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coding assignment

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aaron betz abc university
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write your own irf code suppose you have the following solution to a model from dolo
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is just a

Question

ECON $6305$ : Coding Assignment # $1$ Aaron Betz abc University $1$ Write your own IRF code Suppose you have the following solution to a model from Dolo $($ Problem $1$ is just a Python exercise, you will not actually be using Dolo $) .$ You will generate impulse response functions with these formulas. z $_($ t $) =$ z $_($ t $- 1) \$ rho $+ (1 - \$ rho $) / ($ b $)$ ar $($ z $) +$ lon $_($ t $)$ k $_($ t $) =$ k $_($ t $- 1) (1 - \$ delta $) + 2.16 ($ z $_($ t $- 1) - 1) + 0.017 ($ k $_($ t $- 1) - 9.354) + \$ delta $9.354 1.1$ Capital and Z impulses Where $\$ rho $= 0.9 (,) / ($ b $)$ ar $($ z $) = 1,$ and $\$ delta $= 0.025 .$ You can solve forward to time t $= 1$ from time t $= 0 ($ steady state $)$ when z $_(0) () / ($ b $) =$ ar $($ z $) = 1$ and k $_(0) () / ($ b $) =$ ar $($ k $) = 9.354$ given some initial shock lon $_(0) .$ To continue to solve forward assume lon $_(1) = 0$ and that every following lon $_($ t $)$ is zero. Write a Python function to generate an impulse of length n $,$ given an initial shock, lon $_(0) .$ Show your code and graph impulses $($ for both capital and z $)$ of length $40,$ with an initial shock lon $_(0) = 0.01 . 1.2$ Labor and output impulse Now that you have a $40$ period series of all the necessary state variables, k $_($ t $)$ and z $_($ t $)$ solve for n $_($ t $)$ and then y $_($ t $)$ given the following formulas from Dolo. n $_($ t $) = 0.33 + 0.809 ($ z $_($ t $- 1) - 1) + 0.025 ($ k $_($ t $- 1) - 9.354)$ y $_($ t $) =$ z $_($ t $)$ n $_($ t $)^(0.67)$ k $_($ t $- 1)^(0.33)$ Keep in mind you will need to solve for labor, n $_($ t $),$ before you solve for output, y $_($ t $) .$ Once you have solved all $40$ periods graph both labor and output over time $($ use separate graphs $) . 1$

Julie Jones · Answer

Sure! Let's break down the task and walk through each part step-by-step. We will write Python code to generate impulse response functions for the given model.

Step 1: Define the Model

Given formulas:

$z_t = z_{t-1} ho + (1 - ho)/(b)ar(z) + ext{lon}_t$ $z_{t} = z_{t - 1} ρ + (1 - ρ) / (b) a r (z) + lon_{t}$
- Parameters: $ho = 0.9$ , $(b)ar(z) = 1$
$k_t = k_{t-1}(1 - \delta) + 2.16(z_{t-1} - 1) + 0.017(k_{t-1} - 9.354) + \delta imes 9.354$ $k_{t} = k_{t - 1} (1 - δ) + 2.16 (z_{t - 1} - 1) + 0.017 (k_{t - 1} - 9.354) + δ \times 9.354$
- Parameters: $\delta = 0.025$

Step 2: Write Python Function to Generate Impulses

We'll start with a function to generate the impulse response for $z_t$ and $k_t$ .

import numpy as np
import matplotlib.pyplot as plt

def generate_impulses(n, lon_0):
    # Initialize parameters
    rho = 0.9
    bar_z = 1
    delta = 0.025
    
    # Initial conditions
    z_t = [1]
    k_t = [9.354]
    
    lon = [lon_0] + [0] * (n - 1)
    
    # Generate the series for z_t and k_t
    for t in range(1, n):
        z_t_next = z_t[-1] * rho + (1 - rho) * bar_z + lon[t]
        z_t.append(z_t_next)
        
        k_t_next = k_t[-1] * (1 - delta) + 2.16 * (z_t[-2] - 1) + 0.017 * (k_t[-1] - 9.354) + delta * 9.354
        k_t.append(k_t_next)
    
    return z_t, k_t

# Generate 40 periods with an initial shock lon_0 = 0.01
n = 40
lon_0 = 0.01
z_t, k_t = generate_impulses(n, lon_0)

# Plot the impulses
plt.figure(figsize=(12, 6))

plt.subplot(1, 2, 1)
plt.plot(z_t, label="z_t")
plt.title("Impulse Response of z_t")
plt.xlabel("Periods")
plt.ylabel("z_t")
plt.legend()

plt.subplot(1, 2, 2)
plt.plot(k_t, label="k_t")
plt.title("Impulse Response of k_t")
plt.xlabel("Periods")
plt.ylabel("k_t")
plt.legend()

plt.tight_layout()
plt.show()

Step 3: Calculate $n_t$ and $y_t$

Given formulas:

$n_t = 0.33 + 0.809(z_{t-1} - 1) + 0.025(k_{t-1} - 9.354)$
$y_t = z_t \cdot n_t^{0.67} \cdot k_{t-1}^{0.33}$

We'll calculate these and plot them.

def calculate_labor_and_output(z_t, k_t):
    n_t = []
    y_t = []
    
    for t in range(1, len(z_t)):
        n_t_next = 0.33 + 0.809 * (z_t[t-1] - 1) + 0.025 * (k_t[t-1] - 9.354)
        n_t.append(n_t_next)
        
        y_t_next = z_t[t] * (n_t_next**0.67) * (k_t[t-1]**0.33)
        y_t.append(y_t_next)
    
    return n_t, y_t

# Calculate labor and output
n_t, y_t = calculate_labor_and_output(z_t, k_t)

# Plot labor and output
plt.figure(figsize=(12, 6))

plt.subplot(1, 2, 1)
plt.plot(n_t, label="n_t")
plt.title("Labor over Time")
plt.xlabel("Periods")
plt.ylabel("n_t")
plt.legend()

plt.subplot(1, 2, 2)
plt.plot(y_t, label="y_t")
plt.title("Output over Time")
plt.xlabel("Periods")
plt.ylabel("y_t")
plt.legend()

plt.tight_layout()
plt.show()

Summary

The first part of the code generates the impulse response functions for $z_t$ and $k_t$ .
The second part calculates $n_t$ and $y_t$ using these values and then plots the results.

This process allows us to visualize the impact of an initial shock over time.

Questions