STYLE: quantecon execise/solution style

environment.yml

@@ -7,11 +7,11 @@ dependencies:
   - pip
   - pip:
     - jupyter-book==1.0.4post1
-    - quantecon-book-theme==0.8.3
+    - quantecon-book-theme==0.10.0
     - sphinx-tojupyter==0.4.0
-    - sphinx-proof==0.2.1
+    - sphinx-proof==0.3.0
     - sphinxext-rediraffe==0.2.7
-    - sphinx-exercise==1.0.1
+    - sphinx-exercise==1.2.1
     - sphinxcontrib-youtube==1.4.1
     - sphinx-togglebutton==0.3.2
 

lectures/ergodicity.md

@@ -638,6 +638,12 @@ Let $(P_t)$ be a Markov semigroup.  True or false:
 for this semigroup, every state $x$ is accessible from itself.
 ```
 
+```{solution} ergodicity-ex-1
+:class: dropdown
+
+The statement is true.  With $t=0$ we have $P_t(x,x) = I(x,x) = 1 > 0$.
+```
+
 ```{exercise}
 :label: ergodicity-ex-2
 Let $(\lambda_k)$ be a bounded non-increasing sequence in $(0, \infty)$.
@@ -658,41 +664,30 @@ Show that $(P_t)$, the corresponding Markov semigroup, has no stationary
 distribution.
 ```
 
-
-```{exercise}
-:label: ergodicity-ex-3
-Confirm that {prf:ref}`sdrift` implies {prf:ref}`sfinite`.
-```
-
-## Solutions
-
-```{solution} ergodicity-ex-1
-The statement is true.  With $t=0$ we have $P_t(x,x) = I(x,x) = 1 > 0$.
-```
-
-
 ```{solution} ergodicity-ex-2
+:class: dropdown
+
 Suppose to the contrary that $\phi \in \dD$ and $\phi Q = 0$.
 
 Then, for any $j \geq 1$,
 
 $$
-    (\phi Q)(j) 
-    = \sum_{i \geq 0} \phi(i) Q(i, j)
-    = - \lambda_j \phi(j) + \lambda_{j-1} \phi(j-1) 
-    = 0
+(\phi Q)(j) 
+= \sum_{i \geq 0} \phi(i) Q(i, j)
+= - \lambda_j \phi(j) + \lambda_{j-1} \phi(j-1) 
+= 0
 $$
 
 Since $(\lambda_k)$ is non-increasing, it follows that
 
 $$
-    \frac{\phi(j)}{\phi(j-1)} = \frac{\lambda_{j-1}}{\lambda_j} \geq 1
+\frac{\phi(j)}{\phi(j-1)} = \frac{\lambda_{j-1}}{\lambda_j} \geq 1
 $$
 
 Therefore, for any $j\geq 1$, it must be:
 
 $$
-    \phi(j) \geq \phi(j-1)
+\phi(j) \geq \phi(j-1)
 $$
 
 It follows that $\phi$ is non-decreasing on $\ZZ_+$.
@@ -704,17 +699,24 @@ Contradiction.
 ```
 
 
+```{exercise}
+:label: ergodicity-ex-3
+Confirm that {prf:ref}`sdrift` implies {prf:ref}`sfinite`.
+```
+
 ```{solution} ergodicity-ex-3
+:class: dropdown
+
 Let $(P_t)$ be an irreducible UC Markov semigroup and let $S$ be finite.
 
 Pick any positive constants $M, \epsilon$ and set $v = M$ and $F = S$.
 
 We then have
 
 $$
-    \sum_y Q(x, y) v(y)
-    = M \sum_y Q(x, y)
-    = 0
+\sum_y Q(x, y) v(y)
+= M \sum_y Q(x, y)
+= 0
 $$
 
 Hence the drift condition in {prf:ref}`sdrift` holds and $(P_t)$ is

lectures/generators.md

@@ -432,14 +432,33 @@ example, Chapter 7 of {cite}`bobrowski2005functional`.
 Prove that {eq}`expdiffer` holds for all $A \in \linop$.
 ```
 
+```{solution} ergodicity-ex-1
+:class: dropdown
+
+To show the first equality, fix $t \in \RR_+$, take $h > 0$ and observe that
+
+$$
+e^{(t+h)A} - e^{tA} - e^{tA} A
+= e^{tA} (e^{hA} - I - A)
+$$
+
+Since the norm on $\linop$ is submultiplicative, it suffices to show that 
+$\| e^{hA} - I - A \| = o(h)$ as $h \to 0$.
+
+Using the definition of the exponential, this is easily verified,
+completing the proof of the first equality in {eq}`expdiffer`.
+
+The proof of the second equality is similar.
+```
+
 ```{exercise}
 :label: generators-ex-2
 
 In many texts, a $C_0$ semigroup is defined as an evolution semigroup $(U_t)$
 such that
 
 $$
-    U_t g \to g \text{ as } t \to 0 \text{ for any } g \in \BB
+U_t g \to g \text{ as } t \to 0 \text{ for any } g \in \BB
 $$ (czsg2)
 
 Our aim is to show that {eq}`czsg2` implies continuity at every point $t$, as
@@ -448,54 +467,16 @@ in the definition we used above.
 The [Banach--Steinhaus Theorem](https://en.wikipedia.org/wiki/Uniform_boundedness_principle) can be used to show that, for an evolution semigroup $(U_t)$ satisfying {eq}`czsg2`, there exist finite constants $\omega$ and $M$ such that
 
 $$
-    \| U_t \| \leq e^{t\omega} M
-    \quad \text{for all } \; t \geq 0
+\| U_t \| \leq e^{t\omega} M
+\quad \text{for all } \; t \geq 0
 $$ (sgbound)
 
 Using this and {eq}`czsg2`, show that, for any $g \in \BB$, the map $t \mapsto
 U_t g$ is continuous at all $t$.
 ```
 
-```{exercise}
-:label: generators-ex-3
-
-Following on from the previous exercise, 
-a UC semigroup is often defined as an evolution semigroup $(U_t)$
-such that 
-
-$$
-    \| U_t - I \| \to 0 \text{ as } t \to 0 
-$$ (czsg3)
-
-Show that {eq}`czsg3` implies norm continuity at every point $t$, as
-in the definition we used above.
-
-In particular, show that, for any $t_n \to t$, we have
-$\| U_{t_n} - U_t \| \to 0$  as $n \to \infty$.
-```
-
-
-## Solutions
-
-```{solution} ergodicity-ex-1
-
-To show the first equality, fix $t \in \RR_+$, take $h > 0$ and observe that
-
-$$
-    e^{(t+h)A} - e^{tA} - e^{tA} A
-    = e^{tA} (e^{hA} - I - A)
-$$
-
-Since the norm on $\linop$ is submultiplicative, it suffices to show that 
-$\| e^{hA} - I - A \| = o(h)$ as $h \to 0$.
-
-Using the definition of the exponential, this is easily verified,
-completing the proof of the first equality in {eq}`expdiffer`.
-
-The proof of the second equality is similar.
-```
-
 ```{solution} ergodicity-ex-2
+:class: dropdown
 
 Let $(U_t)$ be an evolution semigroup satisfying {eq}`czsg2` and let
 $\omega$ and $M$ be as in {eq}`sgbound`.
@@ -507,16 +488,36 @@ On one hand, $U_{t+ h_n} g = U_{h_n} U_t g \to U_t g$ by {eq}`czsg2`.
 On the other hand, from {eq}`sgbound` and the definition of the operator norm,
 
 $$
-    \| U_{t-h_n} g - U_t g\|
-    = \|  U_{t-h_n} ( g - U_{h_n} g) \|
-    \leq e^{(t-h_n)\omega} M \| g - U_{h_n} g\|
-    \to 0
+\| U_{t-h_n} g - U_t g\|
+= \|  U_{t-h_n} ( g - U_{h_n} g) \|
+\leq e^{(t-h_n)\omega} M \| g - U_{h_n} g\|
+\to 0
 $$
 
 as $n \to \infty$.  This completes the proof.
 ```
 
+
+```{exercise}
+:label: generators-ex-3
+
+Following on from the previous exercise, 
+a UC semigroup is often defined as an evolution semigroup $(U_t)$
+such that 
+
+$$
+\| U_t - I \| \to 0 \text{ as } t \to 0 
+$$ (czsg3)
+
+Show that {eq}`czsg3` implies norm continuity at every point $t$, as
+in the definition we used above.
+
+In particular, show that, for any $t_n \to t$, we have
+$\| U_{t_n} - U_t \| \to 0$  as $n \to \infty$.
+```
+
 ```{solution} ergodicity-ex-3
+:class: dropdown
 
 The solution is similar to that of the previous exercise.
 
@@ -529,9 +530,9 @@ On the other hand, from the submultiplicative property of the operator norm
 and {eq}`sgbound`,
 
 $$
-    \| U_{t-h_n} - U_t \|
-    = \|  U_{t-h_n} ( I - U_{h_n}) \|
-    \leq e^{(t-h_n)\omega} M  \| I - U_{h_n} \|
+\| U_{t-h_n} - U_t \|
+= \|  U_{t-h_n} ( I - U_{h_n}) \|
+\leq e^{(t-h_n)\omega} M  \| I - U_{h_n} \|
 $$
 
 This converges to 0 as $n \to \infty$, completing our proof.