98. The Credibility Problem#

98.1. Overview#

This lecture describes a basic expectational Phillips curve model of the sort studied by Kydland and Prescott [1977] and Barro and Gordon [1983].

It is the first modeling lecture in a suite based on chapters of Sargent [1999], following The Rise and Fall of U.S. Inflation, which lays out the two stories and reviews the Lucas Critique.

This lecture follows chapter 3 of the book, and Credible Government Policies follows chapter 4.

Together the two lectures formalize

  • the temptation to inflate that is unleashed by the discovery of a Phillips curve,

  • the value of a commitment technology for resisting that temptation, and

  • the fragility of reputational mechanisms as substitutes for commitment.

Throughout, rational expectations is the only equilibrium concept that we use.

Alterations in the timing of decisions by a government and a private sector induce different economies with distinct outcomes.

A government faces a credibility problem whenever it wishes to make decisions sooner than it must.

We shall compare outcomes under two timing protocols:

  • In one, the government chooses inflation before the private sector sets its expectations, so the government takes into account how its choice shapes those expectations.

  • In the other, the government chooses inflation after the private sector has set its expectations.

The deterioration in outcomes under the second protocol measures the loss from an inability to commit.

Time Inconsistency of Ramsey Plans studies the same time-inconsistency problem in Calvo’s dynamic monetary model.

We also study two out-of-equilibrium dynamics that converge to the no-commitment (Nash) outcome:

  • best response dynamics, and

  • least squares learning.

Let’s start with some standard imports:

import matplotlib.pyplot as plt
import numpy as np

98.2. A one-period economy#

Although credibility problems are intrinsically dynamic, it is possible to describe them in a one-period model under different patterns of within-period timing.

This prepares the way for the multi-period analyses in subsequent lectures.

We describe a version of the one-period model of Kydland and Prescott [1977] in the terms used by Nancy Stokey [Stokey, 1989].

Let \((U, y, x)\) be the unemployment rate, the inflation rate, and the public’s expectation of the inflation rate, respectively.

The government’s one-period payoff is

(98.1)#\[- \frac{1}{2} \left( U^2 + y^2 \right) .\]

Unemployment is determined by an expectations-augmented Phillips curve

(98.2)#\[U = U^* - \theta (y - x), \qquad \theta > 0 .\]

The equation asserts that unemployment deviates from \(U^*\), the natural rate, only when there is surprise inflation or deflation.

Substituting (98.2) into (98.1) expresses the government’s payoff as a function \(r(x, y)\):

(98.3)#\[r(x, y) = - \frac{1}{2} \left[ \left(U^* - \theta (y - x)\right)^2 + y^2 \right] .\]

We work with the following objects.

Rational expectations equilibrium: a triple \((U, x, y)\) satisfying (98.2) and \(y = x\).

Government (one-period) best response: given the public’s expectation \(x\), a decision rule \(B(x) = \operatorname{argmax}_y r(x, y)\) for setting \(y\).

Nash equilibrium: a pair \((x, y)\) satisfying (i) \(x = y\), and (ii) \(y = B(x)\).

Ramsey problem: \(\max_y r(y, y)\).

The Ramsey outcome is the value of \(y\) that attains the maximum.

Best response dynamics: the dynamical system \(y_t = B(y_{t-1})\), \(y_0\) given.

A rational expectations equilibrium is a \((U, x, y)\) triple that lies on the Phillips curve and for which private agents are not fooled, given \(x\).

Substituting \(x = y\) into the Phillips curve (98.2) shows that \(U = U^*\) in any rational expectations equilibrium.

This identifies \(U^*\) as the natural rate of unemployment.

98.3. Nash and Ramsey outcomes#

A Nash equilibrium builds in a best response by the government, taking the state of expectations \(x\) as given, together with a response \(x = y\) by the market, i.e., rational expectations for a given \(y\).

Maximizing (98.3) with respect to \(y\) for a fixed \(x\) gives the government’s best response function

(98.4)#\[y = B(x) = \frac{\theta}{\theta^2 + 1} U^* + \frac{\theta^2}{\theta^2 + 1} x .\]

The Nash equilibrium sets \(x = y = B(x)\), which gives

\[ y^N = x^N = \theta U^*, \qquad U = U^* . \]

The Ramsey problem instead imposes \(x = y\) before maximizing, so it maximizes \(r(y, y) = -\tfrac{1}{2}(U^{*2} + y^2)\), which yields the Ramsey outcome

\[ y^R = x^R = 0, \qquad U = U^* . \]

Thus \(r(x^R, y^R) = -\tfrac{1}{2} U^{*2}\) while \(r(x^N, y^N) = -\tfrac{1}{2}(1 + \theta^2) U^{*2}\).

Both outcomes deliver the natural rate \(U^*\), but the Nash equilibrium delivers it with positive inflation and hence a strictly lower payoff.

Let’s collect these formulas in a class.

class CredibilityModel:
    """
    A one-period expectational Phillips curve economy.
    """

    def __init__(self, θ=1.0, U_star=5.0):
        self.θ, self.U_star = θ, U_star

    def phillips(self, y, x):
        "Unemployment implied by inflation y and expected inflation x."
        return self.U_star - self.θ * (y - x)

    def r(self, x, y):
        "Government one-period payoff."
        U = self.phillips(y, x)
        return -0.5 * (U**2 + y**2)

    def B(self, x):
        "Government best response to expected inflation x."
        θ = self.θ
        return θ / (θ**2 + 1) * self.U_star + θ**2 / (θ**2 + 1) * x

    def nash(self):
        "Nash equilibrium inflation (= expected inflation)."
        return self.θ * self.U_star

    def ramsey(self):
        "Ramsey inflation (= expected inflation)."
        return 0.0

The default calibration is \(\theta = 1\) and \(U^* = 5\).

cm = CredibilityModel()

y_N, y_R = cm.nash(), cm.ramsey()
print(f"Nash inflation      y^N = {y_N:.2f}")
print(f"Ramsey inflation    y^R = {y_R:.2f}")
print(f"Nash payoff         r   = {cm.r(y_N, y_N):.3f}")
print(f"Ramsey payoff       r   = {cm.r(y_R, y_R):.3f}")

Nash inflation      y^N = 5.00
Ramsey inflation    y^R = 0.00
Nash payoff         r   = -25.000
Ramsey payoff       r   = -12.500

The Nash payoff is worse than the Ramsey payoff.

The government would prefer the Ramsey outcome, but it cannot attain it without a technology for committing to \(y = 0\) before the public forms its expectations.

The Nash equilibrium is supported by a timing protocol in which the government decides after the private sector sets its expectations.

The Ramsey outcome is associated with a timing protocol in which the government chooses first, knowing that in a rational expectations equilibrium the public’s expectations will move with its choice because \(y = x\).

98.3.1. A picture of the two outcomes#

The government’s indifference curves are circles centered at the origin in \((U, y)\) space, because the payoff (98.1) depends only on \(U^2 + y^2\).

For a given expectation \(x\), the Phillips curve (98.2) is a downward-sloping line in \((U, y)\) space.

The government’s best response for \(y\), given \(x\), occurs where an indifference curve is tangent to the Phillips curve indexed by \(x\).

fig, ax = plt.subplots(figsize=(7, 6))

U_grid = np.linspace(0, 12, 200)

# a family of Phillips curves indexed by expected inflation x
for x in [0.0, y_N / 2, y_N]:
    # U = U_star - θ (y - x)  =>  y = x + (U_star - U) / θ
    y_line = x + (cm.U_star - U_grid) / cm.θ
    ax.plot(U_grid, y_line, 'C0', lw=1)

# government indifference curves (circles U^2 + y^2 = const)
ξ = np.linspace(0, 2 * np.pi, 200)
for R in [np.hypot(cm.U_star, y_R), np.hypot(cm.U_star, y_N)]:
    if R > 0:
        ax.plot(R * np.cos(ξ), R * np.sin(ξ), 'C1--', lw=1)

ax.plot(cm.U_star, y_N, 'ko')
ax.annotate('Nash', (cm.U_star, y_N), (cm.U_star + 0.4, y_N + 0.4))
ax.plot(cm.U_star, y_R, 'ko')
ax.annotate('Ramsey', (cm.U_star, y_R), (cm.U_star + 0.4, y_R + 0.4))

ax.set_xlim(0, 12)
ax.set_ylim(0, 10)
ax.set_xlabel('unemployment $U$')
ax.set_ylabel('inflation $y$')
plt.show()
_images/d5a628ed6008412e7539b10796c3c349b734c5a63e7ba94340635bd8871c4d50.png

Fig. 98.1 Phillips curves indexed by expected inflation, government indifference curves, and the Nash and Ramsey outcomes#

Solid lines are Phillips curves for expected inflation \(x \in \{0, y^N/2, y^N\}\); dashed circles are government indifference curves.

The Nash outcome \((U^*, y^N)\) lies on a larger circle (lower payoff) than the Ramsey outcome \((U^*, 0)\).

98.4. Best response dynamics#

Best response dynamics convert the one-period model into a dynamic one by positing an adaptive mechanism in which expected inflation equals last period’s inflation, \(x_t = y_{t-1}\).

This leads to the dynamics

\[ y_t = B(y_{t-1}), \qquad y_0 \text{ given} . \]

Because \(B\) is an affine map with slope \(\theta^2 / (\theta^2 + 1) \in (0, 1)\), iterating on it converges to the fixed point \(y^N = \theta U^*\) from any starting point.

Let’s plot the best response function against the 45-degree line and simulate the dynamics.

def best_response_path(cm, y0, T=20):
    "Iterate y_{t+1} = B(y_t)."
    y = np.empty(T + 1)
    y[0] = y0
    for t in range(T):
        y[t + 1] = cm.B(y[t])
    return y

y_path = best_response_path(cm, y0=0.0, T=20)

The figure traces this path as a cobweb between the best response function and the 45-degree line.

fig, ax = plt.subplots(figsize=(6, 6))

x_grid = np.linspace(0, y_N + 1, 100)
ax.plot(x_grid, cm.B(x_grid), 'C0', label='$B(x)$')
ax.plot(x_grid, x_grid, 'k--', lw=1, label='45 degrees')

# cobweb of the best response dynamics
for t in range(len(y_path) - 1):
    ax.plot([y_path[t], y_path[t]], [y_path[t], y_path[t + 1]], 'C1', lw=0.8)
    ax.plot([y_path[t], y_path[t + 1]], [y_path[t + 1], y_path[t + 1]],
            'C1', lw=0.8)

ax.plot(y_N, y_N, 'ko')
ax.annotate('Nash', (y_N, y_N), (y_N - 1.5, y_N + 0.3))
ax.set_xlabel('$x$')
ax.set_ylabel('$B(x)$')
ax.legend()
plt.show()
_images/88145999e7404a5e9b3e330358e2e37009dc27ce31110c08351c7a8d7ed0a132.png

Fig. 98.2 Best response dynamics cobwebbing up to the Nash inflation rate#

Starting from \(x = 0\) (the Ramsey inflation rate), the government sets \(y = B(0) > 0\).

This provokes the public to raise its expectation, which leads the government to raise inflation further.

The limit of this process is the Nash outcome \(y = x = y^N\), a self-confirming situation.

Thus best response dynamics converge to the Nash equilibrium and reinforce \((U^*, y^N)\) as the prediction of the model without a commitment technology.

98.5. Least squares learning converges to Nash#

A version of the best response dynamics also emerges from least squares learning.

Least squares learning plays a key role throughout this suite of lectures, and this simple example introduces analytical elements that reappear later in more complex settings.

Bray [1982] studied a competitive market in which traders formed the expected price by averaging past prices.

In the same spirit, assume that for \(t \geq 2\) expected inflation \(x_t\) is the average of past inflation rates,

\[ x_t = \frac{1}{t - 1} \sum_{s=1}^{t-1} y_s , \]

which can be represented recursively as

(98.5)#\[x_t = x_{t-1} + \frac{1}{t-1} (y_{t-1} - x_{t-1}), \qquad x_1 = 0 .\]

Actual inflation is a disturbed version of the best response mapping evaluated at \(x_t\),

(98.6)#\[y_t = B(x_t) + \eta_t ,\]

where \(\eta_t\) is an IID mean-zero term that represents the government’s imperfect control of inflation.

Substituting (98.6), dated \(t-1\), into (98.5) gives the stochastic recursion

(98.7)#\[x_t = x_{t-1} + \frac{1}{t-1} \left[ B(x_{t-1}) - x_{t-1} + \eta_{t-1} \right] .\]

The forecast \(x_t\) depends only on shocks dated \(t-1\) and earlier, as it must, because the public forms it before it sees \(y_t\).

(The corresponding equation (8) of Sargent [1999] dates this shock \(t\), a slip.)

This is Bray’s market of From relaxation to learning: Bray’s market, written in the notation of this lecture.

  • The government’s best response \(B\) plays the role of the firms’ best response \(h\).

  • The forecast \(x_t\) plays the role of the firms’ belief \(X^*_{t-1}\), formed from data through \(t-1\).

  • Inflation \(y_t\) plays the role of realized average output \(X_t\).

The feedback from beliefs to outcomes is

\[ h' = B'(x) = \frac{\theta^2}{1 + \theta^2} \in (0, 1), \]

so Bray’s condition \(h' < 1\) for least squares learners to converge holds for every \(\theta\).

Least Squares Learning in Self-Referential Models analyzes Bray’s model (its Example 2) in the general framework of Marcet and Sargent [1989].

By the theory of stochastic approximation, the limiting behavior of \(x_t\) is described by the associated ordinary differential equation (ODE)

(98.8)#\[\frac{d x}{d t} = B(x) - x .\]

The rest point of this ODE satisfies \(x = B(x)\), which is the Nash equilibrium inflation rate \(x = \theta U^*\).

Because the map \(B\) is affine, the ODE is linear with slope

\[ \mathcal{M} = \frac{d}{d x}\left( B(x) - x \right) = B'(x) - 1 = - \frac{1}{\theta^2 + 1} . \]

Since \(\mathcal{M} < 0\), the ODE is stable about its rest point.

In the language of One recursion, many learners, the Nash equilibrium is E-stable (expectationally stable): \(B\) maps perceived into actual inflation, and the derivative of that map, \(h' = B'\), is less than one.

\(\mathcal{M} = h' - 1\) is the E-stability eigenvalue of this one-dimensional system.

E-stability is what brings to bear the theorems of Marcet and Sargent [1989], which give conditions under which \(x_t\) converges to \(y^N\) globally.

The same number recurs throughout this suite.

When the government itself estimates a Phillips curve by least squares, the mean dynamics of its beliefs have eigenvalues \(-1\) and \(-1/(1+\theta^2)\) (Uniqueness and E-stability, Mean dynamics near the self-confirming equilibrium).

Let’s simulate the recursion (98.7) and confirm convergence to the Nash outcome.

def ls_learning(cm, T=2000, σ_η=1.0, seed=0, ε=None):
    """
    Simulate learning of expected inflation.

    x[t] is the forecast formed after observing y_1, ..., y_t, which is the
    x_{t+1} of pc_expect1; x[0] = x_1 = 0, and y[0] is unused.

    With ε=None the gain is 1/t (least squares: the 1/(t-1) gain of
    pc_expect1, reindexed); otherwise it is the constant ε.
    """
    rng = np.random.default_rng(seed)
    x = np.empty(T + 1)
    y = np.full(T + 1, np.nan)
    x[0] = 0.0
    for t in range(1, T + 1):
        y[t] = cm.B(x[t - 1]) + σ_η * rng.standard_normal()
        gain = 1.0 / t if ε is None else ε
        x[t] = x[t - 1] + gain * (y[t] - x[t - 1])
    return x, y

We simulate 2000 periods at the default calibration.

x, y = ls_learning(cm, T=2000, σ_η=1.0)

fig, ax = plt.subplots(figsize=(9, 5))
ax.plot(x, 'C0', lw=1, label='expected inflation $x_t$')
ax.axhline(y_N, color='k', ls='--', lw=1, label='Nash $y^N$')
ax.axhline(y_R, color='C2', ls=':', lw=1, label='Ramsey $y^R$')
ax.set_xlabel('$t$')
ax.set_ylabel('inflation')
ax.legend()
plt.show()
_images/ba05d6f4e41e17558c8632bf6aa247a105a87b22faf82dc3f3e4ca95a02cbf5d.png

Fig. 98.3 Least squares learning of expected inflation converges to the Nash outcome#

The least squares dynamics confirm the pessimism of the best response dynamics.

As Sargent [1999] puts it, this pessimism undermines the triumph of natural-rate theory story of The Rise and Fall of U.S. Inflation, especially its triumph aspect.

Given an initial condition in the form of a low, gold-standard or Bretton Woods value of \(x\), the best response or least squares dynamics can explain an acceleration of inflation.

But they cannot explain a Volcker-style stabilization that brings inflation back down.

Later lectures reformulate versions of least squares learning in ways designed to moderate this pessimism, and in doing so they move us away from the triumph story and toward the vindication story.

98.5.1. How fast?#

\(\mathcal{M}\) also governs the speed of convergence (footnote 3 of chapter 3 of Sargent [1999]; Evans and Honkapohja [2001]).

Taking expectations of (98.7) shows that the mean error \(E x_t - y^N\) obeys the noise-free version of the recursion, so it shrinks like \(t^{\mathcal{M}}\).

A necessary condition for convergence at the usual rate, with \(\sqrt{t}\,(x_t - y^N)\) settling into a normal distribution, is \(\mathcal{M} < -1/2\), which here requires \(\theta < 1\).

When \(-1/2 \leq \mathcal{M} < 0\), \(x_t\) still converges to \(y^N\), but more slowly, at a rate governed by \(|\mathcal{M}|\).

At \(\theta = 1\), the value used here and in most later lectures, \(\mathcal{M} = -1/2\) sits exactly on the boundary: the standard deviation of \(x_t\) shrinks like \(\sqrt{\log t / t}\) instead of \(1/\sqrt{t}\).

In the market of From relaxation to learning: Bray’s market, by contrast, \(h' = -1.5\) gives \(\mathcal{M} = -2.5\), and errors shrink at the usual rate \(t^{-1/2}\).

Exercise Exercise 98.1 measures these rates.

The \(-1/2\) boundary concerns the convergence rate of a learner whose gain decreases like \(1/t\).

It is not what governs how often a constant-gain learner escapes from the Nash outcome in later lectures; a large-deviation rate does that (A primer on recursive algorithms).

98.6. More foresight#

Best response and least squares learning are out-of-equilibrium dynamics tacked onto a one-period economy.

They force all movement through expectation formation: in choosing inflation, the government forgets that the economy lasts more than one period.

Better outcomes can occur if the government plans for the future.

Subsequent lectures describe three ways of modeling foresight, which impute varying amounts of rationality and predict different qualities of outcomes:

  1. A reputational approach that attributes rational expectations to both the government and the public, the subject of Credible Government Policies, which comes next.

    • Many outcomes are sustainable, ranging from repetition of the Ramsey outcome to paths worse than repetition of the Nash outcome, a multiplicity that turns out to be the theory’s chief lesson.

  2. An approach that keeps the government rational but gives the public adaptive expectations in the original Cagan-Friedman sense, the subject of Adaptive Expectations and the Phelps Problem.

    • Depending on a comparison between a discount factor and an adaptation parameter, this setup can improve outcomes and possibly sustain repetition of the Ramsey outcome.

  3. Approaches that make the government boundedly rational as well.

98.7. Appendix: stochastic approximation#

Here we sketch why the ODE (98.8) governs the tail behavior of the stochastic difference equation (98.7).

The argument, due to Kushner and Clark [1978], has two key components: a shift in time scale and a liberal application of averaging.

The subsection “Stochastic approximation and E-stability” of One recursion, many learners presents the same tools from the standpoint of the Bounded Rationality suite, and Least Squares Learning in Self-Referential Models states the Marcet–Sargent convergence theorems for vectors of beliefs.

Let \(\{a_n\}_{n \geq 0}\) be a positive sequence of gains satisfying

\[ \lim_{n \to \infty} a_n = 0, \qquad \sum_n a_n = +\infty, \qquad \sum_n a_n^2 < +\infty . \]

The choice \(a_n = 1 / (n + 1)\) satisfies these assumptions.

Relabel (98.7) so that its \(x_{n+1}\) and \(\eta_{n+1}\) become \(x_n\) and \(\eta_n\).

Then the gain is \(a_n = 1/(n+1)\) and the recursion reads

(98.9)#\[x_{n+1} = x_n + a_n \left[ B(x_n) - x_n + \eta_n \right] ,\]

where \(\eta_n\) is IID with mean zero and finite variance.

Introduce the transformed time scale \(t_0 = 0\), \(t_n = \sum_{i=0}^{n-1} a_i\), and interpolate the discrete sequence \(\{x_n\}\) into a continuous-time process \(x^0(t)\).

The map \(m(t) = \max\{n : t_n \leq t\}\) takes transformed time back to the original index.

Kushner and Clark show that on this transformed time scale the interpolated process is well approximated by the integral equation

\[ x^0(t) = x^0(0) + \int_0^t \left[ B(x^0(s)) - x^0(s) \right] d s + R(t) , \]

where the approximation error \(R(t)\) has two components — one from approximating a distributed lag in \(B(x) - x\) by an integral, and one from a distributed lag in \(\eta_s\).

Studying a sequence of left-shifted versions of the process, they show that both error components can be driven to zero as \(n \to \infty\).

For the noise component, the key step is to note that over the stretch of transformed time from \(t_n\) to \(t_n + t\) the noise enters through the martingale \(W_n(t) = \sum_{i=n}^{m(t_n + t)} a_i \eta_i\).

Its variance is \(\mathbb{V}[\eta] \sum_{i=n}^{m(t_n + t)} a_i^2\), a tail sum of the convergent series \(\sum_i a_i^2\), and so it goes to zero as \(n \to \infty\).

The remaining error is sent to zero because \(a_i \to 0\) shrinks the mesh of the Riemann sum used to approximate the integral.

In the limit, the stochastic difference equation (98.9) shares the behavior of the non-stochastic ODE

\[ \frac{d}{d t} \tilde x(t) = B(\tilde x(t)) - \tilde x(t) , \]

which is said to describe the mean dynamics of the original system.

Later lectures study systems like (98.9) in which \(a_i\) does not approach zero as \(i\) grows — so-called constant gain algorithms.

Exercise Exercise 98.2 gives a first look at one.

The mean dynamics (98.8) and these constant-gain algorithms become the central tools of Adaptive Learning and Escape Dynamics, Escaping Nash Inflation, and Priors, Escapes, and Learning Cycles, where the scalar expectation \(x\) studied here grows into a whole vector of drifting Phillips-curve coefficients.

98.8. Exercises#

Exercise 98.1

This exercise measures how the speed of least squares learning depends on \(\mathcal{M} = -1/(1 + \theta^2)\).

Let \(e_t = x_t - \theta U^*\) be the forecast error relative to Nash.

  1. Show that the mean error \(E e_t\) obeys the noise-free version of (98.7), and that it shrinks like \(t^{\mathcal{M}}\).

  2. For \(\theta \in \{0.5, 1, 2\}\) and \(U^* = 5\), simulate 200 paths of length \(T = 3000\) (seeds \(0, \ldots, 199\)). On log–log axes, plot \(|\bar e_t| / (\theta U^*)\), where \(\bar e_t\) is the average error across paths, and compare it with the exact mean error. Dividing by \(\theta U^*\), the initial error, makes the three cases comparable.

  3. Plot \(\sqrt{t}\) times the standard deviation of \(x_t\) across paths, and explain why it levels off only for \(\theta < 1\).

  4. What fraction of the \(\theta = 2\) paths is still below the Nash rate at \(t = 3000\)? Does the \(\theta = 2\) learner wander around its limit?

Exercise 98.2

Replace the decreasing gain \(1/(t-1)\) in (98.5) by a constant gain \(\varepsilon \in (0, 1)\), so that \(x_{t+1} = x_t + \varepsilon (y_t - x_t)\).

This is the adaptive-expectations rule of Adaptive Expectations and the Phelps Problem with \(\varepsilon = 1 - \lambda\), but here the government keeps best-responding one period at a time.

  1. Show that \(e_t = x_t - y^N\) is a first-order autoregression with coefficient \(1 + \varepsilon \mathcal{M}\). Show that, when it is stationary, its variance is

    \[ V(\varepsilon) = \frac{\varepsilon \sigma_\eta^2}{(1 - h')\bigl(2 - \varepsilon (1 - h')\bigr)}, \qquad h' = \frac{\theta^2}{1 + \theta^2} . \]

    This is the formula of Exercise 90.5 for Bray’s market, with \(h' = B'\). For which gains is the recursion stable?

  2. With \(\theta = 1\) and \(\sigma_\eta = 1\), use the option ε of ls_learning to simulate 100,000 periods for \(\varepsilon \in \{0.01, 0.025, 0.05, 0.1, 0.2\}\). Drop the first 5,000 periods, and compare the sample variance of \(e_t\) with \(V(\varepsilon)\).

  3. Plot 3000 periods of \(x_t\) under \(\varepsilon = 0.025\) and under the least squares gain, using the same shocks. What does a constant gain buy, and what does it cost?

Exercise 98.3

This exercise studies the inflation bias \(y^N - y^R\) and two remedies discussed in Remedies.

Replace the government’s payoff (98.1) by \(-\frac{1}{2}\left[(U - k U^*)^2 + \alpha y^2\right]\), with \(k \in [0, 1]\) and \(\alpha > 0\).

The lecture’s case is \(k = 0\), \(\alpha = 1\).

  1. Show that the best response is \(B(x) = [\theta(1 - k) U^* + \theta^2 x] / (\theta^2 + \alpha)\) and that Nash inflation is \(y^N = \theta (1 - k) U^* / \alpha\). Check the formula by iterating the best response.

  2. Alan Blinder’s remedy is an authority who knows the natural rate and never wants unemployment to differ from it (\(k = 1\)). Rogoff [1985]’s remedy is a conservative central banker who weights inflation more heavily than society does (\(\alpha > 1\)). What does each do to the inflation bias?

  3. Now let the Phillips curve be \(U = U^* - \theta(y - x) + \zeta\). Here \(\zeta\) is an IID shock with mean zero and variance \(\sigma_\zeta^2\), which the policymaker sees before setting \(y\) but the public does not see when it forms \(x\). Show that in the Nash equilibrium \(x = \theta(1-k)U^*/\alpha\), \(y = x + \theta\zeta/(\theta^2 + \alpha)\), and \(U = U^* + \alpha \zeta / (\theta^2 + \alpha)\).

  4. Society keeps the payoff (98.1). Set \(\theta = 1\) and \(U^* = 5\). For a Rogoff banker (\(k = 0\)) and \(\sigma_\zeta \in \{1, 2.5, 5\}\), plot society’s expected loss \(\frac{1}{2} E(U^2 + y^2)\) against \(\alpha\) and find the best \(\alpha\). Compare with Blinder’s authority (\(k = 1\), \(\alpha = 1\)).