Code should execute sequentially if run in a Jupyter notebook

- See the set up page to install Jupyter, Julia (1.0+) and all necessary libraries
- Please direct feedback to contact@quantecon.org or the discourse forum
- For some notebooks, enable content with "Trust" on the command tab of Jupyter lab
- If using QuantEcon lectures for the first time on a computer, execute
`] add InstantiateFromURL`inside of a notebook or the REPL

# Multiplicative Functionals¶

Co-authored with Chase Coleman and Balint Szoke

## Overview¶

This lecture is a sequel to the lecture on additive functionals

That lecture

- defined a special class of
**additive functionals**driven by a first-order vector VAR - 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¶

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

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 $

### Setup¶

```
using InstantiateFromURL
activate_github("QuantEcon/QuantEconLecturePackages", tag = "v0.9.7");
```

```
using LinearAlgebra, Statistics, Compat
using Distributions, Parameters, Plots, QuantEcon
import Distributions: loglikelihood
gr(fmt = :png);
```

```
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
```

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

```
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
```

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

```
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))
```

### 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 $

```
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 = "")
```

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

```
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)")
```

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

```
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)")
```

### 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

```
# 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)")
```

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

```
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 = "")
```

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

```
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)]")
```

### 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

```
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)))")
```

#### 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

```
plot(seriestype = :histogram, amcT, bin = 25, normed = true, label = "")
plot!(title = "Histogram of Additive Martingale Component")
```

Here’s a histogram of the multiplicative martingale component

```
plot(seriestype = :histogram, mmcT, bin = 25, normed = true, label = "")
plot!(title = "Histogram of Multiplicative Martingale Component")
```

### 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

```
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, fill_between = ([0], pdf), label = "")
end
plot(plots)
```

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

```
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)
```

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

```
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
```

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