,3 +653,5 @@ ax.set_xticklabels(("downstream firms", "upstream firms"))
 plt.show()
 ```
 
+```{solution-end}
+```