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20 changes: 10 additions & 10 deletions lectures/calvo_abreu.md
Original file line number Diff line number Diff line change
Expand Up @@ -221,7 +221,7 @@ described in equation {eq}`eq_old6` in quantecon lecture {cite}`Calvo1978` has
$\theta$

$$
- s(\theta, 0 ) \geq - s(\theta, \mu) \quad
s(\theta, 0) \geq s(\theta, \mu) \quad
$$

This inequality implies that whenever the policy calls for the
Expand Down Expand Up @@ -310,8 +310,8 @@ More precisely, a government plan $\vec \mu^A$ with equilibrium inflation sequen
:label: eq_old10

\begin{aligned}
v_j^A & = - s(\theta^A_j, \mu^A_j) + \beta v_{j+1}^A \\
& \geq - s(\theta^A_j, 0 ) + \beta v_0^A \equiv v_j^{A,D}, \quad j \geq 0
v_j^A & = s(\theta^A_j, \mu^A_j) + \beta v_{j+1}^A \\
& \geq s(\theta^A_j, 0 ) + \beta v_0^A \equiv v_j^{A,D}, \quad j \geq 0
\end{aligned}
```

Expand All @@ -333,15 +333,15 @@ a sufficient condition for another plan $\vec \mu$ associated with inflation $\v
:label: eq_old100a

\begin{aligned}
v_j & = - s( \theta_j, \mu_j) + \beta v_{j+1} \\
& \geq -s( \theta_j, 0) + \beta v_0^A \quad \forall j \geq 0
v_j & = s( \theta_j, \mu_j) + \beta v_{j+1} \\
& \geq s( \theta_j, 0) + \beta v_0^A \quad \forall j \geq 0
\end{aligned}
```

For this condition to be satisfied it is necessary and sufficient that

$$
-s( \theta_j, 0) - ( - s( \theta_j, \mu_j) ) < \beta ( v_{j+1} - v_0^A )
s( \theta_j, 0) - s( \theta_j, \mu_j) < \beta ( v_{j+1} - v_0^A )
$$

The left side of the above inequality is the government's *gain* from deviating from the plan, while the right side is the government's *loss* from deviating
Expand Down Expand Up @@ -389,7 +389,7 @@ $$
The value of $\{\theta_t^A,\mu_t^A \}_{t=0}^\infty$ at time $0$ is

$$
v^A_0 = - \sum_{t=0}^{T_A-1} \beta^t s(\theta_t^A,\mu_t^A) +\beta^{T_A} J(\theta^R_0)
v^A_0 = \sum_{t=0}^{T_A-1} \beta^t s(\theta_t^A,\mu_t^A) +\beta^{T_A} J(\theta^R_0)
$$

For an appropriate $T_A$, this plan can be verified to be self-enforcing and therefore credible.
Expand Down Expand Up @@ -562,7 +562,7 @@ def abreu_plan(clq, T=1000, T_A=10, μ_bar=0.1, T_Plot=20):
# Calculate utility of stick plan
U_A = clq.β ** np.arange(T) * (
clq.u0 + clq.u1 * (-clq.θ_A) - clq.u2 / 2
* (-clq.θ_A) ** 2 - clq.c * clq.μ_A ** 2
* (-clq.θ_A) ** 2 - clq.c / 2 * clq.μ_A ** 2
)

clq.V_A = np.array([np.sum(U_A[t:] / clq.β ** t) for t in range(T)])
Expand Down Expand Up @@ -601,7 +601,7 @@ self-enforcing plan $\vec \mu^A$ by setting $\mu_t = 0$ and
then restarting the plan at $v^A_0$ at $t+1$:

$$
v_t^{A,D} = -s( \theta_j, 0) + \beta v_0^A
v_t^{A,D} = s( \theta_j, 0) + \beta v_0^A
$$

In the above graph $v_t^A > v_t^{A,D}$, which confirms that $\vec \mu^A$ is a self-enforcing plan.
Expand All @@ -617,7 +617,7 @@ Given that plan $\vec \mu^A$ is self-enforcing, we can check that
the Ramsey plan $\vec \mu^R$ is credible by verifying that:

$$
v^R_t \geq - s(\theta^R_t,0) + \beta v^A_0 , \quad \forall t \geq 0
v^R_t \geq s(\theta^R_t,0) + \beta v^A_0 , \quad \forall t \geq 0
$$

```{code-cell} ipython3
Expand Down
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