Humphrey's 20 Mar updates on math aligned syntax

Author: HumphreyYang

lectures/ak2.md

@@ -219,10 +219,10 @@ $$
 To maximize profits a firm equates marginal products to rental rates:
 
 $$
-\begin{align}
+\begin{aligned}
 W_t & = (1-\alpha) K_t^\alpha L_t^{-\alpha} \\
 r_t & = \alpha K_t^\alpha L_t^{1-\alpha}
-\end{align}
+\end{aligned}
 $$  (eq:firmfonc)
 
 Output can either be consumed by old or young households, sold to young households who use it  to augment the capital stock, or  sold to  the government for  uses that do not generate utility for the people in the model  (e.g., ``thrown into the ocean'').  
@@ -275,10 +275,10 @@ $$ (eq:utilfn)
 subject to the following  budget constraints at times $t$ and $t+1$:
 
 $$
-\begin{align}
+\begin{aligned}
 C_{yt} + A_{t+1} & =  W_t (1 - \tau_t) - \delta_{yt} \\
 C_{ot+1} & = (1+ r_{t+1} (1 - \tau_{t+1}))A_{t+1} - \delta_{ot}
-\end{align}
+\end{aligned}
 $$ (eq:twobudgetc)
 
 
@@ -291,9 +291,9 @@ $$ (eq:onebudgetc)
 To solve the young household's choice problem, form a Lagrangian 
 
 $$ 
-\begin{align}
+\begin{aligned}
 {\mathcal L}  & = C_{yt}^\beta C_{o,t+1}^{1-\beta} \\ &  + \lambda \Bigl[ C_{yt} + \frac{C_{ot+1}}{1 + r_{t+1}(1 - \tau_{t+1})} - W_t (1 - \tau_t) + \delta_{yt} + \frac{\delta_{ot}}{1 + r_{t+1}(1 - \tau_{t+1})}\Bigr],
-\end{align}
+\end{aligned}
 $$ (eq:lagC)
 
 where $\lambda$ is a Lagrange multiplier on the intertemporal budget constraint {eq}`eq:onebudgetc`.
@@ -303,10 +303,10 @@ After several lines of algebra, the intertemporal budget constraint {eq}`eq:oneb
 imply that an optimal consumption plan satisfies
 
 $$
-\begin{align}
+\begin{aligned}
 C_{yt} & = \beta \Bigl[ W_t (1 - \tau_t) - \delta_{yt} - \frac{\delta_{ot}}{1 + r_{t+1}(1 - \tau_{t+1})}\Bigr] \\
 \frac{C_{0t+1}}{1 + r_{t+1}(1-\tau_{t+1})  } & = (1-\beta)   \Bigl[ W_t (1 - \tau_t) - \delta_{yt} - \frac{\delta_{ot}}{1 + r_{t+1}(1 - \tau_{t+1})}\Bigr] 
-\end{align}
+\end{aligned}
 $$ (eq:optconsplan)
 
 The first-order condition for minimizing Lagrangian {eq}`eq:lagC` with respect to the Lagrange multipler $\lambda$ recovers the budget constraint {eq}`eq:onebudgetc`,
@@ -352,10 +352,10 @@ As our special case of {eq}`eq:optconsplan`, we compute the following consumptio
 
 
 $$
-\begin{align}
+\begin{aligned}
 C_{yt} & = \beta (1 - \tau_t) W_t \\
 A_{t+1} &= (1-\beta) (1- \tau_t) W_t
-\end{align}
+\end{aligned}
 $$
 
 Using  {eq}`eq:firmfonc` and  $A_t = K_t + D_t$, we obtain the following closed form transition law for capital:
@@ -369,20 +369,20 @@ $$ (eq:Klawclosed)
 From {eq}`eq:Klawclosed` and the government budget constraint {eq}`eq:govbudgetsequence`, we compute **time-invariant** or **steady state values**   $\hat K, \hat D, \hat T$:
 
 $$
-\begin{align}
+\begin{aligned}
 \hat{K} &=\hat{K}\left(1-\hat{\tau}\right)\left(1-\alpha\right)\left(1-\beta\right) - \hat{D} \\
 \hat{D} &= (1 + \hat{r})  \hat{D} + \hat{G} - \hat{T} \\
 \hat{T} &= \hat{\tau} \hat{Y} + \hat{\tau} \hat{r} \hat{D} .
-\end{align}
+\end{aligned}
 $$ (eq:steadystates)
 
 These imply
 
 $$
-\begin{align}
+\begin{aligned}
 \hat{K} &= \left[\left(1-\hat{\tau}\right)\left(1-\alpha\right)\left(1-\beta\right)\right]^{\frac{1}{1-\alpha}} \\
 \hat{\tau} &= \frac{\hat{G} + \hat{r} \hat{D}}{\hat{Y} + \hat{r} \hat{D}}
-\end{align}
+\end{aligned}
 $$
 
 Let's take an example in which
@@ -393,11 +393,11 @@ Let's take an example in which
 Our formulas for steady-state values  tell us that
 
 $$
-\begin{align}
+\begin{aligned}
 \hat{D} &= 0 \\
 \hat{G} &= 0.15 \hat{Y} \\
 \hat{\tau} &= 0.15 \\
-\end{align}
+\end{aligned}
 $$
 
 
@@ -699,12 +699,12 @@ To illustrate the power of `ClosedFormTrans`, let's first experiment with the fo
 The following equations completely characterize the equilibrium transition path originating from the initial steady state
 
 $$
-\begin{align}
+\begin{aligned}
 K_{t+1} &= K_{t}^{\alpha}\left(1-\tau_{t}\right)\left(1-\alpha\right)\left(1-\beta\right) - \bar{D} \\
 \tau_{0} &= (1-\frac{1}{3}) \hat{\tau} \\
 \bar{D} &= \hat{G} - \tau_0\hat{Y} \\
 \quad\tau_{t} & =\frac{\hat{G}+r_{t} \bar{D}}{\hat{Y}+r_{t} \bar{D}}
-\end{align}
+\end{aligned}
 $$
 
 We can simulate the transition of the economy for $20$ periods, after which the economy will be fairly close to the new steady state.

lectures/money_inflation.md

@@ -192,21 +192,21 @@ This is true in the present model.
 
 In a **steady state** equilibrium of the  model we are studying, 
 
-\begin{align}
+\begin{aligned}
 R_t & = \bar R \cr
 b_t & = \bar b
-\end{align}
+\end{aligned}
 
 for $t \geq 0$.  
 
 Notice that both $R_t = \frac{p_t}{p_{t+1}}$ and $b_t = \frac{m_{t+1}}{p_t} $ are **ratios**.
 
 To compute a steady state, we seek gross rates of return on currency  $\bar R, \bar b$ that satisfy steady-state versions of  both the government budget constraint and the demand function for real balances:
 
-\begin{align}
+\begin{aligned}
 g & = \bar b ( 1 - \bar R)  \cr
 \bar b & = \gamma_1- \gamma_2 \bar R^{-1}
-\end{align}
+\end{aligned}
 
 Together these equations  imply
 
@@ -379,10 +379,10 @@ We shall  deploy two distinct computation strategies.
    * set $R_0 \in [\frac{\gamma_2}{\gamma_1}, R_u]$ and compute $b_0 = \gamma_1 - \gamma_2/R_0$.
 
    * compute sequences $\{R_t, b_t\}_{t=1}^\infty$ of rates of return and real balances that are associated with an equilibrium by solving equation {eq}`eq:bmotion` and {eq}`eq:bdemand` sequentially  for $t \geq 1$:  
-   \begin{align}
+   \begin{aligned}
 b_t & = b_{t-1} R_{t-1} + g \cr
 R_t^{-1} & = \frac{\gamma_1}{\gamma_2} - \gamma_2^{-1} b_t 
-\end{align}
+\end{aligned}
 
    * Construct the associated equilibrium $p_0$ from 
   
@@ -393,10 +393,10 @@ R_t^{-1} & = \frac{\gamma_1}{\gamma_2} - \gamma_2^{-1} b_t
    * compute $\{p_t, m_t\}_{t=1}^\infty$  by solving the following equations sequentially
   
   $$
-   \begin{align}
+   \begin{aligned}
    p_t & = R_t p_{t-1} \cr
    m_t & = b_{t-1} p_t 
-   \end{align}
+   \end{aligned}
   $$ (eq:method1) 
    
 **Remark 1:** method 1 uses an indirect approach to computing an equilibrium by first computing an equilibrium  $\{R_t, b_t\}_{t=0}^\infty$ sequence and then using it to back out an equilibrium  $\{p_t, m_t\}_{t=0}^\infty$  sequence.
@@ -458,10 +458,10 @@ Start at $t=0$
 
 Then  for $t \geq 1$ construct $(b_t, R_t)$ by
 iterating  on the system 
-\begin{align}
+\begin{aligned}
 b_t & = b_{t-1} R_{t-1} + g \cr
 R_t^{-1} & = \frac{\gamma_1}{\gamma_2} - \gamma_2^{-1} b_t
-\end{align}
+\end{aligned}
 
 
 When we implement this part of method 1, we shall discover the following  striking 
@@ -605,11 +605,11 @@ $$
 
 where 
 
-\begin{align} H_1 & = \begin{bmatrix} 1 & \gamma_2 \cr
+\begin{aligned} H_1 & = \begin{bmatrix} 1 & \gamma_2 \cr
                  1 & 0 \end{bmatrix} \cr
                 H_2 & = \begin{bmatrix} 0 & \gamma_1 \cr
                  1 & g \end{bmatrix}  
-\end{align}
+\end{aligned}
 
 ```{code-cell} ipython3
 H1 = np.array([[1, msm.γ2], 

lectures/unpleasant.md

@@ -142,19 +142,19 @@ Just before time $0$, the government chooses $(m_0, B_{-1})$  subject to constra
 For $t =0, 1, \ldots, T-1$,
 
 $$
-\begin{align}
+\begin{aligned}
 B_t & = \widetilde R B_{t-1} + g \cr
 m_{t+1} &  = m_0 
-\end{align}
+\end{aligned}
 $$
 
 while for $t \geq T$,
 
 $$
-\begin{align}
+\begin{aligned}
 B_t & = B_{T-1} \cr
 m_{t+1} & = m_t + p_t \overline g
-\end{align}
+\end{aligned}
 $$
 
 where 
@@ -188,21 +188,21 @@ For reasons described at the end of **this lecture**, we select the larger root
 Next, we compute
 
 $$
-\begin{align}
+\begin{aligned}
 R_T & = R_u \cr
 b_T & = \gamma_1 - \gamma_2 R_u^{-1} \cr
 p_T & = \frac{m_0}{\gamma_1 - \overline g - \gamma_2 R_u^{-1}}
-\end{align}
+\end{aligned}
 $$ (eq:LafferTstationary)
 
 
 We can compute continuation sequences $\{R_t, b_t\}_{t=T+1}^\infty$ of rates of return and real balances that are associated with an equilibrium by solving equation {eq}`eq:up_bmotion` and {eq}`eq:up_bdemand` sequentially  for $t \geq 1$:  
-   \begin{align}
+   \begin{aligned}
 b_t & = b_{t-1} R_{t-1} + \overline g \cr
 R_t^{-1} & = \frac{\gamma_1}{\gamma_2} - \gamma_2^{-1} b_t \cr
 p_t & = R_t p_{t-1} \cr
    m_t & = b_{t-1} p_t 
-\end{align}
+\end{aligned}
 
    
 
@@ -219,22 +219,22 @@ Our restrictions that $\gamma_1 > \gamma_2 > 0$ imply that $\lambda \in [0,1)$.
 We want to compute
 
 $$ 
-\begin{align}
+\begin{aligned}
 p_0 &  = \gamma_1^{-1} \left[ \sum_{j=0}^\infty \lambda^j m_{1+j} \right] \cr
 & = \gamma_1^{-1} \left[ \sum_{j=0}^{T-1} \lambda^j m_{0} + \sum_{j=T}^\infty \lambda^j m_{1+j} \right]
-\end{align}
+\end{aligned}
 $$
 
 Thus,
 
 $$
-\begin{align}
+\begin{aligned}
 p_0 & = \gamma_1^{-1} m_0  \left\{ \frac{1 - \lambda^T}{1-\lambda} +  \frac{\lambda^T}{R_u-\lambda}   \right\} \cr
 p_1 & = \gamma_1^{-1} m_0  \left\{ \frac{1 - \lambda^{T-1}}{1-\lambda} +  \frac{\lambda^{T-1}}{R_u-\lambda}   \right\} \cr
 \quad \vdots  & \quad \quad \vdots \cr
 p_{T-1} & = \gamma_1^{-1} m_0  \left\{ \frac{1 - \lambda}{1-\lambda} +  \frac{\lambda}{R_u-\lambda}   \right\}  \cr
 p_T & = \gamma_1^{-1} m_0  \left\{\frac{1}{R_u-\lambda}   \right\}
-\end{align}
+\end{aligned}
 $$ (eq:allts)
 
 We can implement  the preceding formulas by iterating on