stancon2019-intro.Rmd

---
title: "StanCon 2019"
author: "Cambridge, UK"
date: "August 20-23"
output:
  html_document:
    toc: true
    toc_depth: 2
---

## Setup

```{r setup, results="hide", message=FALSE}
knitr::opts_chunk$set(
  echo = TRUE, 
  dev = "png",
  dpi = 150,
  fig.align = "center",
  comment = NA
)
library(rstan)
library(dplyr)
library(lubridate)
library(ggplot2)
library(bayesplot)

theme_set(bayesplot::theme_default())

# seed for R's pseudo-RNGs, not Stan's
set.seed(1123) 

# load data
pest_data <- readRDS('data/pest_data.RDS')
standata_hier <- readRDS('data/standata_hier.RDS')
```

## The problem

### Background

Imagine that you are a statistician or data scientist working as an independent
contractor. One of your clients is a company that owns many residential buildings 
throughout New York City. The property manager explains that they are concerned about the number
of cockroach complaints that they receive from their buildings. Previously
the company has offered monthly visits from a pest inspector as a solution to
this problem. While this is the default solution of many property managers in
NYC, the tenants are rarely home when the inspector visits, and so the manager
reasons that this is a relatively expensive solution that is currently not very
effective.

One alternative to this problem is to deploy long term bait stations. In this
alternative, child and pet safe bait stations are installed throughout the
apartment building. Cockroaches obtain quick acting poison from these stations
and distribute it throughout the colony. The manufacturer of these bait stations
provides some indication of the space-to-bait efficacy, but the manager suspects
that this guidance was not calculated with NYC roaches in mind. NYC roaches, the
manager rationalizes, have more hustle than traditional roaches; and NYC
buildings are built differently than other common residential buildings in the
US. This is particularly important as the uni§t cost for each bait station per
year is quite high.

### The goal

The manager wishes to employ your services to help them to find the optimal
number of roach bait stations they should place in each of their buildings in
order to minimize the number of cockroach complaints while also keeping
expenditure on pest control affordable. 

A subset of the company's buildings have been randomly selected for an experiment: 

* At the beginning of each month, a pest inspector randomly places a number of
bait stations throughout the building, without knowledge of the current
cockroach levels in the building
* At the end of the month, the manager records
the total number of cockroach complaints in that building. 
* The manager would like to determine the optimal number of traps ($\textrm{traps}$) that
balances the lost revenue ($R$) that complaints ($\textrm{complaints}$) generate
with the all-in cost of maintaining the traps ($\textrm{TC}$). 

Fortunately, Bayesian data analysis provides a coherent framework for us to tackle this problem.

Formally, we are interested in finding

$$
\arg\max_{\textrm{traps} \in \mathbb{N}} \mathbb{E}_{\text{complaints}}[R(\textrm{complaints}(\textrm{traps})) - \textrm{TC}(\textrm{traps})]
$$

The property manager would also, if possible, like to learn how these results 
generalize to buildings they haven't treated so they can understand the
potential costs of pest control at buildings they are acquiring as well as for
the rest of their building portfolio.

As the property manager has complete control over the number of traps set, the
random variable contributing to this expectation is the number of complaints
given the number of traps. We will model the number of complaints as a function
of the number of traps.


## The data

The data provided to us is in a file called `pest_data.RDS`. Let's
load the data and see what the structure is:

```{r load-data}
pest_data <- readRDS('data/pest_data.RDS')
str(pest_data)
```

We have access to the following fields: 

* `complaints`: Number of complaints per building per month
* `building_id`: The unique building identifier
* `traps`: The number of traps used per month per building
* `date`: The date at which the number of complaints are recorded
* `live_in_super`: An indicator for whether the building as a live-in super
* `age_of_building`: The age of the building
* `total_sq_foot`: The total square footage of the building
* `average_tenant_age`: The average age of the tenants per building
* `monthly_average_rent`: The average monthly rent per building
* `floors`: The number of floors per building

First, let's see how many buildings we have in the data:

```{r describe-data}
length(unique(pest_data$building_id))
```

And make some plots of the raw data: 

```{r data-plot-1}
ggplot(pest_data, aes(x = complaints)) + 
  geom_bar()
```

```{r data-plot-2}
ggplot(pest_data, aes(x = traps, y = complaints, color = live_in_super == TRUE)) + 
  geom_jitter()
```

```{r data-plot-ts, fig.width = 6, fig.height = 8}
ggplot(pest_data, aes(x = date, y = complaints, color = live_in_super == TRUE)) + 
  geom_line(aes(linetype = "Number of complaints")) + 
  geom_point(color = "black") + 
  geom_line(aes(y = traps, linetype = "Number of traps"), color = "black", size = 0.25) + 
  facet_wrap(~building_id, scales = "free", ncol = 2, labeller = label_both) + 
  scale_x_date(name = "Month", date_labels = "%b") + 
  scale_y_continuous(name = "", limits = range(pest_data$complaints)) + 
  scale_linetype_discrete(name = "") + 
  scale_color_discrete(name = "Live-in super")
```


The first question we'll look at is just whether the number of complaints per
building per month is associated with the number of bait stations per building
per month, ignoring the temporal and across-building variation (we'll get to
that later). This requires only two variables, $\textrm{complaints}$ and
$\textrm{traps}$. How should we model the number of complaints?


## Modeling count data : Poisson distribution

We already know some rudimentary information about what we should expect. The
number of complaints over a month should be either zero or an integer. The
property manager tells us that it is possible but unlikely that number of
complaints in a given month is zero. Occasionally there are a very large number
of complaints in a single month. A common way of modeling this sort of skewed,
single bounded count data is as a Poisson random variable. One concern about
modeling the outcome variable as Poisson is that the data may be
over-dispersed, but we'll start with the Poisson model and then check 
whether over-dispersion is a problem by comparing our model's predictions 
to the data.

### Model 

Given that we have chosen a Poisson regression, we define the likelihood to be
the Poisson probability mass function over the number bait stations placed in
the building, denoted below as `traps`. This model assumes that the mean and
variance of the outcome variable `complaints` (number of complaints) is the
same. We'll investigate whether this is a good assumption after we fit the 
model.

For building $b = 1,\dots,10$ at time (month) $t = 1,\dots,12$, we have

$$
\begin{align*}
\textrm{complaints}_{b,t} & \sim \textrm{Poisson}(\lambda_{b,t}) \\
\lambda_{b,t} & = \exp{(\eta_{b,t})} \\
\eta_{b,t} &= \alpha + \beta \, \textrm{traps}_{b,t}
\end{align*}
$$


### Prior predictive checks

Before we fit the model to real data, we should check that our priors and model
can generate plausible simulated data. 

In R we can simulate from the prior predictive distribution using a 
function like this:

```{r simulate-poisson-data}
# using normal distributions for priors on alpha and beta 
simple_poisson_dpg <- function(traps, alpha_mean, alpha_sd, beta_mean, beta_sd) {
  N <- length(traps)
  alpha <- rnorm(1, mean = alpha_mean, sd = alpha_sd);
  beta <- rnorm(1, mean = beta_mean, sd = beta_sd);
  complaints <- rpois(N, lambda = exp(alpha + beta * traps))
  return(complaints)
}
```

Try with $N(0,10)$ priors on both parameters: 

```{r prior-pred-really-bad}
# you can run this chunk multiple times to keep generating different datasets
simple_poisson_dpg(
  pest_data$traps,
  alpha_mean = 0,
  alpha_sd = 10,
  beta_mean = 0,
  beta_sd = 10
)
```

Try with $N(0,1)$ priors on both parameters: 

```{r prior-pred-bad}
simple_poisson_dpg(
  pest_data$traps,
  alpha_mean = 0,
  alpha_sd = 1,
  beta_mean = 0,
  beta_sd = 1
)
```

Let's calculate some summary statistics of 1000 data sets generated according
to the $N(0,1)$ priors:

```{r prior-pred-check}
prior_preds <- t(replicate(1000, expr = {
  simple_poisson_dpg(
    traps = pest_data$traps,
    alpha_mean = 0,
    alpha_sd = 1,
    beta_mean = 0,
    beta_sd = 1
  )
}))

dim(prior_preds)
```

```{r prior-pred-mean}
summary(apply(prior_preds, 1, mean))
```
```{r prior-pred-min}
summary(apply(prior_preds, 1, min))
```
```{r prior-pred-max}
summary(apply(prior_preds, 1, max))
```


A more reasonable (though still quite flexible) prior:

```{r}
simple_poisson_dpg(
  traps = pest_data$traps,
  alpha_mean = log(7),
  alpha_sd = 0.5,
  beta_mean = -0.25,
  beta_sd = 0.5
)
```

Simulate 1000 times and calculate summary stats:

```{r prior-pred-better}
prior_preds <- t(replicate(1000, expr = {
  simple_poisson_dpg(
    traps = pest_data$traps,
    alpha_mean = log(7),
    alpha_sd = 0.5,
    beta_mean = -0.25,
    beta_sd = 0.5
  )
}))
```

```{r}
summary(apply(prior_preds, 1, mean))

summary(apply(prior_preds, 1, min))

summary(apply(prior_preds, 1, max))
```



### Writing and fitting our first Stan program

Let's encode the model in a Stan program. 

* Write `simple_poisson_regression.stan` together, put in `stan_programs` directory.

```{r compile-simple-poisson}
comp_simple <- stan_model('stan_programs/simple_poisson_regression.stan')
```

To fit the model to the data we'll first create a list to pass
to Stan using the variables in the `pest_data` data frame. The names of the 
list elements must match the names used in the `data` block of the Stan program.

```{r stan-data}
standata_simple <- list(
  N = nrow(pest_data), 
  complaints = pest_data$complaints,
  traps = pest_data$traps
)
str(standata_simple)
```

We have already compiled the model, so we can jump straight to sampling from it.

```{r fit_simple, cache=TRUE}
fit_simple <- sampling(comp_simple, data = standata_simple,
                       # these are the defaults but specifying them anyway
                       # so you can see how to use them: 
                       # posterior sample size = chains * (iter-warmup)
                       chains = 4, iter = 2000, warmup = 1000)
```

Print the summary of the intercept and slope parameters: 

```{r results_simple_P}
print(fit_simple, pars = c('alpha','beta'))
```

We can also plot the posterior distributions: 

```{r mcmc_hist}
# https://mc-stan.org/bayesplot/reference/MCMC-distributions
draws <- as.matrix(fit_simple, pars = c('alpha','beta'))
mcmc_hist(draws) # marginal posteriors of alpha and beta
mcmc_scatter(draws, alpha = 0.2, size = 1) # posterior of (alpha,beta)
```

And compare them to draws from the prior:


```{r compare-prior-posterior}
alpha_prior_post <- cbind(alpha_prior = rnorm(4000, log(7), 1), 
                          alpha_posterior = draws[, "alpha"])
mcmc_hist(alpha_prior_post, facet_args = list(nrow = 2), binwidth = 0.1) + 
  xlim(range(alpha_prior_post))


beta_prior_post <- cbind(beta_prior = rnorm(4000, -0.25, 0.5), 
                         beta_posterior = draws[, "beta"])
mcmc_hist(beta_prior_post, facet_args = list(nrow = 2), binwidth = 0.05) + 
  xlim(range(beta_prior_post))
```


From the posterior of `beta`, it appears the number of bait stations set in a
building is associated with the number of complaints about cockroaches that were
made in the following month. However, we still need to consider how well the
model fits.


### Posterior predictive checking

```{r y_rep_simple}
# see http://mc-stan.org/rstan/articles/stanfit_objects.html for various ways
# of extracting contents from stanfit objects
y_rep <- as.matrix(fit_simple, pars = "y_rep")
```

```{r marginal_PPC}
# https://mc-stan.org/bayesplot/reference/PPC-distributions#plot-descriptions
ppc_dens_overlay(y = standata_simple$complaints, yrep = y_rep[1:200,])
```

In the plot above we have the kernel density estimate of the observed data ($y$,
thicker curve) and 200 simulated data sets ($y_{rep}$, thin curves) from the
posterior predictive distribution. Here the posterior predictive
simulations are not as dispersed as the real data and don't seem to capture the
rate of zeros well at all. This suggests the Poisson model may not be sufficiently
flexible for this data.

Let's explore this further by looking directly at the proportion of zeros in the
real data and predicted data.
```{r ppc_stat-prop_zero}
# calculate the proportion of zeros in a vector
prop_zero <- function(x) mean(x == 0)

# https://mc-stan.org/bayesplot/reference/PPC-test-statistics#plot-descriptions
ppc_stat(
  y = standata_simple$complaints,
  yrep = y_rep,
  stat = "prop_zero",
  binwidth = .01
)
```
The plot above shows the observed proportion of zeros (thick vertical line) and
a histogram of the proportion of zeros in each of the simulated data sets. It is
clear that the model does not capture this feature of the data well at all.


The rootogram is another useful plot to compare the observed vs expected number
of complaints. This is a plot of the square root of the expected counts
(continuous line) vs the observed counts (blue histogram)

```{r rootogram}
# https://mc-stan.org/bayesplot/reference/PPC-discrete#plot-descriptions
ppc_rootogram(standata_simple$complaints, yrep = y_rep)
```

If the model was fitting well these would be relatively similar, however in this
figure we can see the number of complaints is underestimated if there are few
complaints, over-estimated for medium numbers of complaints, and underestimated
if there are a large number of complaints.

The `ppc_bars()` function will make a bar plot of the observed values and
overlay prediction intervals but not on the square root scale (unlike the
rootogram):

```{r ppc_bars}
# https://mc-stan.org/bayesplot/reference/PPC-discrete#plot-descriptions
ppc_bars(standata_simple$complaints, yrep = y_rep)
```


We can also view how the predicted number of complaints varies with the number
of traps:

```{r intervals}
ppc_intervals(y = standata_simple$complaints, yrep = y_rep,
              # jitter number of traps since multiple observations at some values
              x = standata_simple$traps + rnorm(standata_simple$N, 0, 0.2)) +  
  labs(x = "Number of traps", y = "Number of complaints")
```


## Expanding the model: multiple predictors

Currently, our model's mean parameter is a rate of complaints per 30 days, but
we're modeling a process that occurs over an area as well as over time. We have
the square footage of each building, so if we add that information into the
model, we can interpret our parameters as a rate of complaints per square foot
per 30 days.

$$
\begin{align*}
\textrm{complaints}_{b,t} & \sim \textrm{Poisson}(\textrm{sq_foot}_b\,\lambda_{b,t}) \\
\lambda_{b,t} & = \exp{(\eta_{b,t} )} \\
\eta_{b,t} &= \alpha + \beta \, \textrm{traps}_{b,t}
\end{align*}
$$

The term $\text{sq_foot}$ is called an exposure term. If we log the term, we can 
put it in $\eta_{b,t}$:

$$
\begin{align*}
\textrm{complaints}_{b,t} & \sim \textrm{Poisson}(\lambda_{b,t}) \\
\lambda_{b,t} & = \exp{(\eta_{b,t} )} \\
\eta_{b,t} &= \alpha + \beta \, \textrm{traps}_{b,t} + \textrm{log_sq_foot}_b
\end{align*}
$$


We will also include whether there is a live-in super or not as a predictor 
for the number of complaints, which gives us gives us: 

$$
\eta_{b,t} = \alpha + \beta \, \textrm{traps}_{b,t} +
\beta_{\rm super} \textrm{live_in_super}_{b,t} +
\textrm{log_sq_foot}_b
$$
 
Add these new variables to the data list for Stan.

```{r add-predictors-to-standata}
standata_simple$log_sq_foot <- pest_data$log_sq_foot_1e4 # log(total_sq_foot/1e4)
standata_simple$live_in_super <- pest_data$live_in_super
```


### Stan program for Poisson multiple regression

Now we need a new Stan program that uses multiple predictors.

* Write `multiple_poisson_regression.stan`

```{r compile-multi-poisson}
comp_multi <- stan_model('stan_programs/multiple_poisson_regression.stan')
```

### Fit the Poisson multiple regression

```{r fit_multi}
fit_multi <- sampling(comp_multi, data = standata_simple, 
                      refresh = 100) # turn off printed progress updates
print(fit_multi, pars = c("alpha", "beta", "beta_super"))
```


```{r ppc_dens_overlay-random-subset}
y_rep <- as.matrix(fit_multi, pars = "y_rep")
ppc_dens_overlay(standata_simple$complaints, y_rep[1:200,])

# in this case we have very high effective sample sizes, but if there is
# nontrivial autocorrelation then it's better to take a random sample of the draws, 
# for example:
ids <- sample(nrow(y_rep), size = 200)
ppc_dens_overlay(standata_simple$complaints, y_rep[ids, ])
```

This again looks like we haven't captured the smaller counts very well, nor
have we captured the larger counts.

We're still severely underestimating the proportion of zeros in the data:

```{r}
prop_zero <- function(x) mean(x == 0)
ppc_stat(y = standata_simple$complaints, yrep = y_rep, 
         stat = "prop_zero", binwidth = 0.01)
```

Ideally this vertical line would fall somewhere within the histogram.

We can also plot uncertainty intervals for the predicted complaints for different
numbers of traps.

```{r}
ppc_intervals(
  y = standata_simple$complaints, 
  yrep = y_rep,
  x = standata_simple$traps + rnorm(standata_simple$N, 0, 0.2)
) + 
  labs(x = "Number of traps", y = "Number of complaints")
```

We can see that we've increased the tails a bit more at the larger numbers of traps
but we still have some large observed numbers of complaints that the model
would consider extremely unlikely events. 
 

## Modeling count data with the Negative Binomial

When we considered modeling the data using a Poisson, we saw that the model
didn't appear to fit as well to the data as we would like. In particular the
model underpredicted low and high numbers of complaints, and overpredicted the
medium number of complaints. This is one indication of over-dispersion, where
the variance is larger than the mean. A Poisson model doesn't fit over-dispersed
count data very well because the same parameter $\lambda$, controls both the
expected counts and the variance of these counts. The natural alternative to
this is the negative binomial model:

$$
\begin{align*}
\text{complaints}_{b,t} & \sim \text{Neg-Binomial}(\lambda_{b,t}, \phi) \\
\lambda_{b,t} & = \exp{(\eta_{b,t})} \\
\eta_{b,t} &= \alpha + \beta \, {\rm traps}_{b,t} + \beta_{\rm super} \, {\rm super}_{b} + \text{log_sq_foot}_{b}
\end{align*}
$$

In Stan the negative binomial mass function we'll use is called 
$\texttt{neg_binomial_2_log}(\text{ints} \, y, \text{reals} \, \eta, \text{reals} \, \phi)$ 
in Stan. Like the `poisson_log` function, this negative binomial mass function
that is parameterized in terms of its log-mean, $\eta$, but it also has a
precision $\phi$ such that

$$
\mathbb{E}[y] \, = \lambda = \exp(\eta)
$$

$$
\text{Var}[y] = \lambda + \lambda^2/\phi = \exp(\eta) + \exp(\eta)^2 / \phi.
$$ 

As $\phi$ gets larger the term $\lambda^2 / \phi$ approaches zero and so 
the variance of the negative-binomial approaches $\lambda$, i.e., the
negative-binomial gets closer and closer to the Poisson.

### Stan program for negative-binomial regression

* Write `multiple_NB_regression.stan` together

```{r compile-multi-NB, cache=TRUE, results="hide", message=FALSE}
comp_multi_NB <- stan_model('stan_programs/multiple_NB_regression.stan')
```


### Fit to data and check the fit

```{r fit_multi_NB}
fit_multi_NB <- sampling(comp_multi_NB, data = standata_simple, refresh = 250)

# to demonstrate extracting as a list instead of using as.matrix
samps_NB <- rstan::extract(fit_multi_NB) 
```

Let's look at our predictions vs. the data.

```{r ppc_dens_overlay_NB}
ppc_dens_overlay(y = standata_simple$complaints, yrep = samps_NB$y_rep[1:200,])
```

It appears that our model now captures both the number of small counts better
as well as the tails. 

Let's check the proportion of zeros:

```{r prop_zero_NB}
ppc_stat(y = standata_simple$complaints, yrep = samps_NB$y_rep, 
         stat = "prop_zero", binwidth = 0.01)
```


Check predictions by number of traps: 

```{r}
ppc_intervals(
  y = standata_simple$complaints, 
  yrep = samps_NB$y_rep,
  x = standata_simple$traps + rnorm(standata_simple$N, 0, 0.2)
) + 
  labs(x = "Number of traps (jittered)", y = "Number of complaints")
```

We haven't used the fact that the data are clustered by building yet. A posterior 
predictive check might elucidate whether it would be a good idea to add the building
information into the model.

```{r ppc-group-means, fig.width=7, fig.height=6}
ppc_stat_grouped(
  y = standata_simple$complaints, 
  yrep = samps_NB$y_rep, 
  group = pest_data$building_id, 
  stat = 'mean',
  binwidth = 0.2
)
```

We're getting plausible predictions for most building means but some are
estimated better than others and some have larger uncertainties than we might
expect. If we explicitly model the variation across buildings we may be able to
get much better estimates.

## Hierarchical modeling

### Modeling varying intercepts for each building

Let's add a hierarchical intercept parameter, $\alpha_b$ at the building level
to our model.

$$
\text{complaints}_{b,t} \sim \text{Neg-Binomial}(\lambda_{b,t}, \phi) \\
\lambda_{b,t}  = \exp{(\eta_{b,t})} \\
\eta_{b,t} = \mu_b + \beta \, {\rm traps}_{b,t} + \beta_{\rm super}\, {\rm super}_b + \text{log_sq_foot}_b \\
\mu_b \sim \text{Normal}(\alpha, \sigma_{\mu})
$$

In our Stan model, $\mu_b$ is the $b$-th element of the vector
$\texttt{mu}$ which has one element per building.

One of our predictors varies only by building, so we can rewrite the above model
more efficiently like so:

$$
\eta_{b,t} = \mu_b + \beta \, {\rm traps}_{b,t} + \text{log_sq_foot}_b\\
\mu_b \sim \text{Normal}(\alpha +  \beta_{\text{super}} \, \text{super}_b , \sigma_{\mu})
$$


We have more information at the building level as well, like the average age of
the residents, the average age of the buildings, and the average per-apartment
monthly rent so we can add that data into a matrix called
`building_data`, which will have one row per building and four columns:

* `live_in_super`
* `age_of_building`
* `average_tentant_age`
* `monthly_average_rent`

We'll write the Stan model like:

$$
\eta_{b,t} = \mu_b + \beta \, {\rm traps} + \text{log_sq_foot}\\
\mu \sim \text{Normal}(\alpha + \texttt{building_data} \, \zeta, \,\sigma_{\mu})
$$

### Prepare building data for hierarchical model

We'll need to do some more data prep before we can fit our models. Firstly to
use the building variable in Stan we will need to transform it from a factor
variable to an integer variable.

```{r prep-data}
N_months <- length(unique(pest_data$date))
N_buildings <- length(unique(pest_data$building_id))
# Add some IDs for building and month
pest_data <- pest_data %>%
  mutate(
    building_fac = factor(building_id, levels = unique(building_id)),
    building_idx = as.integer(building_fac),
    ids = rep(1:N_months, N_buildings),
    mo_idx = lubridate::month(date)
  )

# Center and rescale the building specific data
building_data <- pest_data %>%
    select(
      building_idx,
      live_in_super,
      age_of_building,
      average_tenant_age,
      monthly_average_rent
    ) %>%
    unique() %>%
    arrange(building_idx) %>%
    select(-building_idx) %>%
    scale(scale=FALSE) %>%
    as.data.frame() %>%
    mutate( # scale by constants to put variables on similar scales
      age_of_building = age_of_building / 10,
      average_tenant_age = average_tenant_age / 10,
      monthly_average_rent = monthly_average_rent / 1000
    ) %>%
    as.matrix()

# Make data list for Stan
standata_hier <-
  with(pest_data,
        list(complaints = complaints,
             N = length(complaints),
             traps = traps,
             log_sq_foot = log(pest_data$total_sq_foot/1e4),
             M = N_months,
             mo_idx = as.integer(as.factor(date)),
             J = N_buildings,
             K = ncol(building_data),
             building_idx = building_idx,
             building_data = building_data
             )
        )
```

### Compile and fit the hierarchical model

Let's compile the model.

```{r comp_hier_NB, results="hide", message=FALSE}
comp_hier_NB <- stan_model('stan_programs/hier_NB_regression.stan')
```

Fit the model to data.

```{r run-NB-hier}
fit_hier_NB <-
  sampling(
    comp_hier_NB,
    data = standata_hier,
    refresh = 1000, # only report progress every 1000 iterations
    # can run markov chains in parallel. you can check how many cores you 
    # have available using parallel::detectCores()
    cores = 4 
  )
```

### Diagnostics

We get a bunch of warnings from Stan about divergent transitions, which is
an indication that there may be regions of the posterior that have not been
explored by the Markov chains.

Divergences are discussed in more detail in the course slides as well as
the **bayesplot** [MCMC diagnostics vignette](http://mc-stan.org/bayesplot/articles/visual-mcmc-diagnostics.html)
and [_A Conceptual Introduction to Hamiltonian Monte Carlo_](https://arxiv.org/abs/1701.02434).

In this example we will see that we have divergent transitions because we need to
reparameterize our model - i.e., we will retain the overall structure of the
model, but transform some of the parameters so that it is easier for Stan to
sample from the parameter space. Before we go through exactly how to do this
reparameterization, we will first go through what indicates that this is
something that reparameterization will resolve. We will go through:

1. Examining the fitted parameter values, including the effective sample size
2. Plots that reveal particular patterns in locations of the divergences.

Then we print the fits for the parameters that are of most interest.

```{r fit_hier_NB-low-ESS}
print(fit_hier_NB, pars = c('sigma_mu','beta','alpha','phi','mu'))
```
You can see that the effective sample sizes are quite low for many of the parameters
relative to the total number of samples. This alone isn't indicative of the need
to reparameterize, but it indicates that we should look further at the trace
plots and pairs plots. First let's look at the traceplots to see if the divergent
transitions form a pattern.

```{r traceplot-with-divs}
# change bayesplot color scheme (to make chains very different colors)
color_scheme_set("brewer-Spectral") 
mcmc_trace(
  # use as.array to keep the markov chains separate for trace plots
  as.array(fit_hier_NB,pars = 'sigma_mu'),
  np = nuts_params(fit_hier_NB),
  window = c(500,1000) # zoom into a window to see more clearly
)
color_scheme_set("blue")
```

Looks like the divergences (marked in red below the traceplot)
correspond to spots where sampler gets stuck.

```{r, scatter-with-divs}
# assign to object so we can compare to another plot later
scatter_with_divs <- mcmc_scatter(
  as.array(fit_hier_NB),
  pars = c("mu[4]", "sigma_mu"),
  # look at sigma on log-scale (what Stan uses under the hood for positive parameters)
  transform = list("sigma_mu" = "log"), 
  size = 1,
  alpha = 0.3,
  np = nuts_params(fit_hier_NB),
  np_style = scatter_style_np(div_size = 1.5) # change style of divergences points
)
scatter_with_divs + coord_equal()
```

What we have here is a cloud-like shape, with most of the divergences clustering
towards the bottom. We'll see a bit later that we actually want this to look more
like a funnel than a cloud, but the divergences are indicating that the sampler
can't explore the narrowing neck of the funnel.

One way to see why we should expect some version of a funnel is to look at some
simulations from the _prior_, which we can do without MCMC and thus with no risk
of sampling problems:

```{r}
N_sims <- 1000
sigma_mu_prior <- rep(NA, N_sims)
mu_prior <- rep(NA, N_sims)
for (j in 1:N_sims) {
  sigma_mu_prior[j] <- abs(rnorm(1, mean = 0, sd = 1))
  mu_prior[j] <- rnorm(1, mean = 0, sd = sigma_mu_prior[j])
}
prior_draws <- cbind("mu" = mu_prior, "log(sigma_mu)" = log(sigma_mu_prior))

mcmc_scatter(prior_draws)
```

Of course, if the data is at all informative we shouldn't expect the posterior
to look exactly like the prior. But unless the data is incredibly informative
about the parameters and the posterior concentrates away from the narrow neck of
the funnel, the sampler is going to have to confront the funnel geometry. 
(See the bayesplot 
[Visual MCMC Diagnostics](https://mc-stan.org/bayesplot/articles/visual-mcmc-diagnostics.html)
vignette for more on this.)

Another way to look at the divergences is via a parallel coordinates plot.

```{r}
# https://mc-stan.org/bayesplot/reference/MCMC-parcoord.html#plot-descriptions
# this plot is a bit computationally intensive
parcoord_with_divs <- mcmc_parcoord(
  as.array(fit_hier_NB, pars = c("sigma_mu", "mu")),
  np = nuts_params(fit_hier_NB)
)
parcoord_with_divs
```

Again, we see evidence that our problems concentrate when $\texttt{sigma_mu}$ is small.

### Reparameterize and recheck diagnostics

Instead, we should use the non-centered parameterization for $\mu_b$. We
define a vector of auxiliary variables in the parameters block,
$\texttt{mu_raw}$ that is given a $\text{Normal}(0, 1)$ prior in the model
block. We then make $\texttt{mu}$ a transformed parameter:
We can reparameterize the random intercept $\mu_b$, which is distributed:

$$
\mu_b \sim \text{Normal}(\alpha + \texttt{building_data} \, \zeta, \sigma_{\mu})
$$

```
transformed parameters {
  vector[J] mu = alpha + building_data * zeta + sigma_mu * mu_raw;
}
```

This implies $\texttt{mu}$ is
$\text{Normal}(\alpha + \texttt{building_data}\, \zeta, \sigma_\mu)$
distribution, but it decouples the dependence of the density of each element of
$\texttt{mu}$ from $\texttt{sigma_mu}$ ($\sigma_\mu$).
hier_NB_regression_ncp.stan uses the non-centered parameterization for
$\texttt{mu}$. We will examine the effective sample size of the fitted model to
see whether we've fixed the problem with our reparameterization.


```{r comp_hier_NB_ncp}
comp_hier_NB_ncp <- stan_model('stan_programs/hier_NB_regression_ncp.stan')
```

Fit the model to the data.

```{r fit_hier_NB_ncp}
fit_hier_NB_ncp <- sampling(comp_hier_NB_ncp, data = standata_hier, cores = 4, 
                            refresh = 1000)
```

You should have no divergences (or only a few) this time and much improved
effective sample sizes:

```{r fit_hier_NB_ncp-ESS}
print(fit_hier_NB_ncp, pars = c('sigma_mu','beta','alpha','phi','mu'))
```


Let's make the same scatter plot we made for the model with many divergences and 
compare: 

```{r compare-scatterplots}
scatter_no_divs <- mcmc_scatter(
  as.array(fit_hier_NB_ncp),
  pars = c("mu[4]", 'sigma_mu'),
  transform = list('sigma_mu' = "log"),
  size = 1,
  alpha = 0.3,
  np = nuts_params(fit_hier_NB_ncp)
)

bayesplot_grid(
  scatter_with_divs, scatter_no_divs,
  grid_args = list(ncol = 2), 
  ylim = c(-10, 1), 
  subtitles = c("Centered parameterization", 
                "Non-centered parameterization")
)
```

Comparing the plots, we see that the divergences were a sign that there was part
of the posterior that the chains were not exploring.

The marginal PPC plot, again.

```{r ppc-full-hier}
y_rep <- as.matrix(fit_hier_NB_ncp, pars = "y_rep")
ppc_dens_overlay(standata_hier$complaints, y_rep[1:200,])
```

This looks quite nice. 

If we've captured the building-level means well, then the
posterior distribution of means by building should match well with the observed
means of the quantity of building complaints by month.

```{r ppc-group_means-hier}
ppc_stat_grouped(
  y = standata_hier$complaints,
  yrep = y_rep,
  group = pest_data$building_id,
  stat = 'mean',
  binwidth = 0.5
)
```

We weren't doing terribly with the building-specific means before, but now they
are all estimated a bit better by the model. The model is also able to do a
decent job estimating within-building variability and, for the most part, 
the probability of zero complaints:

```{r}
ppc_stat_grouped(
  y = standata_hier$complaints,
  yrep = y_rep,
  group = pest_data$building_id,
  stat = 'sd',
  binwidth = 0.5
)
```

```{r}
prop_zero <- function(x) mean(x == 0)
ppc_stat(
  y = standata_hier$complaints,
  yrep = y_rep,
  stat = prop_zero,
  binwidth = 0.02
)

# plot separately for each building
ppc_stat_grouped(
  y = standata_hier$complaints,
  yrep = y_rep,
  group = pest_data$building_id,
  stat = prop_zero,
  binwidth = 0.05
)
```


Check quantiles of predictions: 

```{r}
q90 <- function(x) quantile(x, probs = 0.9)
ppc_stat(
  y = standata_hier$complaints,
  yrep = y_rep,
  stat = "q90",
  binwidth = 1
)
```

Predictions by number of traps:

```{r}
ppc_intervals_grouped(
  y = standata_hier$complaints,
  yrep = y_rep,
  x = standata_hier$traps + rnorm(standata_hier$N, 0, 0.2), # jitter
  group = standata_hier$building_idx
  # difference between grouping by pest_data$building_id and standata_hier$building_idx
  # is that the latter is consecutive from 1 to N_buildings to make things easier 
  # to code in Stan
) +
  labs(
    x = "Number of traps", 
    y = "Predicted complaints", 
    title = "Traps vs predicted complaints by building"
  )
```


### Varying intercepts _and_ varying slopes

We have some new data that extends the number of time points for which we
have observations for each building. This will let us explore how to expand the
model a bit more with varying _slopes_ in addition to the varying intercepts
and later also model temporal variation.

```{r load_longer_data}
standata_hier_long <- readRDS('data/standata_hier_long.RDS')
```

Perhaps if the levels of complaints differ by building, the coefficient for the
effect of traps on building does too. We can add this to our model and observe
the fit.

$$
\text{complaints}_{b,t} \sim \text{Neg-Binomial}(\lambda_{b,t}, \phi)  \\
\lambda_{b,t} = \exp{(\eta_{b,t})}\\
\eta_{b,t} = \mu_b + \kappa_b \, \texttt{traps}_{b,t} + \text{log_sq_foot}_b \\
\mu_b \sim \text{Normal}(\alpha + \texttt{building_data} \, \zeta, \sigma_{\mu}) \\
\kappa_b \sim \text{Normal}(\beta + \texttt{building_data} \, \gamma, \sigma_{\kappa})
$$

Let's compile the model.

```{r comp-NB-hier-slopes, cache=TRUE, results="hide", message=FALSE}
comp_hier_NB_slopes <- stan_model('stan_programs/hier_NB_regression_ncp_slopes_mod.stan')
```

Fit the model to data and extract the posterior draws needed for
our posterior predictive checks.

```{r run-NB-hier-slopes}
fit_hier_NB_slopes <-
  sampling(
    comp_hier_NB_slopes,
    data = standata_hier_long,
    chains = 4, 
    cores = 4,
    control = list(adapt_delta = 0.95)
  )
```

To see if the model infers building-to-building differences in, we can plot a
histogram of our marginal posterior distribution for `sigma_kappa`.

```{r, fig.width = 3, fig.height = 3}
mcmc_hist(
  as.matrix(fit_hier_NB_slopes, pars = "sigma_kappa"),
  binwidth = 0.005
)
```

```{r}
print(fit_hier_NB_slopes, pars = c('kappa','beta','alpha','phi','sigma_mu','sigma_kappa','mu'))
```


While the model can't specifically rule out zero from the posterior, it does
have mass at small non-zero numbers, so we will leave in the hierarchy over
$\texttt{kappa}$. Plotting the marginal data density again, we can see the model
still looks well calibrated.

```{r ppc-full-hier-slopes}
y_rep <- as.matrix(fit_hier_NB_slopes, pars = "y_rep")
ppc_dens_overlay(
  y = standata_hier_long$complaints,
  yrep = y_rep[1:200,]
)
```

Check quantiles of predictions: 

```{r}
q90 <- function(x) quantile(x, probs = 0.9)
ppc_stat(
  y = standata_hier_long$complaints,
  yrep = y_rep,
  stat = "q90"
)
```

## Time varying effects and structured priors

We haven't looked at how cockroach complaints change over time. Let's look
at whether there's any pattern over time.

```{r ppc-group_max-hier-slopes-mean-by-mo}
select_vec <- which(standata_hier_long$mo_idx %in% 1:12)
ppc_stat_grouped(
  y = standata_hier_long$complaints[select_vec],
  yrep = y_rep[,select_vec],
  group = standata_hier_long$mo_idx[select_vec],
  stat = 'mean'
) + xlim(0, 11)
```

We might augment our model with a log-additive monthly effect, $\texttt{mo}_t$.

$$
\eta_{b,t} = \mu_b + \kappa_b \, \texttt{traps}_{b,t} + \texttt{mo}_t + \text{log_sq_foot}_b
$$

We have complete freedom over how to specify the prior for $\texttt{mo}_t$.
There are several competing factors for how the number of complaints might
change over time. It makes sense that there might be more roaches in the
environment during the summer, but we might also expect that there is more roach
control in the summer as well. Given that we're modeling complaints, maybe after
the first sighting of roaches in a building, residents are more vigilant, and
thus complaints of roaches would increase.

This can be a motivation for using an autoregressive prior for our monthly
effects. The model is as follows:

$$
\texttt{mo}_t \sim \text{Normal}(\rho \, \texttt{mo}_{t-1}, \sigma_\texttt{mo}) \\
\equiv \\
\texttt{mo}_t = \rho \, \texttt{mo}_{t-1} +\epsilon_t , \quad \epsilon_t \sim \text{Normal}(0, \sigma_\texttt{mo}) \\
\quad \rho \in [-1,1]
$$

This equation says that the monthly effect in month $t$ is directly related to
the last month's monthly effect. Given the description of the process above,
it seems like there could be either positive or negative associations between
the months, but there should be a bit more weight placed on positive
$\rho$s, so we'll put an informative prior that pushes the parameter $\rho$ towards
0.5.

Before we write our prior, however, we have a problem: Stan doesn't implement any densities
that have support on $[-1,1]$. We can use variable transformation of a raw variable
defined on $[0,1]$ to to give us a density on $[-1,1]$. Specifically,

$$
\rho_{\text{raw}} \in [0, 1] \\
\rho = 2 \times \rho_{\text{raw}} - 1
$$

Then we can use a beta prior on $\rho_\text{raw}$ to push our estimate of $\rho$ towards 0.5.

For example a `rho_raw ~ beta(10, 5)` prior would look like:

```{r, fig.width = 3, fig.height = 3}
suppressPackageStartupMessages(library(ggformula))
gf_dist("beta", shape1 = 10, shape2 = 5) + xlab("rho_raw") 
```

One further wrinkle is that we have a prior for $\texttt{mo}_t$ that depends on
$\texttt{mo}_{t-1}$. That is, we are working with the _conditional_ distribution
of $\texttt{mo}_t$ given $\texttt{mo}_{t-1}$.
But what should we do about the prior for $\texttt{mo}_1$, for which we don't
have a previous time period in the data?

We need to work out the _marginal_ distribution of the first observation.
Thankfully we can use the fact that AR models are stationary, so
$\text{Var}(\texttt{mo}_t) = \text{Var}(\texttt{mo}_{t-1})$ and
$\mathbb{E}(\texttt{mo}_t) = \mathbb{E}(\texttt{mo}_{t-1})$ for all $t$.
Therefore the marginal distribution of $\texttt{mo}_1$ is the same as the
marginal distribution of any $\texttt{mo}_t$.

First we derive the marginal variance of $\texttt{mo}_{t}$.

$$
\text{Var}(\texttt{mo}_t) = \text{Var}(\rho \texttt{mo}_{t-1} + \epsilon_t)  \\
\text{Var}(\texttt{mo}_t) = \text{Var}(\rho \texttt{mo}_{t-1}) + \text{Var}(\epsilon_t)
$$
where the second line holds by independence of $\epsilon_t$ and $\epsilon_{t-1})$.
Then, using the fact that $Var(cX) = c^2Var(X)$ for a constant $c$ and the fact
that, by stationarity, $\textrm{Var}(\texttt{mo}_{t-1}) = \textrm{Var}(\texttt{mo}_{t})$,
we then obtain:

$$
\text{Var}(\texttt{mo}_t)= \rho^2 \text{Var}( \texttt{mo}_{t})  + \sigma_\texttt{mo}^2 \\
\text{Var}(\texttt{mo}_t) = \frac{\sigma_\texttt{mo}^2}{1 - \rho^2}
$$

For the mean of $\texttt{mo}_t$ things are a bit simpler:

$$
\mathbb{E}(\texttt{mo}_t) = \mathbb{E}(\rho \, \texttt{mo}_{t-1} + \epsilon_t) \\
\mathbb{E}(\texttt{mo}_t) = \mathbb{E}(\rho \, \texttt{mo}_{t-1}) + \mathbb{E}(\epsilon_t) \\
$$

Since $\mathbb{E}(\epsilon_t) = 0$ by assumption we have

$$
\mathbb{E}(\texttt{mo}_t) = \mathbb{E}(\rho \, \texttt{mo}_{t-1})  + 0\\
\mathbb{E}(\texttt{mo}_t) = \rho \, \mathbb{E}(\texttt{mo}_{t}) \\
\mathbb{E}(\texttt{mo}_t) - \rho \mathbb{E}(\texttt{mo}_t) = 0  \\
\mathbb{E}(\texttt{mo}_t) = 0/(1 - \rho)
$$

which for $\rho \neq 1$ yields $\mathbb{E}(\texttt{mo}_{t}) = 0$.

We now have the marginal distribution for $\texttt{mo}_{t}$, which, in our case,
we will use for $\texttt{mo}_1$. The full AR(1) specification is then:

$$
\texttt{mo}_1 \sim \text{Normal}\left(0, \frac{\sigma_\texttt{mo}}{\sqrt{1 - \rho^2}}\right) \\
\texttt{mo}_t \sim \text{Normal}\left(\rho \, \texttt{mo}_{t-1}, \sigma_\texttt{mo}\right) \forall t > 1
$$

```{r comp-NB-hier-mos, cache=TRUE, results="hide", message=FALSE}
comp_hier_NB_mos <- stan_model('stan_programs/hier_NB_regression_ncp_slopes_mod_mos.stan')
```

```{r run-NB-hier-slopes-mos}
fit_hier_NB_mos <-
  sampling(
    comp_hier_NB_mos,
    data = standata_hier_long,
    chains = 4,
    cores = 4,
    control = list(adapt_delta = 0.95)
  )
```

In the interest of brevity, we won't go on expanding the model, though we
certainly could. What other information would help us understand the data
generating process better? What other aspects of the data generating process
might we want to capture that we're not capturing now?

As usual, we run through our posterior predictive checks.

```{r ppc-full-hier-mos}
y_rep <- as.matrix(fit_hier_NB_mos, pars = "y_rep")
ppc_dens_overlay(
  y = standata_hier_long$complaints,
  yrep = y_rep[1:200,]
)
```

```{r}
select_vec <- which(standata_hier_long$mo_idx %in% 1:12)
ppc_stat_grouped(
  y = standata_hier_long$complaints[select_vec],
  yrep = y_rep[,select_vec],
  group = standata_hier_long$mo_idx[select_vec],
  stat = 'mean'
)
```

As we can see, our monthly random intercept has captured a monthly pattern
across all the buildings. We can also compare the prior and posterior for the
autoregressive parameter to see how much we've learned. Here are two different
ways of comparing the prior and posterior visually:

```{r}
# 1) compare draws from prior and draws from posterior
rho_draws <- cbind(
  2 * rbeta(4000, 10, 5) - 1, # draw from prior
  as.matrix(fit_hier_NB_mos, pars = "rho")
)
colnames(rho_draws) <- c("prior", "posterior")
mcmc_hist(rho_draws, freq = FALSE, binwidth = 0.025,
          facet_args = list(nrow = 2)) + xlim(-1, 1)
```

```{r}
print(fit_hier_NB_mos, pars = c('rho','sigma_mu','sigma_kappa','gamma'))
```

```{r}
ppc_intervals(
  y = standata_hier_long$complaints,
  yrep = y_rep,
  x = standata_hier_long$traps + rnorm(standata_hier_long$N, 0, 0.2) # jitter
) +
  labs(x = "Number of traps", 
       y = "Number of complaints")
```

It looks as if our model finally generates a reasonable posterior predictive
distribution for all numbers of traps, and appropriately captures the tails of the
data generating process.


## Using our model: loss forecasts for decision making

Our model seems to be fitting well, so now we will go ahead and use the model to
help us make a decision about how many traps to put in our buildings. We'll make
a forecast for 6 months forward.

```{r comp-rev, cache=TRUE, results="hide", message=FALSE}
comp_rev <- stan_model('stan_programs/hier_NB_regression_ncp_slopes_mod_mos_predict.stan')
```

An important input to the revenue model is how much revenue is lost due to each
complaint. The client has a policy that for every 10 complaints, they'll call an
exterminator costing the client \$100, so that'll amount to \$10 per complaint.

```{r run-NB-hier-rev}
rev_model <- sampling(
  comp_rev,
  data = standata_hier_long,
  cores = 4,
  control = list(adapt_delta = 0.9)
)
```

Below we've generated our revenue curves for the buildings. These charts will
give us precise quantification of our uncertainty around our revenue projections
at any number of traps for each building.

A key input to our analysis will be the cost of installing bait stations. We're
simulating the number of complaints we receive over the course of a year, so we
need to understand the cost associated with maintaining each bait station over
the course of a year. There's the cost attributed to the raw bait station, which
is the plastic housing and the bait material, a peanut-buttery substance that's
injected with insecticide. The cost of maintaining one bait station for a year
plus monthly replenishment of the bait material is about \$20.

```{r materials-cost-formulation}
N_traps <- 20
costs <- 20 * (0:N_traps)
```

We'll also need labor for maintaining the bait stations, which need to be
serviced every two months. If there are fewer than five traps, our in-house
maintenance staff can manage the stations (about one hour of work every two
months at \$20/hour), but above five traps we need to hire outside pest control
to help out. They're a bit more expensive, so we've put their cost at \$30 /
hour. Each five traps should require an extra person-hour of work, so that's
factored in as well. The marginal person-person hours above five traps are at
the higher pest-control-labor rate.

```{r labor-cost-formulation}
N_months_forward <- 12
N_months_labor <- N_months_forward / 2
hourly_rate_low <- 20
hourly_rate_high <- 30
costs <- costs +
  (0:N_traps < 5 & 0:N_traps > 0) * (N_months_labor * hourly_rate_low) +
  (0:N_traps >= 5 & 0:N_traps < 10) * (N_months_labor * (hourly_rate_low + 1 * hourly_rate_high)) +
  (0:N_traps >= 10 & 0:N_traps < 15) * (N_months_labor * (hourly_rate_low + 2 * hourly_rate_high)) +
  (0:N_traps >= 15) * (N_months_labor * (hourly_rate_low + 3 * hourly_rate_high))
```

We can now plot curves with number of traps on the x-axis and profit/loss
forecasts and uncertainty intervals on the y-axis. When we do this we see that
the optimal number of bait stations differs by building.

```{r rev-curves, fig.width = 8, fig.height = 8}
# extract as a list instead of matrix for convenience below
samps_rev <- rstan::extract(rev_model)

# get draws of total profits (in this case negative, so losses)
total_profits <- sweep(samps_rev$rev_pred, 3, STATS = costs, FUN = '-')

# calculate medians and central intervals (for example 80% intervals)
median_profit <- t(apply(total_profits, c(2, 3), median))
lower_profit <- t(apply(total_profits, c(2, 3), quantile, 0.1))
upper_profit <- t(apply(total_profits, c(2, 3), quantile, 0.9))

buildings <- seq(1, ncol(median_profit))
traps <- seq(0, nrow(median_profit) - 1)

profit_df <-
  data.frame(
    profit = as.vector(median_profit),
    lower = as.vector(lower_profit),
    upper = as.vector(upper_profit),
    traps = rep(traps, times = max(buildings)),
    building = rep(buildings, each = max(traps) + 1)
  )

p <- ggplot(data = profit_df, aes(x = traps, y = profit)) +
  geom_ribbon(aes(ymin = lower, ymax = upper), fill = "skyblue") +
  geom_line()

# p + facet_wrap(~ building, scales = 'free', ncol = 2)
p + facet_wrap(~ building, scales = 'fixed', ncol = 3)
```


<br>

Left as an exercise for the reader (unless we have time):

* How would we build a revenue curve for a new building?

* Let's say our utility function is revenue. If we wanted to maximize expected
revenue, we can take expectations at each station count for each building, and
choose the trap numbers that maximizes expected revenue. This will be called a
maximum revenue strategy. How can we generate the distribution of portfolio
revenue (i.e. the sum of revenue across all the buildings) under the maximum
revenue strategy from the the draws of `rev_pred` we already have?



## Gaussian process instead of AR(1)

### Joint density for AR(1) process

We can derive the joint distribution for the AR(1) process before we move to the
Gaussian process (GP) which will give us a little more insight into what a GP
is. Remember that we've specified the AR(1) prior as:

$$
\begin{aligned}
\texttt{mo}_1 & \sim \text{Normal}\left(0, \frac{\sigma_\texttt{mo}}{\sqrt{1 - \rho^2}}\right) \\
\texttt{mo}_t & \sim \text{Normal}\left(\rho \, \texttt{mo}_{t-1}, \sigma_\texttt{mo}\right) \forall t > 1
\end{aligned}
$$
Rewriting our process in terms of the errors will make the derivation
of the joint distribution clearer

$$
\begin{aligned}
\texttt{mo}_1 & \sim \text{Normal}\left(0, \frac{\sigma_\texttt{mo}}{\sqrt{1 - \rho^2}}\right) \\
\texttt{mo}_t & = \rho \, \texttt{mo}_{t-1} + \sigma_\texttt{mo}\epsilon_t  \\
\epsilon_t & \sim \text{Normal}\left(0, 1\right)
\end{aligned}
$$

Given that our first term $\texttt{mo}_1$ is normally distributed, and
subsequent terms are sums of normal random variables, this means that jointly
the vector `mo` is multivariate normal. 

More formally, if we have a vector $x \in \mathbb{R}^M$ which is
multivariate normal, $x \sim \text{MultiNormal}(0, \Sigma)$, and we left-multiply $x$
by a nonsingular matrix $L \in \mathbb{R}^{M\times M}$, then
$y = Lx \sim \text{MultiNormal}(0, L\Sigma L^T)$.
We can use this fact to show that our vector `mo` is jointly multivariate
normal.

Just as before with the noncentered parameterization, we'll take a vector
$\texttt{mo_raw} \in \mathbb{R}^M$ (in which each element is univariate
$\text{Normal}(0,1)$) and transform it into `mo`. But instead of doing this with
scalar transformations like we did previously 
(e.g., in the section __Time varying effects and structured priors__) 
we will now do it with linear algebra operations.

The trick is that by specifying each element of $\texttt{mo_raw}$ to be distributed $\text{Normal}(0,1)$ we are implicitly defining 
$\texttt{mo_raw} \sim \text{MultiNormal}(0, I_M)$, where $I_M$ is the 
$M \times M$ identity matrix. Then we do a linear transformation using a matrix 
$L$ and assign the result to `mo`, i.e., $\texttt{mo} = L\times\texttt{mo_raw}$.
This gives us $\texttt{mo} \sim \text{MultiNormal}(0, LI_M L^T)$ and 
$LI_M L^T = LL^T$.

Consider the case where we have three elements in `mo` and we want to figure 
out the form for $L$.

The first element of `mo` is fairly straightforward, because it mirrors our
earlier parameterization of the AR(1) prior. The only difference is that we're
explicitly adding the last two terms of `mo_raw` into the equation so we can use
matrix algebra for our transformation.
$$
\texttt{mo}_1 = \frac{\sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + 0 \times \texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3\\
$$
The second element is a bit more complicated:

$$
\begin{aligned}
\texttt{mo}_2 & = \rho \texttt{mo}_1 + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3  \\
 & = \rho \left(\frac{\sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1\right) + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3  \\[5pt]
 & = \frac{\rho \sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3  \\
\end{aligned}
$$

While the third element will involve all three terms
$$
\begin{aligned}
\texttt{mo}_3 & = \rho \, \texttt{mo}_2 + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_3 \\
 & = \rho \left(\frac{\rho \sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2\right) + \sigma_{\texttt{mo}} \texttt{mo_raw}_3  \\[5pt]
& = \frac{\rho^2 \sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + \rho \, \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 +  \sigma_{\texttt{mo}}\,\texttt{mo_raw}_3  \\
\end{aligned}
$$

Writing this all together:

$$
\begin{aligned}
\texttt{mo}_1 & = \frac{\sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + 0 \times \texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3\\[3pt]
\texttt{mo}_2 & = \frac{\rho \sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 + 0 \times \texttt{mo_raw}_3  \\[3pt]
\texttt{mo}_3 & = \frac{\rho^2 \sigma_{\texttt{mo}}}{\sqrt{1 - \rho^2}} \times \texttt{mo_raw}_1 + \rho \, \sigma_{\texttt{mo}}\,\texttt{mo_raw}_2 +  \sigma_{\texttt{mo}}\,\texttt{mo_raw}_3  \\
\end{aligned}
$$
Separating this into a matrix of coefficients $L$ and the vector `mo_raw`:

$$
\texttt{mo} = \begin{bmatrix} \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & 0 & 0 \\
              \rho \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \sigma_\texttt{mo} & 0 \\
              \rho^2 \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \rho \,\sigma_\texttt{mo}  & \sigma_\texttt{mo}
              \end{bmatrix} \times \texttt{mo_raw}
$$

If we multiply $L$ on the right by its transpose $L^T$, we'll get expressions for the covariance matrix of our multivariate random vector `mo`:

$$
\begin{bmatrix} \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & 0 & 0 \\
\rho \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \sigma_\texttt{mo} & 0 \\
\rho^2 \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \rho \,\sigma_\texttt{mo}  & \sigma_\texttt{mo}
\end{bmatrix} \times
\begin{bmatrix} \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \rho \sigma_\texttt{mo} / \sqrt{1 - \rho^2} & \rho^2 \sigma_\texttt{mo} / \sqrt{1 - \rho^2} \\
0 & \sigma_\texttt{mo} & \rho \,\sigma_\texttt{mo} \\
0 & 0  & \sigma_\texttt{mo}
\end{bmatrix}
$$

which results in:

$$
\begin{bmatrix} \sigma^2_\texttt{mo} / (1 - \rho^2) & \rho \, \sigma^2_\texttt{mo} / (1 - \rho^2) &  \rho^2 \, \sigma^2_\texttt{mo} / (1 - \rho^2)\\
\rho \, \sigma^2_\texttt{mo} / (1 - \rho^2)  & \sigma^2_\texttt{mo} / (1 - \rho^2)  & \rho \, \sigma^2_\texttt{mo} / (1 - \rho^2) \\
\rho^2 \sigma^2_\texttt{mo} / (1 - \rho^2) & \rho \, \sigma^2_\texttt{mo} / (1 - \rho^2) & \sigma^2_\texttt{mo} / (1 - \rho^2)
\end{bmatrix}
$$
We can simplify this result by dividing the matrix by 
$\sigma^2_\texttt{mo} / (1 - \rho^2)$ to get

$$
\begin{bmatrix} 1 & \rho  &  \rho^2 \\
\rho  & 1  & \rho  \\
\rho^2 & \rho & 1
\end{bmatrix}
$$

This should generalize to higher dimensions pretty easily. 

We can replace the Stan code defining `mo` in 
`stan_programs/hier_NB_regression_ncp_slopes_mod_mos.stan` with the following:

```
vector[M] mo;
{
  matrix[M,M] A = rep_matrix(0, M, M);
  A[1,1] = sigma_mo / sqrt(1 - rho^2);
  for (m in 2:M) {
    A[m,1] = rho^(m-1) * sigma_mo / sqrt(1 - rho^2);
  }
  for (m in 2:M) {
    A[m,m] = sigma_mo;
    for (i in (m + 1):M) {
      A[i,m] = rho^(i-m) * sigma_mo;
    }
  }
  mo = A * mo_raw;
}
```

### Cholesky decomposition

If we only knew the covariance matrix of our process, say a matrix called
$\Sigma$, and we had a way of decomposing $\Sigma$ into $L L^T$ we wouldn't need
to write out the equation for the vector. Luckily, there is a matrix
decomposition called the __Cholesky decomposition__ that does just that. The
Stan function for the composition is called `cholesky_decompose`. Instead of
writing out the explicit equation we can just do the following

```
vector[M] mo;
{
  mo = cholesky_decompose(Sigma) * mo_raw;
}
```
provided we've defined `Sigma` appropriately elsewhere in the `transformed
parameters` block. Note that the matrix $L$ is lower-triangular (i.e. all
elements in the upper right triangle of the matrix are zero).

We've already derived the covariance matrix `Sigma` for the three-dimensional
AR(1) process above by explicitly calculating $L L^T$, but we can do so using
the rules of covariance and the way our process is defined. We already know that
each element of $\texttt{mo}_t$ has marginal variance 
$\sigma^2_\texttt{mo} / (1- \rho^2)$, but we don't know the covariance of 
$\texttt{mo}_t$ and $\texttt{mo}_{t+h}$. 
We can do so recursively. First we derive the covariance
for two elements of $\texttt{mo}_t$ separated by one month:

$$
\text{Cov}(\texttt{mo}_{t+1},\texttt{mo}_{t}) = \text{Cov}(\rho \, \texttt{mo}_{t} + \sigma_\texttt{mo}\epsilon_{t+1},\texttt{mo}_{t}) \\
\text{Cov}(\texttt{mo}_{t+1},\texttt{mo}_{t}) = \rho \text{Cov}(\texttt{mo}_{t},\texttt{mo}_{t}) + \sigma_\texttt{mo}\text{Cov}(\epsilon_{t+1},\texttt{mo}_{t}) \\
\text{Cov}(\texttt{mo}_{t+1},\texttt{mo}_{t}) = \rho \text{Var}(\texttt{mo}_t) + 0
$$

Then we define the covariance for $\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t})$ in terms of $\text{Cov}(\texttt{mo}_{t+h-1},\texttt{mo}_{t})$

$$
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \text{Cov}(\rho \, \texttt{mo}_{t+h-1} + \sigma_\texttt{mo}\epsilon_{t+h},\texttt{mo}_{t}) \\
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \rho \, \text{Cov}(\texttt{mo}_{t+h-1},\texttt{mo}_{t}) \\
$$
Which we can use to recursively get at the covariance we need:

$$
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \rho \, \text{Cov}(\texttt{mo}_{t+h-1},\texttt{mo}_{t}) \\
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \rho \,( \rho \, \text{Cov}(\texttt{mo}_{t+h-2},\texttt{mo}_{t}) )\\
\dots \\
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \rho^h \, \text{Cov}(\texttt{mo}_{t},\texttt{mo}_{t}) \\
\text{Cov}(\texttt{mo}_{t+h},\texttt{mo}_{t}) = \rho^h \, \sigma_\texttt{mo}^2/(1 - \rho^2) \\
$$

One way to write this in Stan code is:

```
vector[M] mo;
{
  matrix[M,M] Sigma;
  for (m in 1:M) {
    Sigma[m,m] = 1.0;
    for (i in (m + 1):M) {
      Sigma[i,m] = rho^(i - m);
      Sigma[m,i] = Sigma[i,m];
    }
  }
  Sigma = Sigma * sigma_mo^2 / (1 - rho^2);
  mo = cholesky_decompose(Sigma) * mo_raw;
}
```

### Extension to Gaussian processes

The prior we defined for `mo` is strictly speaking a Gaussian process. It is a
stochastic process that is distributed as jointly multivariate normal for any
finite value of $M$. Formally, we could write the above prior for `mo` like so:

$$
\begin{aligned}
  \sigma_\texttt{mo} & \sim \text{Normal}(0, 1) \\
  \rho & \sim \text{GenBeta}(-1,1,10, 5) \\
  \texttt{mo}_t & \sim \text{GP}\left( 0,
  K(t | \sigma_\texttt{mo},\rho) \right) \\
\end{aligned}
$$
The notation $K(t | \sigma_\texttt{mo},\rho)$ defines the covariance matrix of the process over the domain $t$,
which is months.

In other words:

$$
\text{Cov}(\texttt{mo}_t,\texttt{mo}_{t+h}) = k(t, t+h | \sigma_\texttt{mo}, \rho)
$$

We've already derived the covariance for our process.  What if we want to use a different definition of `Sigma`?

As the above example shows defining a proper covariance matrix will yield a
proper multivariate normal prior on a parameter. We need a way of defining a
proper covariance matrix. These are symmetric positive definite matrices. It
turns out there is a class of functions that define proper covariance matrices,
called __kernel functions__. These functions are applied elementwise to construct a
covariance matrix, $K$:

$$
K_{[t,t+h]} = k(t, t+h | \theta)
$$
where $\theta$ are the hyperparameters that define the behavior of covariance matrix.

One such function is called the __exponentiated quadratic function__, and it happens
to be implemented in Stan as `cov_exp_quad`. The function is defined as:

$$
\begin{aligned}
  k(t, t+h | \theta) & = \alpha^2  \exp \left( - \dfrac{1}{2\ell^2} ((t+h) - t)^2 \right) \\
  & = \alpha^2  \exp \left( - \dfrac{h^2}{2\ell^2} \right)
\end{aligned}
$$

The exponentiated quadratic kernel has two components to theta, $\alpha$, the
marginal standard deviation of the stochastic process $f$, and $\ell$, the
process length-scale.

The length-scale defines how quickly the covariance decays between time points, with
large values of $\ell$ yielding a covariance that decays slowly, and with small values
of $\ell$ yielding a covariance that decays rapidly. It can be seen interpreted as a measure of how
nonlinear the `mo` process is in time.

The marginal standard deviation defines how large the fluctuations are on the output side, which in
our case is the number of roach complaints per month across all buildings. It can be seen as a scale
parameter akin to the scale parameter for our building-level hierarchical intercept, though it now
defines the scale of the monthly deviations.

This kernel's defining quality is its smoothness; the function is infinitely
differentiable. That will present problems for our example, but if we add some
noise the diagonal of our covariance matrix, the model will fit well.

$$
k(t, t+h | \theta) = \alpha^2  \exp \left( - \dfrac{h^2}{2\ell^2} \right) + \text{if } h = 0, \, \sigma^2_\texttt{noise} \text{ else } 0
$$

### Compiling the GP model

```{r comp-NB-hier-gp, cache=TRUE, results="hide", message=FALSE}
comp_hier_NB_gp <- stan_model('stan_programs/hier_NB_regression_ncp_slopes_mod_mos_gp.stan')
```

### Fitting the GP model to data

```{r run-NB-hier-gp}
fit_hier_NB_gp <- sampling(comp_hier_NB_gp, data = standata_hier_long, chains = 4, cores = 4, control = list(adapt_delta = 0.9))
samps_gp <- rstan::extract(fit_hier_NB_gp)
```

### Examining the fit

Let's look at the prior vs. posterior for the GP length scale parameter:

```{r}
length_scale_draws <- cbind(
  prior = rgamma(4000, 10, 2),
  posterior = samps_gp$gp_len
)
mcmc_areas(length_scale_draws)
```

From the plot above it only looks like we learned a small amount, however we can
see a bigger difference between the prior and posterior if we consider how much
we learned about the ratio of `sigma_gp` to the length scale `gp_len`:

```{r}
noise_to_length_scale_ratio_draws <- cbind(
  prior = abs(rnorm(4000)) / rgamma(4000, 10, 2),
  posterior = samps_gp$sigma_gp / samps_gp$gp_len
)
mcmc_areas(noise_to_length_scale_ratio_draws)
```

This is a classic problem with Gaussian processes. Marginally, the length-scale
parameter isn't very well-identified by the data, but jointly the length-scale and
the marginal standard deviation are well-identified.

And let's compare the estimates for the time varying parameters between the
AR(1) and GP. In this case the posterior mean of the time trend is essentially
the same for the AR(1) and GP priors but the 50% uncertainty intervals are
narrower for the AR(1):

```{r, message=FALSE, warning=FALSE}
# visualizing 50% intervals
mo_ar_intervals <- mcmc_intervals_data(as.matrix(fit_hier_NB_mos, pars = "mo"), prob = 0.5)
mo_gp_intervals <- mcmc_intervals_data(as.matrix(fit_hier_NB_gp, pars = "gp"), prob = 0.5)
plot_data <- bind_rows(mo_ar_intervals, mo_gp_intervals)
plot_data$prior <- factor(rep(c("AR1", "GP"), each = 36), levels = c("GP", "AR1"))
plot_data$time <- rep(1:36, times = 2)

ggplot(plot_data, aes(x = time, y = m, ymin = l, ymax = h, fill = prior)) +
  geom_ribbon(alpha = 2/3)
```

The way we coded the GP also lets us plot a decomposition of the GP into a
monthly noise component (`mo_noise` in the Stan code) and the underlying
smoothly varying trend (`gp_exp_quad` in the Stan code):

```{r, message=FALSE, warning=FALSE}
# visualizing 50% intervals
mo_noise_intervals <- mcmc_intervals_data(as.matrix(fit_hier_NB_gp, pars = "mo_noise"), prob = 0.5)
gp_exp_quad_intervals <- mcmc_intervals_data(as.matrix(fit_hier_NB_gp, pars = "gp_exp_quad"), prob = 0.5)

plot_data <- bind_rows(mo_noise_intervals, gp_exp_quad_intervals)
plot_data$time <- rep(1:36, times = 2)
plot_data$term <- factor(rep(c("Monthly Noise", "Smooth Trend"), each = 36),
                         levels = c("Smooth Trend", "Monthly Noise"))

ggplot(plot_data, aes(x = time, y = m, ymin = l, ymax = h, fill = term)) +
  geom_ribbon(alpha = 0.5) +
  geom_line(aes(color = term), size = 0.5) +
  ylab(NULL)
```