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Multiplicative Functionals

Co-authored with Chase Coleman and Balint Szoke

Overview

This lecture is a sequel to the lecture on additive functionals

That lecture

  1. defined a special class of additive functionals driven by a first-order vector VAR
  2. by taking the exponential of that additive functional, created an associated multiplicative functional

This lecture uses this special class to create and analyze two examples

  • A log likelihood process, an object at the foundation of both frequentist and Bayesian approaches to statistical inference
  • A version of Robert E. Lucas’s [Luc03] and Thomas Tallarini’s [Tal00] approaches to measuring the benefits of moderating aggregate fluctuations

A Log-Likelihood Process

Consider a vector of additive functionals $ \{y_t\}_{t=0}^\infty $ described by

$$ \begin{aligned} x_{t+1} & = A x_t + B z_{t+1} \\ y_{t+1} - y_t & = D x_{t} + F z_{t+1}, \end{aligned} $$

where $ A $ is a stable matrix, $ \{z_{t+1}\}_{t=0}^\infty $ is an i.i.d. sequence of $ {\cal N}(0,I) $ random vectors, $ F $ is nonsingular, and $ x_0 $ and $ y_0 $ are vectors of known numbers

Evidently,

$$ x_{t+1} = \left(A - B F^{-1}D \right)x_t + B F^{-1} \left(y_{t+1} - y_t \right), $$

so that $ x_{t+1} $ can be constructed from observations on $ \{y_{s}\}_{s=0}^{t+1} $ and $ x_0 $

The distribution of $ y_{t+1} - y_t $ conditional on $ x_t $ is normal with mean $ Dx_t $ and nonsingular covariance matrix $ FF' $

Let $ \theta $ denote the vector of free parameters of the model

These parameters pin down the elements of $ A, B, D, F $

The log likelihood function of $ \{y_s\}_{s=1}^t $ is

$$ \begin{aligned} \log L_{t}(\theta) = & - {\frac 1 2} \sum_{j=1}^{t} (y_{j} - y_{j-1} - D x_{j-1})'(FF')^{-1}(y_{j} - y_{j-1} - D x_{j-1}) \\ & - {\frac t 2} \log \det (FF') - {\frac {k t} 2} \log( 2 \pi) \end{aligned} $$

Let’s consider the case of a scalar process in which $ A, B, D, F $ are scalars and $ z_{t+1} $ is a scalar stochastic process

We let $ \theta_o $ denote the “true” values of $ \theta $, meaning the values that generate the data

For the purposes of this exercise, set $ \theta_o = (A, B, D, F) = (0.8, 1, 0.5, 0.2) $

Set $ x_0 = y_0 = 0 $

Simulating sample paths

Let’s write a program to simulate sample paths of $ \{ x_t, y_{t} \}_{t=0}^{\infty} $

We’ll do this by formulating the additive functional as a linear state space model and putting the LSS struct to work

Setup

In [1]:
using InstantiateFromURL
github_project("QuantEcon/quantecon-notebooks-julia", version = "0.3.0")
In [2]:
using LinearAlgebra, Statistics
using Distributions, Parameters, Plots, QuantEcon
import Distributions: loglikelihood
gr(fmt = :png);
In [3]:
AMF_LSS_VAR = @with_kw (A, B, D, F = 0.0, ν = 0.0, lss = construct_ss(A, B, D, F, ν))

function construct_ss(A, B, D, F, ν)
    H, g = additive_decomp(A, B, D, F)

    # Build A matrix for LSS
    # Order of states is: [1, t, xt, yt, mt]
    A1 = [1 0 0 0 0]       # Transition for 1
    A2 = [1 1 0 0 0]       # Transition for t
    A3 = [0 0 A 0 0]       # Transition for x_{t+1}
    A4 = [ν 0 D 1 0]       # Transition for y_{t+1}
    A5 = [0 0 0 0 1]       # Transition for m_{t+1}
    Abar = vcat(A1, A2, A3, A4, A5)

    # Build B matrix for LSS
    Bbar = [0, 0, B, F, H]

    # Build G matrix for LSS
    # Order of observation is: [xt, yt, mt, st, tt]
    G1 = [0 0 1 0 0]               # Selector for x_{t}
    G2 = [0 0 0 1 0]               # Selector for y_{t}
    G3 = [0 0 0 0 1]               # Selector for martingale
    G4 = [0 0 -g 0 0]              # Selector for stationary
    G5 = [0 ν 0 0 0]               # Selector for trend
    Gbar = vcat(G1, G2, G3, G4, G5)

    # Build LSS struct
    x0 = [0, 0, 0, 0, 0]
    S0 = zeros(5, 5)
    return LSS(Abar, Bbar, Gbar, mu_0 = x0, Sigma_0 = S0)
end

function additive_decomp(A, B, D, F)
    A_res = 1 / (1 - A)
    g = D * A_res
    H = F + D * A_res * B

    return H, g
end

function multiplicative_decomp(A, B, D, F, ν)
    H, g = additive_decomp(A, B, D, F)
    ν_tilde = ν + 0.5 * H^2

    return ν_tilde, H, g
end

function loglikelihood_path(amf, x, y)
    @unpack A, B, D, F = amf
    T = length(y)
    FF = F^2
    FFinv = inv(FF)
    temp = y[2:end] - y[1:end-1] - D*x[1:end-1]
    obs =  temp .* FFinv .* temp
    obssum = cumsum(obs)
    scalar = (log(FF) + log(2pi)) * (1:T-1)
    return -0.5 * (obssum + scalar)
end

function loglikelihood(amf, x, y)
    llh = loglikelihood_path(amf, x, y)
    return llh[end]
end
Out[3]:
loglikelihood (generic function with 4 methods)

The heavy lifting is done inside the AMF_LSS_VAR struct

The following code adds some simple functions that make it straightforward to generate sample paths from an instance of AMF_LSS_VAR

In [4]:
function simulate_xy(amf, T)
    foo, bar = simulate(amf.lss, T)
    x = bar[1, :]
    y = bar[2, :]
    return x, y
end

function simulate_paths(amf, T = 150, I = 5000)
    # Allocate space
    storeX = zeros(I, T)
    storeY = zeros(I, T)

    for i in 1:I
        # Do specific simulation
        x, y = simulate_xy(amf, T)

        # Fill in our storage matrices
        storeX[i, :] = x
        storeY[i, :] = y
    end

    return storeX, storeY
end

function population_means(amf, T = 150)
    # Allocate Space
    xmean = zeros(T)
    ymean = zeros(T)

    # Pull out moment generator
    moment_generator = moment_sequence(amf.lss)
    for (tt, x) = enumerate(moment_generator)
        ymeans = x[2]
        xmean[tt] = ymeans[1]
        ymean[tt] = ymeans[2]
        if tt == T
            break
        end
    end
    return xmean, ymean
end
Out[4]:
population_means (generic function with 2 methods)

Now that we have these functions in our took kit, let’s apply them to run some simulations

In particular, let’s use our program to generate $ I = 5000 $ sample paths of length $ T = 150 $, labeled $ \{ x_{t}^i, y_{t}^i \}_{t=0}^\infty $ for $ i = 1, ..., I $

Then we compute averages of $ \frac{1}{I} \sum_i x_t^i $ and $ \frac{1}{I} \sum_i y_t^i $ across the sample paths and compare them with the population means of $ x_t $ and $ y_t $

Here goes

In [5]:
F = 0.2
amf = AMF_LSS_VAR(A = 0.8, B = 1.0, D = 0.5, F = F)

T = 150
I = 5000

# Simulate and compute sample means
Xit, Yit = simulate_paths(amf, T, I)
Xmean_t = mean(Xit, dims = 1)
Ymean_t = mean(Yit, dims = 1)

# Compute population means
Xmean_pop, Ymean_pop = population_means(amf, T)

# Plot sample means vs population means
plt_1 = plot(Xmean_t', color = :blue, label = "1/I sum_i x_t^i")
plot!(plt_1, Xmean_pop, color = :black, label = "E x_t")
plot!(plt_1, title = "x_t", xlim = (0, T), legend = :bottomleft)

plt_2 = plot(Ymean_t', color = :blue, label = "1/I sum_i x_t^i")
plot!(plt_2, Ymean_pop, color = :black, label = "E y_t")
plot!(plt_2, title = "y_t", xlim = (0, T), legend = :bottomleft)

plot(plt_1, plt_2, layout = (2, 1), size = (800,500))
Out[5]:

Simulating log-likelihoods

Our next aim is to write a program to simulate $ \{\log L_t \mid \theta_o\}_{t=1}^T $

We want as inputs to this program the same sample paths $ \{x_t^i, y_t^i\}_{t=0}^T $ that we have already computed

We now want to simulate $ I = 5000 $ paths of $ \{\log L_t^i \mid \theta_o\}_{t=1}^T $

  • For each path, we compute $ \log L_T^i / T $
  • We also compute $ \frac{1}{I} \sum_{i=1}^I \log L_T^i / T $

Then we to compare these objects

Below we plot the histogram of $ \log L_T^i / T $ for realizations $ i = 1, \ldots, 5000 $

In [6]:
function simulate_likelihood(amf, Xit, Yit)
    # Get size
    I, T = size(Xit)

    # Allocate space
    LLit = zeros(I, T-1)

    for i in 1:I
        LLit[i, :] = loglikelihood_path(amf, Xit[i, :], Yit[i, :])
    end

    return LLit
end

# Get likelihood from each path x^{i}, Y^{i}
LLit = simulate_likelihood(amf, Xit, Yit)

LLT = 1 / T * LLit[:, end]
LLmean_t = mean(LLT)

plot(seriestype = :histogram, LLT, label = "")
plot!(title = "Distribution of (I/T)log(L_T)|theta_0")
vline!([LLmean_t], linestyle = :dash, color = :black, lw = 2, alpha = 0.6, label = "")
Out[6]:

Notice that the log likelihood is almost always nonnegative, implying that $ L_t $ is typically bigger than 1

Recall that the likelihood function is a pdf (probability density function) and not a probability measure, so it can take values larger than 1

In the current case, the conditional variance of $ \Delta y_{t+1} $, which equals $ FF^T=0.04 $, is so small that the maximum value of the pdf is 2 (see the figure below)

This implies that approximately $ 75\% $ of the time (a bit more than one sigma deviation), we should expect the increment of the log likelihood to be nonnegative

Let’s see this in a simulation

In [7]:
normdist = Normal(0, F)
mult = 1.175
println("The pdf at +/- $mult sigma takes the value: $(pdf(normdist,mult*F))")
println("Probability of dL being larger than 1 is approx: "*
        "$(cdf(normdist,mult*F)-cdf(normdist,-mult*F))")

# Compare this to the sample analogue:
L_increment = LLit[:,2:end] - LLit[:,1:end-1]
r,c = size(L_increment)
frac_nonegative = sum(L_increment.>=0)/(c*r)
print("Fraction of dlogL being nonnegative in the sample is: $(frac_nonegative)")
The pdf at +/- 1.175 sigma takes the value: 1.0001868966924388
Probability of dL being larger than 1 is approx: 0.7600052842019751
Fraction of dlogL being nonnegative in the sample is: 0.7610513513513514

Let’s also plot the conditional pdf of $ \Delta y_{t+1} $

In [8]:
xgrid = range(-1,  1, length = 100)
println("The pdf at +/- one sigma takes the value: $(pdf(normdist, F)) ")
plot(xgrid, pdf.(normdist, xgrid), label = "")
plot!(title = "Conditional pdf f(Delta y_(t+1) | x_t)")
The pdf at +/- one sigma takes the value: 1.2098536225957168 
Out[8]:

An alternative parameter vector

Now consider alternative parameter vector $ \theta_1 = [A, B, D, F] = [0.9, 1.0, 0.55, 0.25] $

We want to compute $ \{\log L_t \mid \theta_1\}_{t=1}^T $

The $ x_t, y_t $ inputs to this program should be exactly the same sample paths $ \{x_t^i, y_t^i\}_{t=0}^T $ that we we computed above

This is because we want to generate data under the $ \theta_o $ probability model but evaluate the likelihood under the $ \theta_1 $ model

So our task is to use our program to simulate $ I = 5000 $ paths of $ \{\log L_t^i \mid \theta_1\}_{t=1}^T $

  • For each path, compute $ \frac{1}{T} \log L_T^i $
  • Then compute $ \frac{1}{I}\sum_{i=1}^I \frac{1}{T} \log L_T^i $

We want to compare these objects with each other and with the analogous objects that we computed above

Then we want to interpret outcomes

A function that we constructed can handle these tasks

The only innovation is that we must create an alternative model to feed in

We will creatively call the new model amf2

We make three graphs

  • the first sets the stage by repeating an earlier graph
  • the second contains two histograms of values of log likelihoods of the two models over the period $ T $
  • the third compares likelihoods under the true and alternative models

Here’s the code

In [9]:
# Create the second (wrong) alternative model
amf2 = AMF_LSS_VAR(A = 0.9, B = 1.0, D = 0.55, F = 0.25) # parameters for θ_1 closer to θ_0

# Get likelihood from each path x^{i}, y^{i}
LLit2 = simulate_likelihood(amf2, Xit, Yit)

LLT2 = 1/(T-1) * LLit2[:, end]
LLmean_t2 = mean(LLT2)

plot(seriestype = :histogram, LLT2, label = "")
vline!([LLmean_t2], color = :black, lw = 2, linestyle = :dash, alpha = 0.6, label = "")
plot!(title = "Distribution of (1/T)log(L_T) | theta_1)")
Out[9]:

Let’s see a histogram of the log-likelihoods under the true and the alternative model (same sample paths)

In [10]:
plot(seriestype = :histogram, LLT, bin = 50, alpha = 0.5, label = "True", normed = true)
plot!(seriestype = :histogram, LLT2, bin = 50, alpha = 0.5, label = "Alternative",
      normed = true)
vline!([mean(LLT)], color = :black, lw = 2, linestyle = :dash, label = "")
vline!([mean(LLT2)], color = :black, lw = 2, linestyle = :dash, label = "")
Out[10]:

Now we’ll plot the histogram of the difference in log likelihood ratio

In [11]:
LLT_diff = LLT - LLT2

plot(seriestype = :histogram, LLT_diff, bin = 50, label = "")
plot!(title = "(1/T)[log(L_T^i | theta_0) - log(L_T^i |theta_1)]")
Out[11]:

Interpretation

These histograms of log likelihood ratios illustrate important features of likelihood ratio tests as tools for discriminating between statistical models

  • The loglikeklihood is higher on average under the true model – obviously a very useful property
  • Nevertheless, for a positive fraction of realizations, the log likelihood is higher for the incorrect than for the true model
  • in these instances, a likelihood ratio test mistakenly selects the wrong model
  • These mechanics underlie the statistical theory of mistake probabilities associated with model selection tests based on likelihood ratio

(In a subsequent lecture, we’ll use some of the code prepared in this lecture to illustrate mistake probabilities)

Benefits from Reduced Aggregate Fluctuations

Now let’s turn to a new example of multiplicative functionals

This example illustrates ideas in the literatures on

  • long-run risk in the consumption based asset pricing literature (e.g., [BY04], [HHL08], [Han07])
  • benefits of eliminating aggregate fluctuations in representative agent macro models (e.g., [Tal00], [Luc03])

Let $ c_t $ be consumption at date $ t \geq 0 $

Suppose that $ \{\log c_t \}_{t=0}^\infty $ is an additive functional described by

$$ \log c_{t+1} - \log c_t = \nu + D \cdot x_t + F \cdot z_{t+1} $$

where

$$ x_{t+1} = A x_t + B z_{t+1} $$

Here $ \{z_{t+1}\}_{t=0}^\infty $ is an i.i.d. sequence of $ {\cal N}(0,I) $ random vectors

A representative household ranks consumption processes $ \{c_t\}_{t=0}^\infty $ with a utility functional $ \{V_t\}_{t=0}^\infty $ that satisfies

$$ \log V_t - \log c_t = U \cdot x_t + {\sf u} \tag{1} $$

where

$$ U = \exp(-\delta) \left[ I - \exp(-\delta) A' \right]^{-1} D $$

and

$$ {\sf u} = {\frac {\exp( -\delta)}{ 1 - \exp(-\delta)}} {\nu} + \frac{(1 - \gamma)}{2} {\frac {\exp(-\delta)}{1 - \exp(-\delta)}} \biggl| D' \left[ I - \exp(-\delta) A \right]^{-1}B + F \biggl|^2, $$

Here $ \gamma \geq 1 $ is a risk-aversion coefficient and $ \delta > 0 $ is a rate of time preference

Consumption as a multiplicative process

We begin by showing that consumption is a multiplicative functional with representation

$$ \frac{c_t}{c_0} = \exp(\tilde{\nu}t ) \left( \frac{\tilde{M}_t}{\tilde{M}_0} \right) \left( \frac{\tilde{e}(x_0)}{\tilde{e}(x_t)} \right) \tag{2} $$

where $ \left( \frac{\tilde{M}_t}{\tilde{M}_0} \right) $ is a likelihood ratio process and $ \tilde M_0 = 1 $

At this point, as an exercise, we ask the reader please to verify the follow formulas for $ \tilde{\nu} $ and $ \tilde{e}(x_t) $ as functions of $ A, B, D, F $:

$$ \tilde \nu = \nu + \frac{H \cdot H}{2} $$

and

$$ \tilde e(x) = \exp[g(x)] = \exp \bigl[ D' (I - A)^{-1} x \bigr] $$

Simulating a likelihood ratio process again

Next, we want a program to simulate the likelihood ratio process $ \{ \tilde{M}_t \}_{t=0}^\infty $

In particular, we want to simulate 5000 sample paths of length $ T=1000 $ for the case in which $ x $ is a scalar and $ [A, B, D, F] = [0.8, 0.001, 1.0, 0.01] $ and $ \nu = 0.005 $

After accomplishing this, we want to display a histogram of $ \tilde{M}_T^i $ for $ T=1000 $

Here is code that accomplishes these tasks

In [12]:
function simulate_martingale_components(amf, T = 1_000, I = 5_000)
    # Get the multiplicative decomposition
    @unpack A, B, D, F, ν, lss = amf
    ν, H, g = multiplicative_decomp(A, B, D, F, ν)

    # Allocate space
    add_mart_comp = zeros(I, T)

    # Simulate and pull out additive martingale component
    for i in 1:I
        foo, bar = simulate(lss, T)
        # Martingale component is third component
        add_mart_comp[i, :] = bar[3, :]
    end

    mul_mart_comp = exp.(add_mart_comp' .- (0:T-1) * H^2 / 2)'

    return add_mart_comp, mul_mart_comp
end

# Build model
amf_2 = AMF_LSS_VAR(A = 0.8, B = 0.001, D = 1.0, F = 0.01, ν = 0.005)

amc, mmc = simulate_martingale_components(amf_2, 1_000, 5_000)

amcT = amc[:, end]
mmcT = mmc[:, end]

println("The (min, mean, max) of additive Martingale component in period T is")
println("\t ($(minimum(amcT)), $(mean(amcT)), $(maximum(amcT)))")

println("The (min, mean, max) of multiplicative Martingale component in period T is")
println("\t ($(minimum(mmcT)), $(mean(mmcT)), $(maximum(mmcT)))")
The (min, mean, max) of additive Martingale component in period T is
	 (-1.613317955324071, 0.006982488586911062, 1.636135037412871)
The (min, mean, max) of multiplicative Martingale component in period T is
	 (0.17804740296993868, 1.003834483098736, 4.5893919265142165)

Comments

  • The preceding min, mean, and max of the cross-section of the date $ T $ realizations of the multiplicative martingale component of $ c_t $ indicate that the sample mean is close to its population mean of 1

    • This outcome prevails for all values of the horizon $ T $
  • The cross-section distribution of the multiplicative martingale component of $ c $ at date $ T $ approximates a log normal distribution well

  • The histogram of the additive martingale component of $ \log c_t $ at date $ T $ approximates a normal distribution well

Here’s a histogram of the additive martingale component

In [13]:
plot(seriestype = :histogram, amcT, bin = 25, normed = true, label = "")
plot!(title = "Histogram of Additive Martingale Component")
Out[13]:

Here’s a histogram of the multiplicative martingale component

In [14]:
plot(seriestype = :histogram, mmcT, bin = 25, normed = true, label = "")
plot!(title = "Histogram of Multiplicative Martingale Component")
Out[14]:

Representing the likelihood ratio process

The likelihood ratio process $ \{\widetilde M_t\}_{t=0}^\infty $ can be represented as

$$ \widetilde M_t = \exp \biggl( \sum_{j=1}^t \biggl(H \cdot z_j -\frac{ H \cdot H }{2} \biggr) \biggr), \quad \widetilde M_0 =1 , $$

where $ H = [F + B'(I-A')^{-1} D] $

It follows that $ \log {\widetilde M}_t \sim {\mathcal N} ( -\frac{t H \cdot H}{2}, t H \cdot H ) $ and that consequently $ {\widetilde M}_t $ is log normal

Let’s plot the probability density functions for $ \log {\widetilde M}_t $ for $ t=100, 500, 1000, 10000, 100000 $

Then let’s use the plots to investigate how these densities evolve through time

We will plot the densities of $ \log {\widetilde M}_t $ for different values of $ t $

Here is some code that tackles these tasks

In [15]:
function Mtilde_t_density(amf, t; xmin = 1e-8, xmax = 5.0, npts = 5000)

    # Pull out the multiplicative decomposition
    νtilde, H, g =
        multiplicative_decomp(amf.A, amf.B, amf.D, amf.F, amf.ν)
    H2 = H*H

    # The distribution
    mdist = LogNormal(-t * H2 / 2, sqrt(t * H2))
    x = range(xmin,  xmax, length = npts)
    p = pdf.(mdist, x)

    return x, p
end

function logMtilde_t_density(amf, t; xmin = -15.0, xmax = 15.0, npts = 5000)

    # Pull out the multiplicative decomposition
    @unpack A, B, D, F, ν = amf
    νtilde, H, g = multiplicative_decomp(A, B, D, F, ν)
    H2 = H * H

    # The distribution
    lmdist = Normal(-t * H2 / 2, sqrt(t * H2))
    x = range(xmin,  xmax, length = npts)
    p = pdf.(lmdist, x)

    return x, p
end

times_to_plot = [10, 100, 500, 1000, 2500, 5000]
dens_to_plot = [Mtilde_t_density(amf_2, t, xmin=1e-8, xmax=6.0) for t in times_to_plot]
ldens_to_plot = [logMtilde_t_density(amf_2, t, xmin=-10.0, xmax=10.0) for t in times_to_plot]

# plot_title = "Densities of M_t^tilda" is required, however, plot_title is not yet
# supported in Plots
plots = plot(layout = (3,2), size = (600,800))

for (it, dens_t) in enumerate(dens_to_plot)
    x, pdf = dens_t
    plot!(plots[it], title = "Density for time (time_to_plot[it])")
    plot!(plots[it], pdf, fillrange = [[0], pdf], label = "")
end
plot(plots)
Out[15]:

These probability density functions illustrate a peculiar property of log likelihood ratio processes:

  • With respect to the true model probabilities, they have mathematical expectations equal to $ 1 $ for all $ t \geq 0 $
  • They almost surely converge to zero

Welfare benefits of reduced random aggregate fluctuations

Suppose in the tradition of a strand of macroeconomics (for example Tallarini [Tal00], [Luc03]) we want to estimate the welfare benefits from removing random fluctuations around trend growth

We shall compute how much initial consumption $ c_0 $ a representative consumer who ranks consumption streams according to (1) would be willing to sacrifice to enjoy the consumption stream

$$ \frac{c_t}{c_0} = \exp (\tilde{\nu} t) $$

rather than the stream described by equation (2)

We want to compute the implied percentage reduction in $ c_0 $ that the representative consumer would accept

To accomplish this, we write a function that computes the coefficients $ U $ and $ u $ for the original values of $ A, B, D, F, \nu $, but also for the case that $ A, B, D, F = [0, 0, 0, 0] $ and $ \nu = \tilde{\nu} $

Here’s our code

In [16]:
function Uu(amf, δ, γ)
    @unpack A, B, D, F, ν = amf
    ν_tilde, H, g = multiplicative_decomp(A, B, D, F, ν)

    resolv = 1 / (1 - exp(-δ) * A)
    vect = F + D * resolv * B

    U_risky = exp(-δ) * resolv * D
    u_risky = exp(-δ) / (1 - exp(-δ)) * (ν + 0.5 * (1 - γ) * (vect^2))

    U_det = 0
    u_det = exp(-δ) / (1 - exp(-δ))  * ν_tilde

    return U_risky, u_risky, U_det, u_det
end

# Set remaining parameters
δ = 0.02
γ = 2.0

# Get coeffs
U_r, u_r, U_d, u_d = Uu(amf_2, δ, γ)
Out[16]:
(4.54129843114712, 0.24220854072375247, 0, 0.25307727077652764)

The values of the two processes are

$$ \begin{aligned} \log V^r_0 &= \log c^r_0 + U^r x_0 + u^r \\ \log V^d_0 &= \log c^d_0 + U^d x_0 + u^d \end{aligned} $$

We look for the ratio $ \frac{c^r_0-c^d_0}{c^r_0} $ that makes $ \log V^r_0 - \log V^d_0 = 0 $

$$ \begin{aligned} \underbrace{ \log V^r_0 - \log V^d_0}_{=0} + \log c^d_0 - \log c^r_0 &= (U^r-U^d) x_0 + u^r - u^d \\ \frac{c^d_0}{ c^r_0} &= \exp\left((U^r-U^d) x_0 + u^r - u^d\right) \end{aligned} $$

Hence, the implied percentage reduction in $ c_0 $ that the representative consumer would accept is given by

$$ \frac{c^r_0-c^d_0}{c^r_0} = 1 - \exp\left((U^r-U^d) x_0 + u^r - u^d\right) $$

Let’s compute this

In [17]:
x0 = 0.0  # initial conditions
logVC_r = U_r * x0 + u_r
logVC_d = U_d * x0 + u_d

perc_reduct = 100 * (1 - exp(logVC_r - logVC_d))
perc_reduct
Out[17]:
1.0809878812017448

We find that the consumer would be willing to take a percentage reduction of initial consumption equal to around 1.081