α = 0.123
β = 0.345
P = [1-α α ; β 1-β]2×2 Matrix{Float64}:
0.877 0.123
0.345 0.655Discrete Dynamic Programming
Markov Chains
A worker’s employment dynamics obey the stochastic matrix
\[P = \begin{bmatrix} 1-\alpha & \alpha \\ \beta & 1-\beta \end{bmatrix}\]
\[P = \begin{bmatrix} 1-\alpha & ... \\ \beta & ... \end{bmatrix}\]
with \(\alpha\in(0,1)\) and \(\beta\in (0,1)\). First line corresponds to employment, second line to unemployment.
Which is the stationary equilibrium? (choose any value for \(\alpha\) and \(\beta\))
α = 0.123
β = 0.345
γ = 0.678
P = [α/2 1-α α/2; 1-β β/2 β/2; γ*0.8 γ*0.2 1-γ]3×3 Matrix{Float64}:
0.0615 0.877 0.0615
0.655 0.1725 0.1725
0.5424 0.1356 0.322α = 0.123
β = 0.345
γ = 0.678
P = [α/2 1-α α/2; 1-β β/2 β/2; γ*0.8 γ*0.2 1-γ]
P3×3 Matrix{Float64}:
0.0615 0.877 0.0615
0.655 0.1725 0.1725
0.5424 0.1356 0.322Pp = P'3×3 adjoint(::Matrix{Float64}) with eltype Float64:
0.0615 0.655 0.5424
0.877 0.1725 0.1356
0.0615 0.1725 0.3223-element Vector{Float64}:
0.41963333333333336
0.39503333333333335
0.18533333333333332μbar = μ0'*P^50 # stupid too many matrix multiplications 1×3 adjoint(::Vector{Float64}) with eltype Float64:
0.400406 0.44903 0.150564μ0 = ones(3)/3 # initial probability vector
for t=1:40
μ1 = Pp * μ0
μ0 = μ1
end
μ03-element Vector{Float64}:
0.4004056827203767
0.4490299796026508
0.1505643376769728size(P, 1)3# refactor as a function to solve any stochastic matrix by simulation
function solve_by_simulation(P; T=1000, τ_η=1e-10)
n = size(P, 1)
Pp = P'
μ0 = ones(n)/n # initial probability vector
for t=1:T
μ1 = Pp * μ0
# check whether we have a fixed point
# ϵ = maximum(abs, μ1 - Pp*μ1)
# not useful because Pp*μ1 will be computed in the next iteration
# check successsive approximation errors
η = maximum(abs,μ1 - μ0)
if η<τ_η
return μ1
end
μ0 = μ1
end
error("No convegence")
endsolve_by_simulation (generic function with 1 method)solve_by_simulation(P)3-element Vector{Float64}:
0.40040568284433287
0.4490299794605776
0.15056433769508987### Solve with linear algebra
# let's solve M μ = μ by defining the appropriate Musing LinearAlgebra: IP' - I3×3 Matrix{Float64}:
-0.9385 0.655 0.5424
0.877 -0.8275 0.1356
0.0615 0.1725 -0.678# using least square solver
F = cat( P' - I, ones(1, 3); dims=1)
R = zeros(4)
R[end] = 1.0
F \ R # here it solves a nonsquare system3-element Vector{Float64}:
0.4004056828214058
0.44902997948685536
0.15056433769173883# using linear solver
M = cat( P' - I,; dims=1)
# replace last line with ones
M[end,:] .= 1.0
D = zeros(3)
D[end] = 1.0
M \ D3-element Vector{Float64}:
0.40040568282140593
0.44902997948685536
0.15056433769173877In the long run, what will the the fraction \(p\) of time spent unemployed? (Denote by \(X_m\) the fraction of dates were one is unemployed)
using Plots
function forecast(P; T=100, μ0=ones(size(P,1))/size(P,1) )
sim = [μ0]
Pp = P'
# initial probability vector
for t=1:T
μ1 = Pp * μ0
push!(sim, μ1)
μ0 = μ1
end
return sim
endforecast (generic function with 1 method)forecast(P)101-element Vector{Vector{Float64}}:
[0.3333333333333333, 0.3333333333333333, 0.3333333333333333]
[0.41963333333333336, 0.39503333333333335, 0.18533333333333332]
[0.3850790833333334, 0.4612928833333334, 0.15362803333333333]
[0.4091570474883334, 0.4381193397783334, 0.15272361273333335]
[0.3949686135219009, 0.4551156386456709, 0.14991574783242834]
[0.4037056146688205, 0.4452234971311626, 0.15107088820001704]
[0.39839013568273324, 0.45133608975960343, 0.15027377455766347]
[0.40163462745710504, 0.4476207483073078, 0.1507446242355873]
[0.3996560039152811, 0.44988911840923734, 0.15045487767548166]
[0.4008629424500215, 0.4485058697720903, 0.1506311877778883]
⋮
[0.4004056828214061, 0.4490299794868556, 0.15056433769173902]
[0.40040568282140615, 0.4490299794868555, 0.15056433769173902]
[0.4004056828214061, 0.4490299794868556, 0.15056433769173902]
[0.40040568282140615, 0.4490299794868555, 0.15056433769173902]
[0.4004056828214061, 0.4490299794868556, 0.15056433769173902]
[0.40040568282140615, 0.4490299794868555, 0.15056433769173902]
[0.4004056828214061, 0.4490299794868556, 0.15056433769173902]
[0.40040568282140615, 0.4490299794868555, 0.15056433769173902]
[0.4004056828214061, 0.4490299794868556, 0.15056433769173902]Illustrate this convergence by generating a simulated series of length 10000 starting at \(X_0=1\). Plot \(X_m-p\) against \(m\). (Take \(\alpha=\beta=0.1\)).
Basic Asset Pricing model
A financial asset yields dividend \((x_t)\), which follows an AR1. It is evaluated using the stochastic discount factor: \(\rho_{0,t} = \beta^t \exp(y_t)\) where \(\beta<1\) and \(y_t\) is an \(AR1\). The price of the asset is given by \(p_0 = \sum_{t\geq 0} \rho_{0,t} U(x_t)\) where \(U(u)=\exp(u)^{0.5}/{0.5}\). Our goal is to find the pricing function \(p(x,y)\), which yields the price of the asset in any state.
Write down the recursive equation which must be satisfied by \(p\).
\[p_t = U(x_t) + \beta E_t \left[ \frac{e^{y_{t+1}}}{e^{y_t}} p_{t+1} \right]\]
Compute the ergodic distribution of \(x\) and \(y\).
Discretize processes \((x_t)\) and \((y_t)\) using 2 states each. How would you represent the unknown \(p()\)?
Solve for \(p()\) using successive approximations
Solve for \(p()\) by solving a linear system (homework)
Asset replacement (from Compecon)
At the beginning of each year, a manufacturer must decide whether to continue to operate an aging physical asset or replace it with a new one.
An asset that is \(a\) years old yields a profit contribution \(p(a)\) up to \(n\) years, at which point, the asset becomes unsafe and must be replaced by law.
The cost of a new asset is \(c\). What replacement policy maximizes profits?
Calibration: profit \(p(a)=50-2.5a-2.5a^2\). Maximum asset age: 5 years. Asset replacement cost: 75, annual discount factor \(\delta=0.9\).
Define kind of problem, the state space, the actions, the reward function, and the Bellman updating equation
Solve the problem using Value Function Iteration
Solve the problem using Policy Iteration. Compare with VFI.