# Writing Stan programs for use with the loo package

#### Aki Vehtari and Jonah Gabry

#### 2019-12-18

# Introduction

This vignette demonstrates how to write a Stan program that computes and stores the pointwise log-likelihood required for using the loo package. The other vignettes included with the package demonstrate additional functionality.

Some sections from this vignette are excerpted from our papers

- Vehtari, A., Gelman, A., and Gabry, J. (2017). Practical Bayesian model evaluation using leave-one-out cross-validation and WAIC. Statistics and Computing. 27(5), 1413–1432. :10.1007/s11222-016-9696-4. Links: published | arXiv preprint.

- Vehtari, A., Gelman, A., and Gabry, J. (2017). Pareto smoothed importance sampling. arXiv preprint arXiv:1507.04544.

which provide important background for understanding the methods implemented in the package.

# Example: Well water in Bangladesh

This example comes from a survey of residents from a small area in Bangladesh that was affected by arsenic in drinking water. Respondents with elevated arsenic levels in their wells were asked if they were interested in getting water from a neighbor’s well, and a series of logistic regressions were fit to predict this binary response given various information about the households (Gelman and Hill, 2007). Here we fit a model for the well-switching response given two predictors: the arsenic level of the water in the resident’s home, and the distance of the house from the nearest safe well.

The sample size in this example is \(N=3020\), which is not huge but is large enough that it is important to have a computational method for LOO that is fast for each data point. On the plus side, with such a large dataset, the influence of any given observation is small, and so the computations should be stable.

## Coding the Stan model

Here is the Stan code for fitting the logistic regression model, which we save in a file called `logistic.stan`:

    data {
      int<lower=0> N;             // number of data points
      int<lower=0> P;             // number of predictors (including intercept)
      matrix[N,P] X;              // predictors (including 1s for intercept)
      int<lower=0,upper=1> y[N];  // binary outcome
    }
    parameters {
      vector[P] beta;
    }
    model {
      beta ~ normal(0, 1);
      y ~ bernoulli_logit(X * beta);
    }
    generated quantities {
      vector[N] log_lik;
      for (n in 1:N) {
        log_lik[n] = bernoulli_logit_lpmf(y[n] | X[n] * beta);
      }
    }

We have defined the log likelihood as a vector named `log_lik` in the generated quantities block so that the individual terms will be saved by Stan. After running Stan, `log_lik` can be extracted (using the `extract_log_lik` function provided in the loo package) as an \(S \times N\) matrix, where \(S\) is the number of simulations (posterior draws) and \(N\) is the number of data points.

## Fitting the model with RStan

Next we fit the model in Stan using the rstan package:

    library("rstan")

    # Prepare data
    url <- "http://stat.columbia.edu/~gelman/arm/examples/arsenic/wells.dat"
    wells <- read.table(url)
    wells$dist100 <- with(wells, dist / 100)
    X <- model.matrix(~ dist100 + arsenic, wells)
    standata <- list(y = wells$switch, X = X, N = nrow(X), P = ncol(X))

    # Fit model
    fit_1 <- stan("logistic.stan", data = standata)
    print(fit_1, pars = "beta")

             mean se_mean   sd  2.5%   25%   50%   75% 97.5% n_eff Rhat
    beta[1]  0.00       0 0.08 -0.16 -0.05  0.00  0.05  0.15  1964    1
    beta[2] -0.89       0 0.10 -1.09 -0.96 -0.89 -0.82 -0.68  2048    1
    beta[3]  0.46       0 0.04  0.38  0.43  0.46  0.49  0.54  2198    1

## Computing approximate leave-one-out cross-validation using PSIS-LOO

We can then use the loo package to compute the efficient PSIS-LOO approximation to exact LOO-CV:

    library("loo")

    # Extract pointwise log-likelihood
    # using merge_chains=FALSE returns an array, which is easier to
    # use with relative_eff()
    log_lik_1 <- extract_log_lik(fit_1, merge_chains = FALSE)

    # as of loo v2.0.0 we can optionally provide relative effective sample sizes
    # when calling loo, which allows for better estimates of the PSIS effective
    # sample sizes and Monte Carlo error
    r_eff <- relative_eff(exp(log_lik_1), cores = 2)

    # preferably use more than 2 cores (as many cores as possible)
    # will use value of 'mc.cores' option if cores is not specified
    loo_1 <- loo(log_lik_1, r_eff = r_eff, cores = 2)
    print(loo_1)

    Computed from 4000 by 3020 log-likelihood matrix

             Estimate   SE
    elpd_loo  -1968.5 15.6
    p_loo         3.2  0.1
    looic      3937.0 31.2
    ------
    Monte Carlo SE of elpd_loo is 0.0.

    All Pareto k estimates are good (k < 0.5).
    See help('pareto-k-diagnostic') for details.

The printed output from the `loo` function shows the estimates \(\widehat{\mbox{elpd}}_{\rm loo}\) (expected log predictive density), \(\widehat{p}_{\rm loo}\) (effective number of parameters), and \({\rm looic} =-2\, \widehat{\mbox{elpd}}_{\rm loo}\) (the LOO information criterion).

The line at the bottom of the printed output provides information about the reliability of the LOO approximation (the interpretation of the \(k\) parameter is explained in `help('pareto-k-diagnostic')` and in greater detail in Vehtari, Gelman, and Gabry (2017)). In this case the message tells us that all of the estimates for \(k\) are fine.

## Comparing models

To compare this model to an alternative model for the same data we can use the `loo_compare` function in the loo package. First we’ll fit a second model to the well-switching data, using `log(arsenic)` instead of `arsenic` as a predictor:

    standata$X[, "arsenic"] <- log(standata$X[, "arsenic"])
    fit_2 <- stan(fit = fit_1, data = standata)

    log_lik_2 <- extract_log_lik(fit_2, merge_chains = FALSE)
    r_eff_2 <- relative_eff(exp(log_lik_2))
    loo_2 <- loo(log_lik_2, r_eff = r_eff_2, cores = 2)
    print(loo_2)

    Computed from 4000 by 3020 log-likelihood matrix

             Estimate   SE
    elpd_loo  -1952.3 16.2
    p_loo         3.1  0.1
    looic      3904.6 32.4
    ------
    Monte Carlo SE of elpd_loo is 0.0.

    All Pareto k estimates are good (k < 0.5).
    See help('pareto-k-diagnostic') for details.

We can now compare the models on LOO using the `loo_compare` function:

    # Compare
    comp <- loo_compare(loo_1, loo_2)

This new object, `comp`, contains the estimated difference of expected leave-one-out prediction errors between the two models, along with the standard error:

    print(comp) # can set simplify=FALSE for more detailed print output

           elpd_diff se_diff
    model2   0.0       0.0
    model1 -16.3       4.4

The first column shows the difference in ELPD relative to the model with the largest ELPD. In this case, the difference in `elpd` and its scale relative to the approximate standard error of the difference) indicates a preference for the second model (`model2`).

# References

Gelman, A., and Hill, J. (2007). Data Analysis Using Regression and Multilevel Hierarchical Models. Cambridge University Press.

Stan Development Team (2017). The Stan C++ Library, Version 2.17.0. http://mc-stan.org

Stan Development Team (2018) RStan: the R interface to Stan, Version 2.17.3. http://mc-stan.org

Vehtari, A., Gelman, A., and Gabry, J. (2017). Practical Bayesian model evaluation using leave-one-out cross-validation and WAIC. Statistics and Computing. 27(5), 1413–1432. :10.1007/s11222-016-9696-4. online, arXiv preprint arXiv:1507.04544.

Vehtari, A., Gelman, A., and Gabry, J. (2017). Pareto smoothed importance sampling. arXiv preprint arXiv:1507.02646.