diff --git a/lectures/calvo_machine_learn.md b/lectures/calvo_machine_learn.md index 282b368d..5cd967f8 100644 --- a/lectures/calvo_machine_learn.md +++ b/lectures/calvo_machine_learn.md @@ -15,7 +15,7 @@ kernelspec: ## Introduction -This lecture studies a problem that we also study in another quantecon lecture +This lecture studies a problem that we shall study from another angle in another quantecon lecture {doc}`calvo`. That lecture used an analytic approach based on ``dynamic programming squared`` to guide computation of a Ramsey plan in a version of a model of Calvo {cite}`Calvo1978`. @@ -172,7 +172,7 @@ U(m_t - p_t) = u_0 + u_1 (m_t - p_t) - \frac{u_2}{2} (m_t - p_t)^2, \quad u_0 > The money demand function {eq}`eq_grad_old1` and the utility function {eq}`eq_grad_old5` imply that $$ -U(-\alpha \theta_t) = u_1 + u_2 (-\alpha \theta_t) -\frac{u_2}{2}(-\alpha \theta_t)^2 . +U(-\alpha \theta_t) = u_0 + u_1 (-\alpha \theta_t) -\frac{u_2}{2}(-\alpha \theta_t)^2 . $$ (eq_grad_old5a) @@ -697,7 +697,8 @@ np.linalg.norm(clq.μ_CR - optimized_μ_CR) ``` ```{code-cell} ipython3 -compute_V(optimized_μ_CR, β=0.85, c=2) +V_CR = compute_V(optimized_μ_CR, β=0.85, c=2) +V_CR ``` ```{code-cell} ipython3 @@ -708,7 +709,14 @@ compute_V(jnp.array([clq.μ_CR]), β=0.85, c=2) By thinking a little harder about the mathematical structure of the Ramsey problem and using some linear algebra, we can simplify the problem that we hand over to a ``machine learning`` algorithm. -The idea here is that the Ramsey problem that chooses $\vec \mu$ to maximize the government's value function {eq}`eq:Ramseyvalue`subject to equation {eq}`eq:inflation101` is actually a quadratic optimum problem whose solution is characterized by a set of simultaneous linear equations in $\vec \mu$. +We start by recalling that the Ramsey problem that chooses $\vec \mu$ to maximize the government's value function {eq}`eq:Ramseyvalue`subject to equation {eq}`eq:inflation101`. + +This is actually an optimization problem with a quadratic objective function and linear constraints. + +First-order conditions for this problem are a set of simultaneous linear equations in $\vec \mu$. + +If we trust that the second-order conditions for a maximum are also satisfied (they are in our problem), +we can compute the Ramsey plan by solving these equations for $\vec \mu$. We'll apply this approach here and compare answers with what we obtained above with the gradient descent approach. @@ -933,7 +941,8 @@ print(f'deviation = {np.linalg.norm(optimized_μ - clq.μ_series)}') ``` ```{code-cell} ipython3 -compute_V(optimized_μ, β=0.85, c=2) +V_R = compute_V(optimized_μ, β=0.85, c=2) +V_R ``` We find that by exploiting more knowledge about the structure of the problem, we can significantly speed up our computation. @@ -1005,32 +1014,31 @@ closed_grad print(f'deviation = {np.linalg.norm(closed_grad - (- grad_J(jnp.ones(T))))}') ``` -## Some Regressions +## Some Exploratory Regressions To help us learn about the structure of the Ramsey plan, we shall compute some least squares linear regressions of particular components of $\vec \theta$ and $\vec \mu$ on others. Our hope is that these regressions will reveal structure hidden within the $\vec \mu^R, \vec \theta^R$ sequences associated with a Ramsey plan. -It is worth pausing here to think about roles played by **human** intelligence and **artificial** intelligence here. +It is worth pausing here to think about roles being played by **human** intelligence and **artificial** intelligence. -Artificial intelligence (AI a.k.a. ML) is running the regressions. +Artificial intelligence, i.e., some Python code and a computer, is running the regressions for us. -But you can regress anything on anything else. +But we are free to regress anything on anything else. -Human intelligence tell us which regressions to run. +Human intelligence tells us what regressions to run. -Even more human intelligence is required fully to appreciate what they reveal about the structure of the Ramsey plan. +Additional inputs of human intelligence will be required fully to appreciate what those regressions reveal about the structure of a Ramsey plan. ```{note} -At this point, it is worthwhile to read how Chang {cite}`chang1998credible` chose +When we eventually get around to trying to understand the regressions below, it will worthwhile to study the reasoning that let Chang {cite}`chang1998credible` to choose $\theta_t$ as his key state variable. ``` We'll begin by simply plotting the Ramsey plan's $\mu_t$ and $\theta_t$ for $t =0, \ldots, T$ against $t$ in a graph with $t$ on the ordinate axis. -These are the data that we'll be running some linear least squares regressions on. - +These are the data that we'll be running some linear least squares regressions on. ```{code-cell} ipython3 # Compute θ using optimized_μ @@ -1040,23 +1048,23 @@ These are the data that we'll be running some linear least squares regressions o # Plot the two sequences Ts = np.arange(T) -plt.plot(Ts, μs, label=r'$\mu_t$') -plt.plot(Ts, θs, label=r'$\theta_t$') +plt.scatter(Ts, μs, label=r'$\mu_t$', alpha=0.7) +plt.scatter(Ts, θs, label=r'$\theta_t$', alpha=0.7) plt.xlabel(r'$t$') plt.legend() plt.show() ``` We notice that $\theta_t$ is less than $\mu_t$for low $t$'s but that it eventually converges to -the same limit that $\mu_t$ does. +the same limit $\bar \mu$ that $\mu_t$ does. -This pattern reflects how formula {eq}`eq_grad_old3` for low $t$'s makes $\theta_t$ makes a weighted average of future $\mu_t$'s. +This pattern reflects how formula {eq}`eq_grad_old3` makes $\theta_t$ be a weighted average of future $\mu_t$'s. We begin by regressing $\mu_t$ on a constant and $\theta_t$. -This might seem strange because, first of all, equation {eq}`eq_grad_old3` asserts that inflation at time $t$ is determined $\{\mu_s\}_{s=t}^\infty$ +This might seem strange because, after all, equation {eq}`eq_grad_old3` asserts that inflation at time $t$ is determined $\{\mu_s\}_{s=t}^\infty$ -Nevertheless, we'll run this regression anyway and provide a justification later. +Nevertheless, we'll run this regression anyway. ```{code-cell} ipython3 # First regression: μ_t on a constant and θ_t @@ -1077,21 +1085,25 @@ $$ fits perfectly. -Let's plot this function and the points $(\theta_t, \mu_t)$ that lie on it for $t=0, \ldots, T$. + +```{note} +Of course, this means that a regression of $\theta_t$ on $\mu_t$ and a constant would also fit perfectly. +``` + +Let's plot the regression line $\mu_t = .0645 + 1.5995 \theta_t$ and the points $(\theta_t, \mu_t)$ that lie on it for $t=0, \ldots, T$. ```{code-cell} ipython3 -plt.scatter(θs, μs) -plt.plot(θs, results1.predict(X1_θ), 'C1', label='$\hat \mu_t$', linestyle='--') +plt.scatter(θs, μs, label=r'$\mu_t$') +plt.plot(θs, results1.predict(X1_θ), 'grey', label='$\hat \mu_t$', linestyle='--') plt.xlabel(r'$\theta_t$') plt.ylabel(r'$\mu_t$') plt.legend() plt.show() ``` -The time $0$ pair $\theta_0, \mu_0$ appears as the point on the upper right. +The time $0$ pair $(\theta_0, \mu_0)$ appears as the point on the upper right. -Points for succeeding times appear further and further to the lower left and eventually converge to -$\bar \mu, \bar \mu$. +Points $(\theta_t, \mu_t)$ for succeeding times appear further and further to the lower left and eventually converge to $(\bar \mu, \bar \mu)$. Next, we'll run a linear regression of $\theta_{t+1}$ against $\theta_t$. @@ -1122,8 +1134,8 @@ that prevails along the Ramsey outcome for inflation. Let's plot $\theta_t$ for $t =0, 1, \ldots, T$ along the line. ```{code-cell} ipython3 -plt.scatter(θ_t, θ_t1) -plt.plot(θ_t, results2.predict(X2_θ), color='C1', label='$\hat θ_t$', linestyle='--') +plt.scatter(θ_t, θ_t1, label=r'$\theta_{t+1}$') +plt.plot(θ_t, results2.predict(X2_θ), color='grey', label='$\hat θ_{t+1}$', linestyle='--') plt.xlabel(r'$\theta_t$') plt.ylabel(r'$\theta_{t+1}$') plt.legend() @@ -1135,8 +1147,174 @@ plt.show() Points for succeeding times appear further and further to the lower left and eventually converge to $\bar \mu, \bar \mu$. +### Continuation Values + +Next, we'll compute a sequence $\{v_t\}_{t=0}^T$ of what we'll call "continuation values" along a Ramsey plan. + +To do so, we'll start at date $T$ and compute -### What has machine learning taught us? +$$ +v_T = \frac{1}{1-\beta} s(\bar \mu, \bar \mu). +$$ + +Then starting from $t=T-1$, we'll iterate backwards on the recursion + +$$ +v_t = s(\theta_t, \mu_t) + \beta v_{t+1} +$$ + +for $t= T-1, T-2, \ldots, 0.$ + +```{code-cell} ipython3 +# Define function for s and U in section 41.3 +def s(θ, μ, u0, u1, u2, α, c): + U = lambda x: u0 + u1 * x - (u2 / 2) * x**2 + return U(-α*θ) - (c / 2) * μ**2 + +# Calculate v_t sequence backward +def compute_vt(μ, β, c, u0=1, u1=0.5, u2=3, α=1): + T = len(μs) + θ = compute_θ(μ, α) + + v_t = np.zeros(T) + μ_bar = μs[-1] + + # Reduce parameters + s_p = lambda θ, μ: s(θ, μ, + u0=u0, u1=u1, u2=u2, α=α, c=c) + + # Define v_T + v_t[T-1] = (1 / (1 - β)) * s_p(μ_bar, μ_bar) + + # Backward iteration + for t in reversed(range(T-1)): + v_t[t] = s_p(θ[t], μ[t]) + β * v_t[t+1] + + return v_t + +v_t = compute_vt(μs, β=0.85, c=2) +``` + +The initial continuation value $v_0$ should equals the optimized value of the Ramsey planner's criterion $V$ defined +in equation {eq}`eq:RamseyV`. + + +Indeed, we find that the deviation is very small: + +```{code-cell} ipython3 +print(f'deviation = {np.linalg.norm(v_t[0] - V_R)}') +``` + +We can also verify approximate equality by inspecting a graph of $v_t$ against $t$ for $t=0, \ldots, T$ along with the value attained by a restricted Ramsey planner $V^{CR}$ and the optimized value of the ordinary Ramsey planner $V^R$ + +```{code-cell} ipython3 +--- +mystnb: + figure: + caption: Continuation values + name: continuation_values +--- +# Plot the scatter plot +plt.scatter(Ts, v_t, label='$v_t$') + +# Plot horizontal lines +plt.axhline(V_CR, color='C1', alpha=0.5) +plt.axhline(V_R, color='C2', alpha=0.5) + +# Add labels +plt.text(max(Ts) + max(Ts)*0.07, V_CR, '$V^{CR}$', color='C1', + va='center', clip_on=False, fontsize=15) +plt.text(max(Ts) + max(Ts)*0.07, V_R, '$V^R$', color='C2', + va='center', clip_on=False, fontsize=15) +plt.xlabel(r'$t$') +plt.ylabel(r'$v_t$') + +plt.tight_layout() +plt.show() +``` + +Figure {numref}`continuation_values` shows several striking patterns: + + * The sequence of continuation values $\{v_t\}_{t=0}^T$ is monotonically decreasing + * Evidently, $v_0 > V^{CR} > v_T$ so that + * the value $v_0$ of the ordinary Ramsey plan exceeds the value $V^{CR}$ of the special Ramsey plan in which the planner is constrained to set $\mu_t = \mu^{CR}$ for all $t$. + * the continuation value $v_T$ of the ordinary Ramsey plan for $t \geq T$ is constant and is less than the value $V^{CR}$ of the special Ramsey plan in which the planner is constrained to set $\mu_t = \mu^{CR}$ for all $t$ + + +```{note} +The continuation value $v_T$ is what some researchers call the "value of a Ramsey plan under a +time-less perspective." A more descriptive phrase is "the value of the worst continuation Ramsey plan." +``` + + +Next we ask Python to regress $v_t$ against a constant, $\theta_t$, and $\theta_t^2$. + +$$ +v_t = g_0 + g_1 \theta_t + g_2 \theta_t^2 . +$$ + +```{code-cell} ipython3 +# Third regression: v_t on a constant, θ_t and θ^2_t +X3_θ = np.column_stack((np.ones(T), θs, θs**2)) +model3 = sm.OLS(v_t, X3_θ) +results3 = model3.fit() + + +# Print regression summary +print("\nRegression of v_t on a constant, θ_t and θ^2_t:") +print(results3.summary(slim=True)) +``` + +The regression has an $R^2$ equal to $1$ and so fits perfectly. + +However, notice the warning about the high condition number. + +As indicated in the printout, this is a consequence of + $\theta_t$ and $\theta_t^2$ being highly correlated along the Ramsey plan. + +```{code-cell} ipython3 +np.corrcoef(θs, θs**2) +``` + +Let's plot $v_t$ against $\theta_t$ along with the nonlinear regression line. + +```{code-cell} ipython3 +θ_grid = np.linspace(min(θs), max(θs), 100) +X3_grid = np.column_stack((np.ones(len(θ_grid)), θ_grid, θ_grid**2)) + +plt.scatter(θs, v_t) +plt.plot(θ_grid, results3.predict(X3_grid), color='grey', + label='$\hat v_t$', linestyle='--') +plt.axhline(V_CR, color='C1', alpha=0.5) + +plt.text(max(θ_grid) - max(θ_grid)*0.025, V_CR, '$V^{CR}$', color='C1', + va='center', clip_on=False, fontsize=15) + +plt.xlabel(r'$\theta_{t}$') +plt.ylabel(r'$v_t$') +plt.legend() + +plt.tight_layout() +plt.show() +``` + +The highest continuation value $v_0$ at $t=0$ appears at the peak of the function quadratic function +$g_0 + g_1 \theta_t + g_2 \theta_t^2$. + + +Subsequent values of $v_t$ for $t \geq 1$ appear to the lower left of the pair $(\theta_0, v_0)$ and converge monotonically from above to $v_T$ at time $T$. + +The value $V^{CR}$ attained by the Ramsey plan that is restricted to be a constant $\mu_t = \mu^{CR}$ sequence appears as a horizontal line. + +Evidently, continuation values $v_t > V^{CR}$ for $t=0, 1, 2$ while $v_t < V^{CR}$ for $t \geq 3$. + + + + + + + +## What has Machine Learning Taught Us? Our regressions tells us that along the Ramsey outcome $\vec \mu^R, \vec \theta^R$, the linear function @@ -1145,10 +1323,14 @@ $$ \mu_t = .0645 + 1.5995 \theta_t $$ -fits perfectly and that so does the regression line +fits perfectly and that so do the regression lines + +$$ +\theta_{t+1} = - .0645 + .4005 \theta_t +$$ $$ -\theta_{t+1} = - .0645 + .4005 \theta_t . +v_t = 6.8052 - .7580 \theta_t - 4.6991 \theta_t^2. $$ @@ -1170,12 +1352,18 @@ that along a Ramsey plan, the following relationships prevail: where the initial value $\theta_0^R$ was computed along with other components of $\vec \mu^R, \vec \theta^R$ when we computed the Ramsey plan, and where $b_0, b_1, d_0, d_1$ are parameters whose values we estimated with our regressions. +In addition, we learned that continuation values are described by the quadratic function + +$$ +v_t = g_0 + g_1 \theta_t + g_2 \theta_t^2 +$$ + -We discovered this representation by running some carefully chosen regressions and staring at the results, noticing that the $R^2$ of unity tell us that the fits are perfect. +We discovered these relationships by running some carefully chosen regressions and staring at the results, noticing that the $R^2$'s of unity tell us that the fits are perfect. We have learned something about the structure of the Ramsey problem. -But it is challenging to say more just by using the methods and ideas that we have deployed in this lecture. +However, it is challenging to say more just by using the methods and ideas that we have deployed in this lecture. There are many other linear regressions among components of $\vec \mu^R, \theta^R$ that would also have given us perfect fits. @@ -1187,7 +1375,7 @@ After all, the Ramsey planner chooses $\vec \mu$, while $\vec \theta$ is an o Isn't it more natural then to expect that we'd learn more about the structure of the Ramsey problem from a regression of components of $\vec \theta$ on components of $\vec \mu$? -To answer such questions, we'll have to deploy more economic theory. +To answer these questions, we'll have to deploy more economic theory. We do that in this quantecon lecture {doc}`calvo`. @@ -1204,7 +1392,6 @@ and the parameters $d_0, d_1$ in the updating rule for $\theta_{t+1}$ in represe First, we'll again use ``ChangLQ`` to compute these objects (along with a number of others). - ```{code-cell} ipython3 clq = ChangLQ(β=0.85, c=2, T=T) ```