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[olg] use function #320

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Nov 17, 2023
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[olg] use function #320

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19 changes: 9 additions & 10 deletions lectures/olg.md
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Original file line number Diff line number Diff line change
Expand Up @@ -4,7 +4,7 @@ jupytext:
extension: .md
format_name: myst
format_version: 0.13
jupytext_version: 1.14.5
jupytext_version: 1.15.2
kernelspec:
display_name: Python 3 (ipykernel)
language: python
Expand Down Expand Up @@ -294,12 +294,17 @@ def capital_demand(R, α):
return (α/R)**(1/(1-α))
```

```{code-cell} ipython3
def capital_supply(R, β, w):
R = np.ones_like(R)
return R * (β / (1 + β)) * w
```

The next figure plots the supply of capital, as in [](saving_log_2_olg), as well as the demand for capital, as in [](aggregate_demand_capital_olg), as functions of the interest rate $R_{t+1}$.

(For the special case of log utility, supply does not depend on the interest rate, so we have a constant function.)

```{code-cell} ipython3

R_vals = np.linspace(0.3, 1)
α, β = 0.5, 0.9
w = 2.0
Expand All @@ -308,7 +313,7 @@ fig, ax = plt.subplots()

ax.plot(R_vals, capital_demand(R_vals, α),
label="aggregate demand")
ax.plot(R_vals, np.ones_like(R_vals) * (β / (1 + β)) * w,
ax.plot(R_vals, capital_supply(R_vals, β, w),
label="aggregate supply")

ax.set_xlabel("$R_{t+1}$")
Expand Down Expand Up @@ -391,7 +396,6 @@ That is,
Let's redo our plot above but now inserting the equilibrium quantity and price.

```{code-cell} ipython3

R_vals = np.linspace(0.3, 1)
α, β = 0.5, 0.9
w = 2.0
Expand All @@ -400,7 +404,7 @@ fig, ax = plt.subplots()

ax.plot(R_vals, capital_demand(R_vals, α),
label="aggregate demand")
ax.plot(R_vals, np.ones_like(R_vals) * (β / (1 + β)) * w,
ax.plot(R_vals, capital_supply(R_vals, β, w),
label="aggregate supply")

R_e = equilibrium_R_log_utility(α, β, w)
Expand Down Expand Up @@ -452,7 +456,6 @@ $$
g(k) := \frac{\beta}{1+\beta} (1-\alpha)(k)^{\alpha}
$$


```{code-cell} ipython3
def k_update(k, α, β):
return β * (1 - α) * k**α / (1 + β)
Expand Down Expand Up @@ -627,7 +630,6 @@ def savings_crra(w, R, model):
```

```{code-cell} ipython3

R_vals = np.linspace(0.3, 1)
model = create_olg_model()
w = 2.0
Expand Down Expand Up @@ -760,7 +762,6 @@ ax.set_ylabel('$k_{t+1}$', fontsize=12)
plt.show()
```


```{solution-end}
```

Expand Down Expand Up @@ -821,7 +822,6 @@ k_star = optimize.newton(h, 0.2, args=(model,))
print(f"k_star = {k_star}")
```


```{solution-end}
```

Expand All @@ -846,7 +846,6 @@ Use initial conditions for $k_0$ of $0.001, 1.2, 2.6$ and time series length 10.

Let's define the constants and three distinct intital conditions


```{code-cell} ipython3
ts_length = 10
k0 = np.array([0.001, 1.2, 2.6])
Expand Down