A course introduction
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We will first set up the stage with an introduction to scientific computing for applied economics
2 parts
I. Numerical methods to solve economic models
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First week or so
Weeks 2 to 10
Weeks 11 to 16
Dates are tentative: This is a “new” course. Your feedback and understanding are appreciated.
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Let’s stop here for a while and introduce ourselves!
Most things you’ve done so far have likely been solvable analytically
Including OLS: \(\hat{\beta} = (X'X)^{-1}X'Y\)
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Not all economic models have closed-form solutions, and others can’t have closed-form solutions with losing important economic content
We can use computation + theory to answer quantitative questions
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Theory can’t give us welfare in dollar terms
Theory can’t tell us the value of economic primitives
Theory often relies on strong assumptions like:
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It can be unclear what the costs of these assumptions are
Suppose we have a constant elasticity demand function:
\[q(p) = p^{-0.2}\]
In equilibrium, quantity demanded is \(q^* = 2\)
Q: What price clears the market in equilibrium?
A: Just invert the demand function
\[2 = p^{-0.2}\]
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\[p^* = 2^{-5} \checkmark\]
Your calculator can do the rest
Suppose the demand function is now
\[q(p) = 0.5p^{-0.2} + 0.5p^{-0.5},\]
a weighted average of two CE demand functions
In equilibrium, quantity demanded is \(q^* = 2\)
Q: What price clears the market in equilibrium?
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First, does a solution exist?
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Yes, why?
\(q(p)\) is continuous and monotonically decreasing
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\(q(p)\) is greater than 2 at \(p=0.1\) and less than 2 at \(p=0.2\)
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\(\Rightarrow\) by intermediate value theorem \(q(p) = 2\) somewhere in \((0.1,0.2)\)
# We know solution is between .1 and .2
x = collect(range(.1, stop = .2, length = 10)); # generate evenly spaced grid
q_d = ones(size(x)).*2; # generate equal length vector of qd=2
# Define price function
price(p) = p.^(-0.2)/2 .+ p.^(-0.5)/2;
# Get corresponding quantity values at these prices
y = price(x);Now plot \(q_d\) and \(q(p)\)
using Plots;
plot(x, [y q_d],
linestyle = [:solid :dot],
linewidth = [3 3],
linecolor = [:red :blue],
tickfontsize = 12,
grid = :no,
xlabel = "p",
ylabel = "q(p)",
label = ["q(p)" "Quantity Demanded"])using Plots;
plot(x, [y q_d],
linestyle = [:solid :dot],
linewidth = [3 3],
linecolor = [:red :blue],
tickfontsize = 12,
grid = :no,
xlabel = "p",
ylabel = "q(p)",
label = ["q(p)" "Quantity Demanded"])A solution exist!
Notice: if we let \(t = p^{-0.1}\) then
\[q(t) = 0.5t^2 + 0.5t^5\]
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Can we solve for \(t\) now?
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No! Closed-form solutions to fifth order polynomials are not guaranteed to exist!
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So how do we solve the problem?
Iteratively do the following:
We will learn more about this and why it works in a later class
# Define demand functions
demand(p) = p^(-0.2)/2 + p^(-0.5)/2 - 2; # demanded quantity minus target
demand_grad(p) = .1*p^(-1.2) + .25*p^(-1.5); # demand gradient. . .
function find_root_newton(demand, demand_grad)
p = .3 # initial guess
deltap = 1e10 # initialize stepsize
while abs(deltap) > 1e-4
deltap = demand(p)/demand_grad(p)
p += deltap
println("Intermediate guess of p = $(round(p,digits=3)).")
end
println("The solution is p = $(round(p,digits=3)).")
return p
end;# Solve for price
find_root_newton(demand, demand_grad)Intermediate guess of p = 0.068.
Intermediate guess of p = 0.115.
Intermediate guess of p = 0.147.
Intermediate guess of p = 0.154.
Intermediate guess of p = 0.154.
Intermediate guess of p = 0.154.
The solution is p = 0.154.0.15419764093200633Consider a two period ag commodity market model
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The farmer’s policy function is:
\[a(E[p]) = \frac{1}{2} + \frac{1}{2}E[p]\]
Yield \(\hat{y}\) is stochastic and only realized after planting period. It follows \(\hat{y} \sim \mathcal{N}(1, 0.1)\)
Crop production is
\[q = a\hat{y}\]
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Demand is given by
\[p(q) = 3-2q\]
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Q: How many acres get planted?
Q: How many acres get planted?
A: We plug in the stochastic quantity and take expectations
\(p(\hat{y}) = 3-2a\hat{y}\)
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\(a = \frac{1}{2} + \frac{1}{2}(3-2aE[\hat{y}])\)
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Rearrange and solve:
\(a^* = 1\)
Now suppose the government implements a price floor such that \(p \geq 1\).
We now have that
\[p(\hat{y}) = \max(1,3-2a\hat{y})\]
Q: How many acres get planted with a price floor?
Q: How many acres get planted with a price floor?
This is analytically intractable
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The max operator is non-linear so we can’t pass the expectation through
\(E[\max(1,3-2a\hat{y})] \neq \max(1,E[3-2a\hat{y}])\)
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We need to solve this numerically
We can solve this using Monte Carlo integration and function iteration
using Statistics
# Function iteration method to find a root
function find_root_fi(mn, variance)
y = randn(1000)*sqrt(variance) .+ mn # draws of the random variable
a = 1. # initial guess
differ = 100. # initialize error
exp_price = 1. # initialize expected price
while differ > 1e-4
a_old = a # save old acreage
p = max.(1, 3 .- 2 .*a.*y) # compute price at all distribution points
exp_price = mean(p) # compute expected price
a = 1/2 + 1/2*exp_price # get new acreage planted given new price
differ= abs(a - a_old) # change in acreage planted
println("Intermediate acreage guess: $(round(a,digits=3))")
end
return a, exp_price
end;acreage, expected_price = find_root_fi(1, 0.1);
println("The optimal number of acres to plant is $(round(acreage, digits = 3)).\nThe expected price is $(round(expected_price, digits = 3)).")Intermediate acreage guess: 1.132
Intermediate acreage guess: 1.093
Intermediate acreage guess: 1.103
Intermediate acreage guess: 1.1
Intermediate acreage guess: 1.101
Intermediate acreage guess: 1.101
Intermediate acreage guess: 1.101
The optimal number of acres to plant is 1.101.
The expected price is 1.202.Things to do before next class