Assignment 2

Author

Jesse Perla, UBC

Student Name/Number: (doubleclick to edit)

Instructions

Edit the above cell to include your name and student number.
Submit just this ipynb to Canvas. Do not rename, it associates your student number with the submission automatically.

using Distributions, Plots, LaTeXStrings, LinearAlgebra, Statistics, Random

Question 1

Using the code in Keynesian Multiplier example or looking at the lecture notes on Geometric Series.

Consider if the true government expenditures are \(g + \sigma \epsilon\) where \(\epsilon \sim N(0, 1)\) i.e., a unit normal and with \(\sigma = 0.01\). Consequently the law of motion for y_t becomes

\[ y_t = b y_{t-1} + i + (g + \sigma \epsilon_{t}) \]

Part (a)

Take the code (copied below) which generates the time-path of aggregate output from the initial condition (i.e., the calculate_y function) and change it code so that it implements the new process with the random \(\epsilon\)

# modify the original code below

# Function that calculates a path of y
function calculate_y(i, b, g, T, y_init)
    y = zeros(T+1)
    y[1] = y_init
    for t = 2:(T+1)
        y[t] = b * y[t-1] + i + g
    end
  return y
end

# Initial values
i_0 = 0.3
g_0 = 0.3
b = 2/3 # proportion of income to consumption
y_init = 0
T = 100

plt = plot(0:T, calculate_y(i_0, b, g_0, T, y_init),
           title= "Path of Aggregate Output Over Time",
           ylim= (0.5, 1.9), xlabel = L"t", ylabel = L"y_t")
hline!([i_0 / (1 - b) + g_0 / (1 - b)], linestyle=:dash, seriestype="hline",
       legend = false)

Part (b)

Redo the “Changing Consumption as a Fraction of Income” figure with the random simulations by modifying the original code below

# modify the original code below:
# Changing fraction of consumption
bs = round.([1/3, 2/3, 5/6, 0.9], digits = 2)

plt = plot(title= "Changing Consumption as a Fraction of Income",
           xlabel = L"t", ylabel = L"y_t", legend = :topleft)
[plot!(plt, 0:T, calculate_y(i_0, b, g_0, T, y_init), label=L"b = %$b")
 for b in bs]
plt

Part (c)

Redo the “An Increase in Investment on Output” and “An Increase in Government Spending on Output” figures by modifying the original code below

# modify the original code below:
x = 0:T
y_0 = calculate_y(i_0, b, g_0, T, y_init)
l = @layout [a ; b]

# Changing initial investment:
i_1 = 0.4
y_1 = calculate_y(i_1, b, g_0, T, y_init)
plt_1 = plot(x,y_0, label = L"i=0.3", linestyle= :dash,
             title= "An Increase in Investment on Output",
             xlabel = L"t", ylabel = L"y_t", legend = :bottomright)
plot!(plt_1, x, y_1, label = L"i=0.4")

# Changing government spending
g_1 = 0.4
y_1 = calculate_y(i_0, b, g_1, T, y_init)
plt_2 = plot(x,y_0, label = L"g=0.3", linestyle= :dash,
             title= "An Increase in Government Spending on Output",
             xlabel = L"t", ylabel = L"y_t", legend = :bottomright)
plot!(plt_2, x, y_1, label=L"g=0.4")
plot(plt_1, plt_2, layout = l)

Reusable functions, do not modify without clearly annotating your changes.

# Iterates a function from an initial condition 
function iterate_map(f, x0, T)
    x = zeros(T + 1)
    x[1] = x0
    for t in 2:(T + 1)
        x[t] = f(x[t - 1])
    end
    return x
end

function plot45(f, xmin, xmax, x0, T; num_points = 100, label = L"h(k)",
                xlabel = "k", size = (600, 500))
    # Plot the function and the 45 degree line
    x_grid = range(xmin, xmax, num_points)
    plt = plot(x_grid, f.(x_grid); xlim = (xmin, xmax), ylim = (xmin, xmax),
               linecolor = :black, lw = 2, label, size)
    plot!(x_grid, x_grid; linecolor = :blue, lw = 2, label = nothing)

    # Iterate map and add ticks
    x = iterate_map(f, x0, T)
    if !isnothing(xlabel) && T > 1
      xticks!(x, [L"%$(xlabel)_{%$i}" for i in 0:T])
      yticks!(x, [L"%$(xlabel)_{%$i}" for i in 0:T])
    end

    # Plot arrows and dashes
    for i in 1:T
        plot!([x[i], x[i]], [x[i], x[i + 1]], arrow = :closed, linecolor = :black,
              alpha = 0.5, label = nothing)
        plot!([x[i], x[i + 1]], [x[i + 1], x[i + 1]], arrow = :closed,
              linecolor = :black, alpha = 0.5, label = nothing)
        plot!([x[i + 1], x[i + 1]], [0, x[i + 1]], linestyle = :dash,
              linecolor = :black, alpha = 0.5, label = nothing)
    end
    plot!([x[1], x[1]], [0, x[1]], linestyle = :dash, linecolor = :black,
          alpha = 0.5, label = nothing)
end


function ts_plot(f, x0, T; xlabel=L"t", label=L"k_t")
    x = iterate_map(f, x0, T)
    plot(0:T, x; xlabel, label)
    plot!(0:T, x; seriestype=:scatter, mc=:blue, alpha=0.7, label=nothing)
end

ts_plot (generic function with 1 method)

Question 2

Take the Solow growth model in our lectures with the code above.

In this question we will explore the depreciation parameter, \(\delta\). The baseline parameters are the same and given in the code below.

Recall that the steady state is

\[ \bar{k} = \left(\frac{s \bar{z}}{g_N + \delta}\right)^{\frac{1}{1-\alpha}} \]

Part (a)

With k_0=0.25 use the plot45 contrasting the delta = 0.1 and delta = 0.001 to the existing delta = 0.4. The plots should be done in separate cells. Adapt the range and domain as required.

h(k; p) = (1 / (1 + p.g_N)) * (
   p.s * p.z_bar * k^p.alpha
   + (1 - p.delta) * k)
k_bar(p) = (p.s * p.z_bar /
      (p.g_N + p.delta))^(1/(1-p.alpha))

p_baseline = (;z_bar = 2.0, s = 0.3, alpha = 0.3,
     delta = 0.4, g_N = 0.0)
k_0 = 0.25
k_min = 0.0
k_max = 3.0
T = 5

# add code here and new cells as required.

Part (b)

Plot the time series with ts_plot for those same cases

k_0 = 0.25

p_baseline = (;z_bar = 2.0, s = 0.3, alpha = 0.3,
     delta = 0.4, g_N = 0.0)
k_min = 0.0
k_max = 4.0

# add code here and new cells as required.

4.0

Part (c)

Can you interpret the results? What is happening as \(\delta \to 0\) and why?

(double click to edit your answer)

Question 3

Take the Solow growth model in our lectures.

In this question we will explore the \(\alpha\) parameter. The baseline parameters are the same and given in the code below.

Part (a)

Find the new \(k^*\) using the formula for the steady state for the case of alpha = 0.8 and alpha = 0.99 contrasting with the alpha = 0.3 default. Hint: it might diverge

p_baseline = (;z_bar = 2.0, s = 0.3, alpha = 0.3,
     delta = 0.4, g_N = 0.0)
k_min = 0.0
k_max = 4.0
# Add code here

4.0

Part (b)

Plot ts_plot and plot45 for these cases, starting at k_0=0.25 as before for alpha = 0.8 and alpha = 0.99. The plots should be done in separate cells.

k_0 = 0.25
k_min = 0.0
k_max = 3.0
p_baseline = (;z_bar = 2.0, s = 0.3, alpha = 0.3,
     delta = 0.4, g_N = 0.0)
# Add code and create new cells as required

(z_bar = 2.0, s = 0.3, alpha = 0.3, delta = 0.4, g_N = 0.0)

Part (c)

What is your interpretation? What is happening to the steady state and convergence?

(double click to edit your answer)

Part (d)

Now take the same case with alpha=0.8 but now have a higher depreciation rate, delta=0.8. Plot ts_plot and plot45 as before

k_0 = 0.25
p_baseline = (;z_bar = 2.0, s = 0.3, alpha = 0.3,
     delta = 0.4, g_N = 0.0)  # for reference
# Add code and create new cells as required

(z_bar = 2.0, s = 0.3, alpha = 0.3, delta = 0.4, g_N = 0.0)

Try to guess what would happen as \(\alpha \to 1\), and how it depends on \(\delta\).

(double click to edit your answer)

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Student Name/Number: (doubleclick to edit)

Instructions

Question 1

Part (a)

Part (b)

Part (c)

Question 2

Part (a)

Part (b)

Part (c)

Question 3

Part (a)

Part (b)

Part (c)

Part (d)

Reuse