Update lecture-python sources with latest updates

lectures/ar1_processes.md

@@ -109,7 +109,7 @@ series $\{ X_t\}$.
 
 To see this, we first note that $X_t$ is normally distributed for each $t$.
 
-This is immediate form {eq}`ar1_ma`, since linear combinations of independent
+This is immediate from {eq}`ar1_ma`, since linear combinations of independent
 normal random variables are normal.
 
 Given that $X_t$ is normally distributed, we will know the full distribution
@@ -212,7 +212,7 @@ In fact it's easy to show that such convergence will occur, regardless of the in
 To see this, we just have to look at the dynamics of the first two moments, as
 given in {eq}`dyn_tm`.
 
-When $|a| < 1$, these sequence converge to the respective limits
+When $|a| < 1$, these sequences converge to the respective limits
 
 ```{math}
 :label: mu_sig_star

lectures/cake_eating_numerical.md

@@ -100,7 +100,7 @@ Let's write this a bit more mathematically.
 ### The Bellman Operator
 
 We introduce the **Bellman operator** $T$ that takes a function v as an
-argument and returns a new function $Tv$ defined by.
+argument and returns a new function $Tv$ defined by
 
 $$
 Tv(x) = \max_{0 \leq c \leq x} \{u(c) + \beta v(x - c)\}
@@ -118,7 +118,7 @@ v$ converges to the solution to the Bellman equation.
 
 ### Fitted Value Function Iteration
 
-Both consumption $c$ and the state variable $x$ are continous.
+Both consumption $c$ and the state variable $x$ are continuous.
 
 This causes complications when it comes to numerical work.
 
@@ -419,7 +419,7 @@ ax.legend()
 plt.show()
 ```
 
-The fit is reasoable but not perfect.
+The fit is reasonable but not perfect.
 
 We can improve it by increasing the grid size or reducing the
 error tolerance in the value function iteration routine.
@@ -509,7 +509,7 @@ modification in the exercise above).
 
 ### Exercise 1
 
-We need to create a class to hold our primitives and return the right hand side of the bellman equation.
+We need to create a class to hold our primitives and return the right hand side of the Bellman equation.
 
 We will use [inheritance](https://en.wikipedia.org/wiki/Inheritance_%28object-oriented_programming%29) to maximize code reuse.
 

lectures/cass_koopmans_1.md

@@ -26,14 +26,14 @@ kernelspec:
 
 ## Overview
 
-This lecture and in {doc}`Cass-Koopmans Competitive Equilibrium <cass_koopmans_2>` describe a model that Tjalling Koopmans {cite}`Koopmans`
+This lecture and lecture {doc}`Cass-Koopmans Competitive Equilibrium <cass_koopmans_2>` describe a model that Tjalling Koopmans {cite}`Koopmans`
 and David Cass {cite}`Cass` used to analyze optimal growth.
 
 The model can be viewed as an extension of the model of Robert Solow
 described in [an earlier lecture](https://lectures.quantecon.org/py/python_oop.html)
 but adapted to make the saving rate the outcome of an optimal choice.
 
-(Solow assumed a constant saving rate determined outside the model).
+(Solow assumed a constant saving rate determined outside the model.)
 
 We describe two versions of the model, one in this lecture and the other in {doc}`Cass-Koopmans Competitive Equilibrium <cass_koopmans_2>`.
 
@@ -696,7 +696,7 @@ its steady state value most of the time.
 plot_paths(pp, 0.3, k_ss/3, [250, 150, 50, 25], k_ss=k_ss);
 ```
 
-Different colors in the above graphs are associated
+Different colors in the above graphs are associated with
 different horizons $T$.
 
 Notice that as the horizon increases, the planner puts $K_t$

lectures/cass_koopmans_2.md

@@ -397,7 +397,7 @@ verify** approach.
 In this lecture {doc}`Cass-Koopmans Planning Model <cass_koopmans_1>`, we  computed an allocation $\{\vec{C}, \vec{K}, \vec{N}\}$
 that solves the planning problem.
 
-(This allocation will constitute the **Big** $K$  to be in the presence instance of the *Big** $K$ **, little** $k$ trick
+(This allocation will constitute the **Big** $K$  to be in the present instance of the *Big** $K$ **, little** $k$ trick
 that we'll apply to  a competitive equilibrium in the spirit of [this lecture](https://lectures.quantecon.org/py/rational_expectations.html#)
 and  [this lecture](https://lectures.quantecon.org/py/dyn_stack.html#).)
 
@@ -597,7 +597,7 @@ representative household living in a competitive equilibrium.
 We now turn to  the problem faced by a firm in a competitive
 equilibrium:
 
-If we plug in {eq}`eq-pl` into {eq}`Zero-profits` for all t, we
+If we plug {eq}`eq-pl` into {eq}`Zero-profits` for all t, we
 get
 
 $$

lectures/finite_markov.md

@@ -603,7 +603,7 @@ We'll come back to this a bit later.
 
 ### Aperiodicity
 
-Loosely speaking, a Markov chain is called periodic if it cycles in a predictible way, and aperiodic otherwise.
+Loosely speaking, a Markov chain is called periodic if it cycles in a predictable way, and aperiodic otherwise.
 
 Here's a trivial example with three states
 
@@ -771,7 +771,7 @@ with the unit eigenvalue $\lambda = 1$.
 
 A more stable and sophisticated algorithm is implemented in [QuantEcon.py](http://quantecon.org/quantecon-py).
 
-This is the one we recommend you use:
+This is the one we recommend you to use:
 
 ```{code-cell} python3
 P = [[0.4, 0.6],
@@ -1023,7 +1023,7 @@ A topic of interest for economics and many other disciplines is *ranking*.
 Let's now consider one of the most practical and important ranking problems
 --- the rank assigned to web pages by search engines.
 
-(Although the problem is motivated from outside of economics, there is in fact a deep connection between search ranking systems and prices in certain competitive equilibria --- see {cite}`DLP2013`)
+(Although the problem is motivated from outside of economics, there is in fact a deep connection between search ranking systems and prices in certain competitive equilibria --- see {cite}`DLP2013`.)
 
 To understand the issue, consider the set of results returned by a query to a web search engine.
 

lectures/heavy_tails.md

@@ -186,7 +186,7 @@ where $\mu := \mathbb E X_i = \int x F(x)$ is the common mean of the sample.
 The condition $\mathbb E | X_i | = \int |x| F(x) < \infty$ holds
 in most cases but can fail if the distribution $F$ is very heavy tailed.
 
-For example, it fails for the Cauchy distribution
+For example, it fails for the Cauchy distribution.
 
 Let's have a look at the behavior of the sample mean in this case, and see
 whether or not the LLN is still valid.
@@ -590,7 +590,7 @@ $$
 2^{1/\alpha} = \exp(\mu)
 $$
 
-which we solve for $\mu$ and $\sigma$ given $\alpha = 1.05$
+which we solve for $\mu$ and $\sigma$ given $\alpha = 1.05$.
 
 Here is code that generates the two samples, produces the violin plot and
 prints the mean and standard deviation of the two samples.

lectures/ifp.md

@@ -48,7 +48,7 @@ model <optgrowth>` and yet differs in important ways.
 
 For example, the choice problem for the agent includes an additive income term that leads to an occasionally binding constraint.
 
-Moreover, in this and the following lectures, we will inject more realisitic
+Moreover, in this and the following lectures, we will inject more realistic
 features such as correlated shocks.
 
 To solve the model we will use Euler equation based time iteration, which proved
@@ -194,7 +194,7 @@ strict inequality $u' (c_t) > \beta R \,  \mathbb{E}_t  u'(c_{t+1})$
 can occur because $c_t$ cannot increase sufficiently to attain equality.
 
 (The lower boundary case $c_t = 0$ never arises at the optimum because
-$u'(0) = \infty$)
+$u'(0) = \infty$.)
 
 With some thought, one can show that {eq}`ee00` and {eq}`ee01` are
 equivalent to
@@ -409,8 +409,7 @@ Next we provide a function to compute the difference
 ```{math}
 :label: euler_diff_eq
 
-u'(c)
-- \max \left\{
+u'(c) - \max \left\{
            \beta R \, \mathbb E_z (u' \circ \sigma) \,
            [R (a - c) + \hat Y, \, \hat Z]
            \, , \;
@@ -629,7 +628,7 @@ shocks.
 Your task is to investigate how this measure of aggregate capital varies with
 the interest rate.
 
-Following tradition, put the price (i.e., interest rate) is on the vertical axis.
+Following tradition, put the price (i.e., interest rate) on the vertical axis.
 
 On the horizontal axis put aggregate capital, computed as the mean of the
 stationary distribution given the interest rate.

lectures/ifp_advanced.md

@@ -250,7 +250,7 @@ It can be shown that
 
 We now have a clear path to successfully approximating the optimal policy:
 choose some $\sigma \in \mathscr C$ and then iterate with $K$ until
-convergence (as measured by the distance $\rho$)
+convergence (as measured by the distance $\rho$).
 
 ### Using an Endogenous Grid
 
@@ -325,7 +325,7 @@ $$
 L(z, \hat z) := P(z, \hat z) \int R(\hat z, x) \phi(x) dx
 $$
 
-This indentity is proved in {cite}`ma2020income`, where $\phi$ is the
+This identity is proved in {cite}`ma2020income`, where $\phi$ is the
 density of the innovation $\zeta_t$ to returns on assets.
 
 (Remember that $\mathsf Z$ is a finite set, so this expression defines a matrix.)
@@ -618,7 +618,7 @@ For example, we will pass in the solutions `a_star, σ_star` along with
 `ifp`, even though it would be more natural to just pass in `ifp` and then
 solve inside the function.
 
-The reason we do this is because `solve_model_time_iter` is not
+The reason we do this is that `solve_model_time_iter` is not
 JIT-compiled.
 
 ```{code-cell} python3

lectures/inventory_dynamics.md

@@ -34,7 +34,7 @@ follow so-called s-S inventory dynamics.
 Such firms
 
 1. wait until inventory falls below some level $s$ and then
-1. order sufficent quantities to bring their inventory back up to capacity $S$.
+1. order sufficient quantities to bring their inventory back up to capacity $S$.
 
 These kinds of policies are common in practice and also optimal in certain circumstances.
 
@@ -176,7 +176,7 @@ fixed $T$.
 We will do this by generating many draws of $X_T$ given initial
 condition $X_0$.
 
-With these draws of $X_T$ we can build up a picture of its distribution $\psi_T$
+With these draws of $X_T$ we can build up a picture of its distribution $\psi_T$.
 
 Here's one visualization, with $T=50$.
 

lectures/jv.md

@@ -223,7 +223,7 @@ class JVWorker:
 ```
 
 The function `operator_factory` takes an instance of this class and returns a
-jitted version of the Bellman operator `T`, ie.
+jitted version of the Bellman operator `T`, i.e.
 
 $$
 Tv(x)

lectures/kalman.md

@@ -499,7 +499,7 @@ Conditions under which a fixed point exists and the sequence $\{\Sigma_t\}$ conv
 
 A sufficient (but not necessary) condition is that all the eigenvalues $\lambda_i$ of $A$ satisfy $|\lambda_i| < 1$ (cf. e.g., {cite}`AndersonMoore2005`, p. 77).
 
-(This strong condition assures that the unconditional  distribution of $x_t$  converges as $t \rightarrow + \infty$)
+(This strong condition assures that the unconditional  distribution of $x_t$  converges as $t \rightarrow + \infty$.)
 
 In this case, for any initial choice of $\Sigma_0$ that is both non-negative and symmetric, the sequence $\{\Sigma_t\}$ in {eq}`kalman_sdy` converges to a non-negative symmetric matrix $\Sigma$ that solves {eq}`kalman_dare`.
 

lectures/kesten_processes.md

@@ -496,7 +496,7 @@ s_{t+1} = e_{t+1} \mathbb{1}\{s_t < \bar s\} +
 
 Here
 
-* the state variable $s_t$ is represents productivity (which is a proxy
+* the state variable $s_t$ represents productivity (which is a proxy
   for output and hence firm size),
 * the IID sequence $\{ e_t \}$ is thought of as a productivity draw for a new
   entrant and

lectures/likelihood_ratio_process.md

@@ -254,7 +254,7 @@ But it would be too challenging for us to that  here simply by applying a standa
 
 The reason is that the distribution of $L\left(w^{t}\right)$ is extremely skewed for large values of  $t$.
 
-Because the probabilty density in the right tail is close to $0$,  it just takes too much computer time to sample enough points from the right tail.
+Because the probability density in the right tail is close to $0$,  it just takes too much computer time to sample enough points from the right tail.
 
 Instead, the following code just illustrates that the unconditional means of $l(w_t)$ are $1$.
 
@@ -498,7 +498,7 @@ Notice that as $t$ increases, we are assured a larger probability
 of detection and a smaller probability of false alarm associated with
 a given discrimination threshold $c$.
 
-As $t \rightarrow + \infty$, we approach the the perfect detection
+As $t \rightarrow + \infty$, we approach the perfect detection
 curve that is indicated by a right angle hinging on the green dot.
 
 For a given sample size $t$, a value discrimination threshold $c$ determines a point on the receiver operating

lectures/linear_algebra.md

@@ -1290,7 +1290,7 @@ $$
 (Q + B'PB)u + B'PAx = 0
 $$
 
-which is the first-order condition for maximizing L w.r.t. u.
+which is the first-order condition for maximizing $L$ w.r.t. $u$.
 
 Thus, the optimal choice of u must satisfy
 

lectures/lln_clt.md

@@ -385,7 +385,7 @@ To this end, we now perform the following simulation
 Here's some code that does exactly this for the exponential distribution
 $F(x) = 1 - e^{- \lambda x}$.
 
-(Please experiment with other choices of $F$, but remember that, to conform with the conditions of the CLT, the distribution must have a finite second moment)
+(Please experiment with other choices of $F$, but remember that, to conform with the conditions of the CLT, the distribution must have a finite second moment.)
 
 (sim_one)=
 ```{code-cell} python3
@@ -437,7 +437,7 @@ random variable, the distribution of $Y_n$ will smooth out into a bell-shaped cu
 The next figure shows this process for $X_i \sim f$, where $f$ was
 specified as the convex combination of three different beta densities.
 
-(Taking a convex combination is an easy way to produce an irregular shape for $f$)
+(Taking a convex combination is an easy way to produce an irregular shape for $f$.)
 
 In the figure, the closest density is that of $Y_1$, while the furthest is that of
 $Y_5$
@@ -650,7 +650,7 @@ n \to \infty
 
 This theorem is used frequently in statistics to obtain the asymptotic distribution of estimators --- many of which can be expressed as functions of sample means.
 
-(These kinds of results are often said to use the "delta method")
+(These kinds of results are often said to use the "delta method".)
 
 The proof is based on a Taylor expansion of $g$ around the point $\mu$.
 
@@ -741,7 +741,7 @@ n \| \mathbf Q ( \bar{\mathbf X}_n - \boldsymbol \mu ) \|^2
 where $\chi^2(k)$ is the chi-squared distribution with $k$ degrees
 of freedom.
 
-(Recall that $k$ is the dimension of $\mathbf X_i$, the underlying random vectors)
+(Recall that $k$ is the dimension of $\mathbf X_i$, the underlying random vectors.)
 
 Your second exercise is to illustrate the convergence in {eq}`lln_ctc` with a simulation.
 

lectures/mccall_fitted_vfi.md

@@ -320,7 +320,7 @@ The exercises ask you to explore the solution and how it changes with parameters
 Use the code above to explore what happens to the reservation wage when the wage parameter $\mu$
 changes.
 
-Use the default parameters and $\mu$ in `mu_vals = np.linspace(0.0, 2.0, 15)`
+Use the default parameters and $\mu$ in `mu_vals = np.linspace(0.0, 2.0, 15)`.
 
 Is the impact on the reservation wage as you expected?
 
@@ -338,7 +338,7 @@ support.
 
 Use `s_vals = np.linspace(1.0, 2.0, 15)` and `m = 2.0`.
 
-State how you expect the reservation wage vary with $s$.
+State how you expect the reservation wage to vary with $s$.
 
 Now compute it.  Is this as you expected?
 

lectures/mccall_model.md

@@ -296,7 +296,7 @@ Step 4: if the deviation is larger than some fixed tolerance, set $v = v'$ and g
 
 Step 5: return $v$.
 
-Let $\{ v_k \}$ denote the sequence genererated by this algorithm.
+Let $\{ v_k \}$ denote the sequence generated by this algorithm.
 
 This sequence converges to the solution
 to {eq}`odu_pv2` as $k \to \infty$, which is the value function $v^*$.
@@ -321,7 +321,7 @@ itself via
 ```
 
 (A new vector $Tv$ is obtained from given vector $v$ by evaluating
-the r.h.s. at each $i$)
+the r.h.s. at each $i$.)
 
 The element $v_k$ in the sequence $\{v_k\}$ of successive
 approximations corresponds to $T^k v$.

lectures/mccall_model_with_separation.md

@@ -124,7 +124,7 @@ If he rejects, then he receives unemployment compensation $c$.
 
 The process then repeats.
 
-(Note: we do not allow for job search while employed---this topic is taken up in a {doc}`later lecture <jv>`)
+(Note: we do not allow for job search while employed---this topic is taken up in a {doc}`later lecture <jv>`.)
 
 ## Solving the Model
 

lectures/multi_hyper.md

@@ -77,7 +77,7 @@ $$
 To evaluate whether the selection procedure is **color blind** the administrator wants to  study whether the particular realization of $X$ drawn can plausibly
 be said to be a random draw from the probability distribution that is implied by the **color blind** hypothesis.
 
-The appropriate probability distribution is the one described [here](https://en.wikipedia.org/wiki/Hypergeometric_distribution)
+The appropriate probability distribution is the one described [here](https://en.wikipedia.org/wiki/Hypergeometric_distribution).
 
 Let's now instantiate the administrator's problem, while continuing to use the colored balls metaphor.