diff --git a/environment.yml b/environment.yml
index fbc5d16..5955384 100644
--- a/environment.yml
+++ b/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
 
diff --git a/lectures/ergodicity.md b/lectures/ergodicity.md
index 83e45e9..047f392 100644
--- a/lectures/ergodicity.md
+++ b/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 $\ZZ_+$.
@@ -704,7 +699,14 @@ 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 $F = S$.
@@ -712,9 +714,9 @@ 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
diff --git a/lectures/generators.md b/lectures/generators.md
index 278129c..91a8506 100644
--- a/lectures/generators.md
+++ b/lectures/generators.md
@@ -432,6 +432,25 @@ 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
 
@@ -439,7 +458,7 @@ 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.
diff --git a/lectures/kolmogorov_bwd.md b/lectures/kolmogorov_bwd.md
index b632aa8..23a66cf 100644
--- a/lectures/kolmogorov_bwd.md
+++ b/lectures/kolmogorov_bwd.md
@@ -522,34 +522,11 @@ Try to generate the same figure using {eq}`psolq` instead, modifying code from
 {doc}`our lecture <markov_prop>` on the Markov property.
 ````
 
-```{exercise}
-:label: kolmogorov-bwd-2
-
-Prove that differentiating {eq}`kbinteg` at each $(x, y)$ yields {eq}`kolbackeq`.
-```
-
-```{exercise}
-:label: kolmogorov-bwd-3
-
-We claimed above that the solution $P_t = e^{t Q}$ is the unique
-Markov semigroup satisfying the backward equation $P'_t = Q P_t$.
-
-Try to supply a proof.
-
-(This is not an easy exercise but worth thinking about in any case.)
+```{solution-start} kolmogorov-bwd-1
+:class: dropdown
 ```
 
-## Solutions
-
-```{note}
-code is currently not supported in `sphinx-exercise`
-so code-cell solutions are immediately after this
-solution block.
-```
-
-```{solution} kolmogorov-bwd-1
 Here is one solution:
-```
 
 ```{code-cell} ipython3
 α = 0.6
@@ -594,55 +571,75 @@ ax.set_xlabel("inventory", fontsize=14)
 plt.show()
 ```
 
+```{solution-end}
+```
+
+```{exercise}
+:label: kolmogorov-bwd-2
+
+Prove that differentiating {eq}`kbinteg` at each $(x, y)$ yields {eq}`kolbackeq`.
+```
 
 ```{solution} kolmogorov-bwd-2
+:class: dropdown
 
 One can easily verify that, when $f$ is a differentiable function and $\alpha >
 0$, we have
 
 $$
-    g(t) = e^{- t \alpha} f(t)
-    \quad \implies \quad
-    g'(t) = e^{- t \alpha} f'(t) - \alpha g(t)
+g(t) = e^{- t \alpha} f(t)
+\quad \implies \quad
+g'(t) = e^{- t \alpha} f'(t) - \alpha g(t)
 $$ (gdiff)
 
 Note also that, with the change of variable $s = t - \tau$, we can rewrite
 {eq}`kbinteg` as
 
 $$
-    P_t(x, y) =
-    e^{-t \lambda(x)}
-    \left\{
-        I(x, y)
-        + \lambda(x)
-        \int_0^t (K P_s)(x, y) e^{s \lambda(x)} d s
-    \right\}
+P_t(x, y) =
+e^{-t \lambda(x)}
+\left\{
+    I(x, y)
+    + \lambda(x)
+    \int_0^t (K P_s)(x, y) e^{s \lambda(x)} d s
+\right\}
 $$ (kbinteg2)
 
 Applying {eq}`gdiff` yields
 
 $$
-    P'_t(x, y)
-    = e^{-t \lambda(x)}
-        \left\{ 
-             \lambda(x)
-             (K P_t)(x, y) e^{t \lambda(x)}
-        \right\}
-        - \lambda(x) P_t(x, y)
+P'_t(x, y)
+= e^{-t \lambda(x)}
+    \left\{ 
+            \lambda(x)
+            (K P_t)(x, y) e^{t \lambda(x)}
+    \right\}
+    - \lambda(x) P_t(x, y)
 $$
 
 After minor rearrangements this becomes
 
 $$
-    P'_t(x, y)
-    = \lambda(x) [ (K - I)  P_t](x, y)
+P'_t(x, y)
+= \lambda(x) [ (K - I)  P_t](x, y)
 $$
 
 which is identical to {eq}`kolbackeq`.
 ```
 
+```{exercise}
+:label: kolmogorov-bwd-3
+
+We claimed above that the solution $P_t = e^{t Q}$ is the unique
+Markov semigroup satisfying the backward equation $P'_t = Q P_t$.
+
+Try to supply a proof.
+
+(This is not an easy exercise but worth thinking about in any case.)
+```
 
 ```{solution} kolmogorov-bwd-3
+:class: dropdown
 
 Here is one proof of uniqueness.
 
@@ -657,15 +654,14 @@ Note that $V_0 = \hat P_t$ and $V_t = P_t$.
 Note also that $s \mapsto V_s$ is differentiable, with derivative
 
 $$
-    V'_s 
-    = P'_s \hat P_{t-s} - P_s \hat P'_{t-s}
-    = P_s Q \hat P_{t-s} - P_s Q \hat P_{t-s}
-    = 0
+V'_s 
+= P'_s \hat P_{t-s} - P_s \hat P'_{t-s}
+= P_s Q \hat P_{t-s} - P_s Q \hat P_{t-s}
+= 0
 $$
 
 where, in the second last equality, we used {eq}`expoderiv`.
 
-
 Hence $V_s$ is constant, so our previous observations $V_0 = \hat P_t$ and $V_t = P_t$
 now yield $\hat P_t = P_t$.
 
diff --git a/lectures/kolmogorov_fwd.md b/lectures/kolmogorov_fwd.md
index 54c249b..0cb1fd0 100644
--- a/lectures/kolmogorov_fwd.md
+++ b/lectures/kolmogorov_fwd.md
@@ -497,52 +497,22 @@ differentiable at all $t \geq 0$ and $(x, y) \in S \times S$.
 Define (pointwise, at each $(x,y)$),
 
 $$
-    Q := P'_0 = \lim_{h \downarrow 0} \frac{P_h - I}{h}
+Q := P'_0 = \lim_{h \downarrow 0} \frac{P_h - I}{h}
 $$ (genfl)
 
 Assuming that this limit exists, and hence $Q$ is well-defined, show that
 
 $$
-    P'_t = P_t Q
-    \quad \text{and} \quad
-    P'_t = Q P_t
+P'_t = P_t Q
+\quad \text{and} \quad
+P'_t = Q P_t
 $$
 
 both hold. (These are the Kolmogorov forward and backward equations.)
 ```
 
-```{exercise}
-:label: kolmogorov-fwd-2
-
-Recall {ref}`our model <sdji>` of jump chains with state-dependent jump
-intensities given by rate function $x \mapsto \lambda(x)$.
-
-After a wait time with exponential rate $\lambda(x) \in (0, \infty)$, the
-state transitions from $x$ to $y$ with probability $K(x,y)$.
-
-We found that the associated semigroup $(P_t)$ satisfies the Kolmogorov
-backward equation $P'_t = Q P_t$ with
-
-$$
-    Q(x, y) := \lambda(x) (K(x, y) - I(x, y))
-$$ (qeqagain)
-
-Show that $Q$ is an intensity matrix and that {eq}`genfl` holds.
-```
-
-```{exercise}
-:label: kolmogorov-fwd-3
-
-Prove {prf:ref}`intvsmk` by adapting the arguments in {prf:ref}`jctosg`.
-(This is nontrivial but worth at least trying.)
-
-Hint: The constant $m$ in the proof can be set to $\max_x |Q(x, x)|$.
-```
-
-
-## Solutions
-
 ```{solution} kolmogorov-fwd-1
+:class: dropdown
 
 Let $(P_t)$ be a Markov semigroup and let $Q$ be as defined in the statement
 of the exercise.
@@ -552,9 +522,9 @@ Fix $t \geq 0$ and $h > 0$.
 Combining the semigroup property and linearity with the restriction $P_0 = I$, we get
 
 $$
-    \frac{P_{t+h} - P_t}{h}
-    = \frac{P_t P_h - P_t}{h}
-    = \frac{P_t (P_h - I)}{h}
+\frac{P_{t+h} - P_t}{h}
+= \frac{P_t P_h - P_t}{h}
+= \frac{P_t (P_h - I)}{h}
 $$
 
 Taking $h \downarrow 0$ and using the definition of $Q$ give $P_t' = P_t Q$,
@@ -563,16 +533,35 @@ which is the Kolmogorov forward equation.
 For the backward equation we observe that
 
 $$
-    \frac{P_{t+h} - P_t}{h}
-    = \frac{P_h P_t - P_t}{h}
-    = \frac{(P_h - I) P_t}{h}
+\frac{P_{t+h} - P_t}{h}
+= \frac{P_h P_t - P_t}{h}
+= \frac{(P_h - I) P_t}{h}
 $$
 
 also holds.  Taking $h \downarrow 0$ gives the Kolmogorov backward equation.
 ```
 
+```{exercise}
+:label: kolmogorov-fwd-2
+
+Recall {ref}`our model <sdji>` of jump chains with state-dependent jump
+intensities given by rate function $x \mapsto \lambda(x)$.
+
+After a wait time with exponential rate $\lambda(x) \in (0, \infty)$, the
+state transitions from $x$ to $y$ with probability $K(x,y)$.
+
+We found that the associated semigroup $(P_t)$ satisfies the Kolmogorov
+backward equation $P'_t = Q P_t$ with
+
+$$
+Q(x, y) := \lambda(x) (K(x, y) - I(x, y))
+$$ (qeqagain)
+
+Show that $Q$ is an intensity matrix and that {eq}`genfl` holds.
+```
 
 ```{solution} kolmogorov-fwd-2
+:class: dropdown
 
 Let $Q$ be as defined in {eq}`qeqagain`.
 
@@ -585,15 +574,24 @@ For the second, we use the fact that $K$ is a Markov matrix, so that, with $1$
 as a column vector of ones,
 
 $$
-    Q 1
-    = \lambda (K 1 - 1)
-    = \lambda (1 - 1)
-    = 0
+Q 1
+= \lambda (K 1 - 1)
+= \lambda (1 - 1)
+= 0
 $$
 ```
 
+```{exercise}
+:label: kolmogorov-fwd-3
+
+Prove {prf:ref}`intvsmk` by adapting the arguments in {prf:ref}`jctosg`.
+(This is nontrivial but worth at least trying.)
+
+Hint: The constant $m$ in the proof can be set to $\max_x |Q(x, x)|$.
+```
 
 ```{solution} kolmogorov-fwd-3
+:class: dropdown
 
 Suppose that $Q$ is an intensity matrix, fix $t \geq 0$ and set $P_t = e^{tQ}$.
 
@@ -618,12 +616,12 @@ Because $P_t$ has unit row sums and differentiation is linear,
 we can employ the Kolmogorov backward equation to obtain
 
 $$
-    Q 1
-      = Q P_t 1
-      = \left( \frac{d}{d t} P_t \right) 1
-      = \frac{d}{d t} (P_t 1)
-      = \frac{d}{d t} 1
-      = 0
+Q 1
+    = Q P_t 1
+    = \left( \frac{d}{d t} P_t \right) 1
+    = \frac{d}{d t} (P_t 1)
+    = \frac{d}{d t} 1
+    = 0
 $$
 
 Hence $Q$ has zero row sums.
@@ -632,7 +630,7 @@ We can use the definition of the matrix exponential to obtain,
  for any $x, y$ and $t \geq 0$,
 
 $$
-    P_t(x, y) = \mathbb 1\{x = y\} + t Q(x, y) + o(t)
+P_t(x, y) = \mathbb 1\{x = y\} + t Q(x, y) + o(t)
 $$ (otp)
 
 From this equality and the assumption that $P_t$ is a Markov matrix for all
diff --git a/lectures/markov_prop.md b/lectures/markov_prop.md
index 0d96371..72de0d4 100644
--- a/lectures/markov_prop.md
+++ b/lectures/markov_prop.md
@@ -913,6 +913,18 @@ probability $0.5$.
 Construct two different random variables with this distribution.
 ```
 
+```{solution} markov-prop-1
+:class: dropdown
+
+One example is to take $U$ to be uniform on $(0, 1)$ and set $X=0$ if $U <
+0.5$ and $1$ otherwise.
+
+Then $X$ has the desired distribution.
+
+Alternatively, we could take $Z$ to be standard normal and set $X=0$ if $Z <
+0$ and $1$ otherwise.
+```
+
 
 ```{exercise}
 :label: markov-prop-2
@@ -926,57 +938,9 @@ Hints
 * Consider using the [binomial formula](https://en.wikipedia.org/wiki/Binomial_theorem).
 ```
 
-
-```{exercise}
-:label: markov-prop-3
-
-Consider the distribution over $S^{n+1}$ previously shown in {eq}`mathjointd`, which is
-
-$$
-    \mathbf P_\psi^n(x_0, x_1, \ldots, x_n)
-        = \psi(x_0)
-        P(x_0, x_1)
-        \times \cdots \times
-        P(x_{n-1}, x_n)
-$$ 
-
-Show that, for any Markov chain $(X_t)$ satisfying {eq}`markovpropd`
-and $X_0 \sim \psi$, the restriction $(X_0, \ldots, X_n)$ has joint
-distribution $\mathbf P_\psi^n$.
-```
-
-
-```{exercise}
-:label: markov-prop-4
-
-Try to produce your own version of the figure {ref}`flow_fig`
-
-The initial condition is ``ψ_0 = binom.pmf(states, n, 0.25)`` where ``n = b + 1``.
-```
-
-## Solutions
-
-```{note}
-code is currently not supported in `sphinx-exercise`
-so code-cell solutions are immediately after this
-solution block.
-```
-
-```{solution} markov-prop-1
-
-This is easy.
-
-One example is to take $U$ to be uniform on $(0, 1)$ and set $X=0$ if $U <
-0.5$ and $1$ otherwise.
-
-Then $X$ has the desired distribution.
-
-Alternatively, we could take $Z$ to be standard normal and set $X=0$ if $Z <
-0$ and $1$ otherwise.
-```
-
-
 ```{solution} markov-prop-2
+:class: dropdown
+
 Fixing $s, t \in \RR_+$ and $j \leq k$, we have 
 
 $$
@@ -1015,8 +979,27 @@ $$
 Hence {eq}`chapkol_ct2` holds, and the semigroup property is satisfied.
 ```
 
+```{exercise}
+:label: markov-prop-3
+
+Consider the distribution over $S^{n+1}$ previously shown in {eq}`mathjointd`, which is
+
+$$
+\mathbf P_\psi^n(x_0, x_1, \ldots, x_n)
+    = \psi(x_0)
+    P(x_0, x_1)
+    \times \cdots \times
+    P(x_{n-1}, x_n)
+$$ 
+
+Show that, for any Markov chain $(X_t)$ satisfying {eq}`markovpropd`
+and $X_0 \sim \psi$, the restriction $(X_0, \ldots, X_n)$ has joint
+distribution $\mathbf P_\psi^n$.
+```
 
 ```{solution} markov-prop-3
+:class: dropdown
+
 Let $(X_t)$ be a Markov chain satisfying {eq}`markovpropd` and $X_0 \sim \psi$.
 
 When $n=0$, we have $\mathbf P_\psi^n = \mathbf P_\psi^0 = \psi$, and this
@@ -1029,35 +1012,45 @@ defined above.
 Then 
 
 $$
-    \PP \{X_0 = x_0, \ldots, X_n = x_n\}
-    = \PP \{X_n = x_n \,|\, X_0 = x_0, \ldots, X_{n-1} = x_{n-1}  \}
-    \\
-        \times \PP \{X_0 = x_0, \ldots, X_{n-1} = x_{n-1}\}
+\PP \{X_0 = x_0, \ldots, X_n = x_n\}
+= \PP \{X_n = x_n \,|\, X_0 = x_0, \ldots, X_{n-1} = x_{n-1}  \}
+\\
+    \times \PP \{X_0 = x_0, \ldots, X_{n-1} = x_{n-1}\}
 $$
 
 From the Markov property and the induction hypothesis, the right hand side is
 
 $$
+P (x_{n-1}, x_n )
+\mathbf P_\psi^{n-1}(x_0, x_1, \ldots, x_{n-1})
+=
     P (x_{n-1}, x_n )
-    \mathbf P_\psi^{n-1}(x_0, x_1, \ldots, x_{n-1})
-    =
-        P (x_{n-1}, x_n )
-        \psi(x_0)
-        P(x_0, x_1)
-        \times \cdots \times
-        P(x_{n-2}, x_{n-1})
+    \psi(x_0)
+    P(x_0, x_1)
+    \times \cdots \times
+    P(x_{n-2}, x_{n-1})
 $$
 
 The last expression equals $\mathbf P_\psi^n$, which concludes the proof.
 ```
 
 
-```{solution} markov-prop-4
+```{exercise}
+:label: markov-prop-4
+
+Try to produce your own version of the figure {ref}`flow_fig`
+
+The initial condition is ``ψ_0 = binom.pmf(states, n, 0.25)`` where ``n = b + 1``.
+```
+
+```{solution-start} markov-prop-4
+:class: dropdown
+```
+
 Here is one approach.
 
 (The statements involving ``glue`` are specific to this book and can be deleted
 by most readers.  They store the output so it can be displayed elsewhere.)
-```
 
 ```{code-cell} ipython3
 α = 0.6
@@ -1106,3 +1099,6 @@ plt.savefig("_static/lecture_specific/markov_prop/flow_fig.png")
 
 plt.show()
 ```
+
+```{solution-end}
+```
\ No newline at end of file
diff --git a/lectures/memoryless.md b/lectures/memoryless.md
index f92532f..249a9b6 100644
--- a/lectures/memoryless.md
+++ b/lectures/memoryless.md
@@ -413,30 +413,8 @@ In particular, consider the random variable $X$ defined as follows:
 Show that $X \sim \Exp(\lambda)$.
 ```
 
-```{exercise}
-:label: memoryless-ex-2
-
-Fix $\lambda = 0.5$ and $s=1.0$.
-
-Simulate 1,000 draws of $X$ using the algorithm above.
-
-Plot the fraction of the sample exceeding $t$ for each $t \geq 0$ (on a grid)
-and compare to $t \mapsto e^{-\lambda t}$.
-
-Is the fit good?  How about if the number of draws is increased?
-
-Are the results in line with those of the previous exercise?
-```
-
-## Solutions
-
-```{note}
-code is currently not supported in `sphinx-exercise`
-so code-cell solutions are immediately after this
-solution block.
-```
-
 ```{solution} memoryless-ex-1
+:class: dropdown
 
 Let $X$ be constructed as in the statement of the exercise and fix $t > 0$.
 
@@ -445,27 +423,43 @@ Notice that $X > s + t$ if and only if $Y > s$ and $Z > t$.
 As a result of this fact and independence,
 
 $$
-    \PP\{X > s + t\}
-    = \PP\{Y > s \} \PP\{Z > t\}
-    = e^{-\lambda(s + t)}
+\PP\{X > s + t\}
+= \PP\{Y > s \} \PP\{Z > t\}
+= e^{-\lambda(s + t)}
 $$
 
 At the same time, $X > s-t$ if and only if $Y > s-t$, so
 
 $$
-    \PP\{X > s - t\}
-    = \PP\{Y > s - t \} 
-    = e^{-\lambda(s - t)}
+\PP\{X > s - t\}
+= \PP\{Y > s - t \} 
+= e^{-\lambda(s - t)}
 $$
 
 Either way, we have $X \sim \Exp(\lambda)$, as was to be shown.
 ```
 
+```{exercise}
+:label: memoryless-ex-2
 
-```{solution} memoryless-ex-2
-Here's one solution, starting with 1,000 draws.
+Fix $\lambda = 0.5$ and $s=1.0$.
+
+Simulate 1,000 draws of $X$ using the algorithm above.
+
+Plot the fraction of the sample exceeding $t$ for each $t \geq 0$ (on a grid)
+and compare to $t \mapsto e^{-\lambda t}$.
+
+Is the fit good?  How about if the number of draws is increased?
+
+Are the results in line with those of the previous exercise?
 ```
 
+```{solution-start} memoryless-ex-2
+:class: dropdown
+```
+
+Here's one solution, starting with 1,000 draws.
+
 ```{code-cell} ipython3
 λ = 0.5 
 np.random.seed(1234)
@@ -494,24 +488,19 @@ ax.legend()
 plt.show()
 ```
 
-```{solution} memoryless-ex-2
-**Solution Continued:**
-
-The fit is already very close, which matches with the theory in Exercise 1.
+The fit is already very close, which matches with the theory in [](memoryless-ex-1).
 
 The two lines become indistinguishable as $n$ is increased further.
-```
 
 ```{code-cell} ipython3
-
 fig, ax = plt.subplots()
 draws = draw_X(n=10_000)
 empirical_exceedance = [np.mean(draws > t) for t in t_grid]
 ax.plot(t_grid, np.exp(- λ * t_grid), label='exponential exceedance')
 ax.plot(t_grid, empirical_exceedance, label='empirical exceedance')
 ax.legend()
-
 plt.show()
 ```
 
-
+```{solution-end}
+```
diff --git a/lectures/poisson.md b/lectures/poisson.md
index db4aa0a..5ae0dce 100644
--- a/lectures/poisson.md
+++ b/lectures/poisson.md
@@ -433,32 +433,15 @@ distribution with rate $T \lambda$.
 Try first with $\lambda = 0.5$ and $T=10$.
 ```
 
-
-```{exercise}
-:label: poisson-ex-2
-
-In the lecture we used the fact that $\Binomial(n, \theta) \approx \Poisson(n \theta)$ when $n$ is large and $\theta$ is small.
-
-Investigate this relationship by plotting the distributions side by side.
-
-Experiment with different values of $n$ and $\theta$.
-```
-
-## Solutions
-
-```{note}
-code is currently not supported in `sphinx-exercise`
-so code-cell solutions are immediately after this
-solution block.
+```{solution-start} poisson-ex-1
+:class: dropdown
 ```
 
-```{solution} poisson-ex-1
 Here is one solution.
 
 The figure shows that the fit is already good with a modest sample size.
 
 Increasing the sample size will further improve the fit.
-```
 
 ```{code-cell} ipython3
 λ = 0.5
@@ -503,11 +486,26 @@ ax.legend(fontsize=12)
 plt.show()
 ```
 
+```{solution-end}
+```
+
+
+```{exercise}
+:label: poisson-ex-2
+
+In the lecture we used the fact that $\Binomial(n, \theta) \approx \Poisson(n \theta)$ when $n$ is large and $\theta$ is small.
+
+Investigate this relationship by plotting the distributions side by side.
+
+Experiment with different values of $n$ and $\theta$.
+```
+
+```{solution-start} poisson-ex-2
+:class: dropdown
+```
 
-```{solution} poisson-ex-2
 Here is one solution.  It shows that the approximation is good when $n$ is
 large and $\theta$ is small.
-```
 
 ```{code-cell} ipython3
 def binomial(k, n, p):
@@ -534,4 +532,6 @@ fig.tight_layout()
 plt.show()
 ```
 
+```{solution-end}
+```
 
diff --git a/lectures/uc_mc_semigroups.md b/lectures/uc_mc_semigroups.md
index b3a0f51..2fcc02f 100644
--- a/lectures/uc_mc_semigroups.md
+++ b/lectures/uc_mc_semigroups.md
@@ -19,8 +19,6 @@ kernelspec:
 In our previous lecture we covered some of the general theory of operator
 semigroups. 
 
-
-
 Next we translate these results into the setting of Markov semigroups.
 
 The Markov semigroups are defined on a countable set $S$.
@@ -506,52 +504,19 @@ Let $P$ be a Markov matrix on $S$ and identify it with the linear operator in
 {eq}`mmismo`.  Verify the claims in {eq}`propp`.
 ```
 
-```{exercise}
-:label: uc-mc-semigroups-ex-2
-
-Prove the claim in {prf:ref}`scintcon`.
-```
-
-```{exercise}
-:label: uc-mc-semigroups-ex-3
-
-Confirm that $Q$ defined in {eq}`poissonq` induces a bounded linear operator on
-$\ell_1$ via {eq}`imislo`.
-```
-
-```{exercise}
-:label: uc-mc-semigroups-ex-4
-
-Let $K$ be defined on $\ZZ_+ \times \ZZ_+$ by $K(i, j) = \mathbb 1\{j = i + 1\}$. 
-
-Show that, with $K^m$ representing the $m$-th matrix product of $K$ with itself, 
-we have $K^m(i, j) = \mathbb 1\{j = i + m\}$ for any $i, j \in \ZZ_+$.
-```
-
-```{exercise}
-:label: uc-mc-semigroups-ex-5
-
-Let $Q$ be any intensity matrix on $S$.
-
-Prove that the jump chain decomposition of $Q$ is in fact a jump chain pair.
-
-Prove that, in addition, this decomposition $(\lambda, K)$ satisfies {eq}`jcinmat`.
-```
-
-
-## Solutions
-
 ```{solution} uc-mc-semigroups-ex-1
+:class: dropdown
+
 To determine the norm of $P$, we use the definition in {eq}`norml`.
 
 If $f \in \ell_1$ and $\| f \| \leq 1$, then 
 
 $$
-    \| f P \| 
-    \leq \sum_y \sum_x |f(x)| P(x, y)
-    = \sum_x |f(x)| \sum_y P(x, y)
-    = \sum_x |f(x)| 
-    = \| f \|
+\| f P \| 
+\leq \sum_y \sum_x |f(x)| P(x, y)
+= \sum_x |f(x)| \sum_y P(x, y)
+= \sum_x |f(x)| 
+= \| f \|
 $$
 
 Hence $\| P \| \leq 1$.  
@@ -564,17 +529,24 @@ Now pick any $\phi \in \dD$.
 Clearly $\phi P \geq 0$, and 
 
 $$
-    \sum_y (\phi P)(y)
-    =\sum_y \sum_x \phi (x) P(x, y)
-    =\sum_x \phi (x) \sum_y P(x, y)
-    = 1
+\sum_y (\phi P)(y)
+=\sum_y \sum_x \phi (x) P(x, y)
+=\sum_x \phi (x) \sum_y P(x, y)
+= 1
 $$
 
 Hence $\phi P \in \dD$ as claimed.
 ```
 
+```{exercise}
+:label: uc-mc-semigroups-ex-2
+
+Prove the claim in {prf:ref}`scintcon`.
+```
 
 ```{solution} uc-mc-semigroups-ex-2
+:class: dropdown
+
 Here is one solution.
 
 Let $Q$ be an intensity matrix on $S$.
@@ -603,27 +575,34 @@ Let $f \in \ell_1$ be defined by $f(z) = \mathbb 1\{z = x\}$.
 Since $\|f\| = 1$, we have
 
 $$
-    \| Q \| 
-    \geq \| f Q \|
-    = \sum_y \left| \sum_z f(z) Q(z, y) \right|
-    = \sum_y | Q(x, y) |
-    \geq | Q(x, x) |
+\| Q \| 
+\geq \| f Q \|
+= \sum_y \left| \sum_z f(z) Q(z, y) \right|
+= \sum_y | Q(x, y) |
+\geq | Q(x, x) |
 $$
 
 Contradiction.
 ```
 
+```{exercise}
+:label: uc-mc-semigroups-ex-3
+
+Confirm that $Q$ defined in {eq}`poissonq` induces a bounded linear operator on
+$\ell_1$ via {eq}`imislo`.
+```
 
 ```{solution} uc-mc-semigroups-ex-3
+:class: dropdown
 
 Linearity is obvious so we focus on boundedness.
 
 For any $f \in \ell_1$ and this choice of $Q$, we have
 
 $$
-    \sum_y |(fQ)(y)|
-    \leq \sum_y \sum_x |f(x) Q(x, y)|
-    \leq \lambda \sum_y \sum_x |f(y) - f(y+1)|
+\sum_y |(fQ)(y)|
+\leq \sum_y \sum_x |f(x) Q(x, y)|
+\leq \lambda \sum_y \sum_x |f(y) - f(y+1)|
 $$
 
 Applying the triangle inequality, we see that the right hand side is dominated
@@ -634,7 +613,17 @@ as required.
 ```
 
 
+```{exercise}
+:label: uc-mc-semigroups-ex-4
+
+Let $K$ be defined on $\ZZ_+ \times \ZZ_+$ by $K(i, j) = \mathbb 1\{j = i + 1\}$. 
+
+Show that, with $K^m$ representing the $m$-th matrix product of $K$ with itself, 
+we have $K^m(i, j) = \mathbb 1\{j = i + m\}$ for any $i, j \in \ZZ_+$.
+```
+
 ```{solution} uc-mc-semigroups-ex-4
+:class: dropdown
 
 The statement $K^m(i, j) = \mathbb 1\{j = i + m\}$ holds by definition when
 $m=1$.
@@ -653,8 +642,18 @@ $$
 Applying the definition $K(i, j) = \mathbb 1\{j = i + 1\}$ completes verification of the claim.
 ```
 
+```{exercise}
+:label: uc-mc-semigroups-ex-5
+
+Let $Q$ be any intensity matrix on $S$.
+
+Prove that the jump chain decomposition of $Q$ is in fact a jump chain pair.
+
+Prove that, in addition, this decomposition $(\lambda, K)$ satisfies {eq}`jcinmat`.
+```
 
 ```{solution} uc-mc-semigroups-ex-5
+:class: dropdown
 
 Let $Q$ be an intensity matrix and let $(\lambda, K)$ be the jump chain
 decomposition of $Q$.
@@ -668,10 +667,10 @@ $\lambda(x) > 0$.
 Then 
 
 $$
-    \sum_y K(x, y) 
-    = \sum_{y \not= x} K(x,y) 
-    = \sum_{y \not= x} \frac{Q(x,y)}{\lambda(x)}
-    = 1
+\sum_y K(x, y) 
+= \sum_{y \not= x} K(x,y) 
+= \sum_{y \not= x} \frac{Q(x,y)}{\lambda(x)}
+= 1
 $$
 
 If, on the other hand, $\lambda(x) = 0$, then