# Introduction

This vignette demonstrates how to do leave-one-out cross-validation for large data using the loo package and Stan. There are two approaches covered: LOO with subsampling and LOO using approximations to posterior distributions. Some sections from this vignette are excerpted from the papers

• Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2020). Leave-One-Out Cross-Validation for Model Comparison in Large Data. Proceedings of the 23rd International Conference on Artificial Intelligence and Statistics (AISTATS), in PMLR 108. arXiv preprint arXiv:2001.00980.

• Magnusson, M., Andersen, M., Jonasson, J. & Vehtari, A. (2019). Bayesian leave-one-out cross-validation for large data. Proceedings of the 36th International Conference on Machine Learning, in PMLR 97:4244-4253 online, arXiv preprint arXiv:1904.10679.

• 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., Simpson, D., Gelman, A., Yao, Y., and Gabry, J. (2019). Pareto smoothed importance sampling. arXiv preprint arXiv:1507.04544.

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

# Setup

In addition to the loo package, we’ll also be using rstan:

library("rstan")
library("loo")
set.seed(4711)

# Example: Well water in Bangladesh

We will use the same example as in the vignette Writing Stan programs for use with the loo package. See that vignette for a description of the problem and data.

The sample size in this example is only $$N=3020$$, which is not large enough to require the special methods for large data described in this vignette, but is sufficient for demonstration purposes in this tutorial.

## Coding the Stan model

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

// save in 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);
}

Importantly, unlike the general approach recommended in Writing Stan programs for use with the loo package, we do not compute the log-likelihood for each observation in the generated quantities block of the Stan program. Here we are assuming we have a large data set (larger than the one we’re actually using in this demonstration) and so it is preferable to instead define a function in R to compute the log-likelihood for each data point when needed rather than storing all of the log-likelihood values in memory.

The log-likelihood in R can be coded as follows:

# we'll add an argument log to toggle whether this is a log-likelihood or
# likelihood function. this will be useful later in the vignette.
llfun_logistic <- function(data_i, draws, log = TRUE) {
x_i <- as.matrix(data_i[, which(grepl(colnames(data_i), pattern = "X")), drop=FALSE])
logit_pred <- draws %*% t(x_i)
dbinom(x = data_i$y, size = 1, prob = 1/(1 + exp(-logit_pred)), log = log) } The function llfun_logistic() needs to have arguments data_i and draws. Below we will test that the function is working by using the loo_i() function. ## Fitting the model with RStan Next we fit the model in Stan using the rstan package: # Prepare data url <- "http://stat.columbia.edu/~gelman/arm/examples/arsenic/wells.dat" wells <- read.table(url) wells$dist100 <- with(wells, dist / 100)
X <- model.matrix(~ dist100 + arsenic, wells)
standata <- list(y = wells$switch, X = X, N = nrow(X), P = ncol(X)) # Compile stan_mod <- stan_model("logistic.stan") # Fit model fit_1 <- sampling(stan_mod, data = standata, seed = 4711) print(fit_1, pars = "beta")  mean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat beta[1] 0.00 0 0.08 -0.15 -0.05 0.00 0.06 0.16 1933 1 beta[2] -0.89 0 0.10 -1.09 -0.96 -0.89 -0.82 -0.69 2332 1 beta[3] 0.46 0 0.04 0.38 0.43 0.46 0.49 0.54 2051 1 Before we move on to computing LOO we can now test that the log-likelihood function we wrote is working as it should. The loo_i() function is a helper function that can be used to test a log-likelihood function on a single observation. # used for draws argument to loo_i parameter_draws_1 <- extract(fit_1)$beta

# used for data argument to loo_i
stan_df_1 <- as.data.frame(standata)

# compute relative efficiency (this is slow and optional but is recommended to allow
# for adjusting PSIS effective sample size based on MCMC effective sample size)
r_eff <- relative_eff(llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_1,
draws = parameter_draws_1,
cores = 2)

loo_i(i = 1, llfun_logistic, r_eff = r_eff, data = stan_df_1, draws = parameter_draws_1)
$pointwise elpd_loo mcse_elpd_loo p_loo looic influence_pareto_k 1 -0.3314552 0.0002887608 0.0003361772 0.6629103 -0.05679886 ... # Approximate LOO-CV using PSIS-LOO and subsampling We can then use the loo_subsample() function to compute the efficient PSIS-LOO approximation to exact LOO-CV using subsampling: set.seed(4711) loo_ss_1 <- loo_subsample( llfun_logistic, observations = 100, # take a subsample of size 100 cores = 2, # these next objects were computed above r_eff = r_eff, draws = parameter_draws_1, data = stan_df_1 ) print(loo_ss_1) Computed from 4000 by 100 subsampled log-likelihood values from 3020 total observations. Estimate SE subsampling SE elpd_loo -1968.5 15.6 0.3 p_loo 3.1 0.1 0.4 looic 3936.9 31.2 0.6 ------ 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 loo_subsample() function creates an object of class psis_loo_ss, that inherits from psis_loo, loo (the classes of regular loo objects). The printed output above 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). Unlike when using loo(), when using loo_subsample() there is an additional column giving the “subsampling SE”, which reflects the additional uncertainty due to the subsampling used. 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, Simpson, Gelman, Yao, and Gabry (2019)). In this case, the message tells us that all of the estimates for $$k$$ are fine for this given subsample. ## Adding additional subsamples If we are not satisfied with the subsample size (i.e., the accuracy) we can simply add more samples until we are satisfied using the update() method. set.seed(4711) loo_ss_1b <- update( loo_ss_1, observations = 200, # subsample 200 instead of 100 r_eff = r_eff, draws = parameter_draws_1, data = stan_df_1 ) print(loo_ss_1b) Computed from 4000 by 200 subsampled log-likelihood values from 3020 total observations. Estimate SE subsampling SE elpd_loo -1968.3 15.6 0.2 p_loo 3.2 0.1 0.4 looic 3936.7 31.2 0.5 ------ 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. ## Specifying estimator and sampling method The performance relies on two components: the estimation method and the approximation used for the elpd. See the documentation for loo_subsample() more information on which estimators and approximations are implemented. The default implementation is using the point log predictive density evaluated at the mean of the posterior (loo_approximation="plpd") and the difference estimator (estimator="diff_srs"). This combination has a focus on fast inference. But we can easily use other estimators as well as other elpd approximations, for example: set.seed(4711) loo_ss_1c <- loo_subsample( x = llfun_logistic, r_eff = r_eff, draws = parameter_draws_1, data = stan_df_1, observations = 100, estimator = "hh_pps", # use Hansen-Hurwitz loo_approximation = "lpd", # use lpd instead of plpd loo_approximation_draws = 100, cores = 2 ) print(loo_ss_1c) Computed from 4000 by 100 subsampled log-likelihood values from 3020 total observations. Estimate SE subsampling SE elpd_loo -1968.9 15.4 0.5 p_loo 3.5 0.2 0.5 looic 3937.9 30.7 1.1 ------ 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. See the documentation and references for loo_subsample() for details on the implemented approximations. # Approximate LOO-CV using PSIS-LOO with posterior approximations Using posterior approximations, such as variational inference and Laplace approximations, can further speed-up LOO-CV for large data. Here we demonstrate using a Laplace approximation in Stan. fit_laplace <- optimizing(stan_mod, data = standata, draws = 2000, importance_resampling = TRUE) parameter_draws_laplace <- fit_laplace$theta_tilde # draws from approximate posterior
log_p <- fit_laplace$log_p # log density of the posterior log_g <- fit_laplace$log_g # log density of the approximation

Using the posterior approximation we can then do LOO-CV by correcting for the posterior approximation when we compute the elpd. To do this we use the loo_approximate_posterior() function.

set.seed(4711)
loo_ap_1 <-
loo_approximate_posterior(
x = llfun_logistic,
draws = parameter_draws_laplace,
data = stan_df_1,
log_p = log_p,
log_g = log_g,
cores = 2
)
print(loo_ap_1)

The function creates a class, psis_loo_ap that inherits from psis_loo, loo.

Computed from 2000 by 3020 log-likelihood matrix

Estimate   SE
elpd_loo  -1968.4 15.6
p_loo         3.2  0.2
looic      3936.8 31.2
------
Posterior approximation correction used.
Monte Carlo SE of elpd_loo is 0.0.

Pareto k diagnostic values:
Count Pct.    Min. n_eff
(-Inf, 0.5]   (good)     2989  99.0%   1827
(0.5, 0.7]   (ok)         31   1.0%   1996
(0.7, 1]   (bad)         0   0.0%   <NA>
(1, Inf)   (very bad)    0   0.0%   <NA>

All Pareto k estimates are ok (k < 0.7).
See help('pareto-k-diagnostic') for details.

## Combining the posterior approximation method with subsampling

The posterior approximation correction can also be used together with subsampling:

set.seed(4711)
loo_ap_ss_1 <-
loo_subsample(
x = llfun_logistic,
draws = parameter_draws_laplace,
data = stan_df_1,
log_p = log_p,
log_g = log_g,
observations = 100,
cores = 2
)
print(loo_ap_ss_1)
Computed from 2000 by 100 subsampled log-likelihood
values from 3020 total observations.

Estimate   SE subsampling SE
elpd_loo  -1968.2 15.6            0.4
p_loo         2.9  0.1            0.5
looic      3936.4 31.1            0.8
------
Posterior approximation correction used.
Monte Carlo SE of elpd_loo is 0.0.

Pareto k diagnostic values:
Count Pct.    Min. n_eff
(-Inf, 0.5]   (good)     97    97.0%   1971
(0.5, 0.7]   (ok)        3     3.0%   1997
(0.7, 1]   (bad)       0     0.0%   <NA>
(1, Inf)   (very bad)  0     0.0%   <NA>

All Pareto k estimates are ok (k < 0.7).
See help('pareto-k-diagnostic') for details.

The object created is of class psis_loo_ss, which inherits from the psis_loo_ap class previously described.

## Comparing models

To compare this model to an alternative model for the same data we can use the loo_compare() function just as we would if using loo() instead of loo_subsample() or loo_approximate_posterior(). 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 <- sampling(stan_mod, data = standata)
parameter_draws_2 <- extract(fit_2)\$beta
stan_df_2 <- as.data.frame(standata)

# recompute subsampling loo for first model for demonstration purposes

# compute relative efficiency (this is slow and optional but is recommended to allow
# for adjusting PSIS effective sample size based on MCMC effective sample size)
r_eff_1 <- relative_eff(
llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_1,
draws = parameter_draws_1,
cores = 2
)

set.seed(4711)
loo_ss_1 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_1,
draws = parameter_draws_1,
data = stan_df_1,
observations = 200,
cores = 2
)

# compute subsampling loo for a second model (with log-arsenic)

r_eff_2 <- relative_eff(
llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_2,
draws = parameter_draws_2,
cores = 2
)
loo_ss_2 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = 200,
cores = 2
)

print(loo_ss_2)
Computed from 4000 by 100 subsampled log-likelihood
values from 3020 total observations.

Estimate   SE subsampling SE
elpd_loo  -1952.0 16.2            0.2
p_loo         2.6  0.1            0.3
looic      3903.9 32.4            0.4
------
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_ss_1, loo_ss_2)
print(comp)
Warning: Different subsamples in 'model2' and 'model1'. Naive diff SE is used.

elpd_diff se_diff subsampling_se_diff
model2  0.0       0.0     0.0
model1 16.5      22.5     0.4               

This new object comp contains the estimated difference of expected leave-one-out prediction errors between the two models, along with the standard error. As the warning indicates, because different subsamples were used the comparison will not take the correlations between different observations into account. Here we see that the naive SE is 22.5 and we cannot see any difference in performance between the models.

To force subsampling to use the same observations for each of the models we can simply extract the observations used in loo_ss_1 and use them in loo_ss_2 by supplying the loo_ss_1 object to the observations argument.

loo_ss_2 <-
loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = loo_ss_1,
cores = 2
)

We could also supply the subsampling indices using the obs_idx() helper function:

idx <- obs_idx(loo_ss_1)
loo_ss_2 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = idx,
cores = 2
)
Simple random sampling with replacement assumed.

This results in a message indicating that we assume these observations to have been sampled with simple random sampling, which is true because we had used the default "diff_srs" estimator for loo_ss_1.

We can now compare the models and estimate the difference based on the same subsampled observations.

comp <- loo_compare(loo_ss_1, loo_ss_2)
print(comp)
       elpd_diff se_diff subsampling_se_diff
model2  0.0       0.0     0.0
model1 16.1       4.4     0.1               

First, notice that now the se_diff is now around 4 (as opposed to 20 when using different subsamples). 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). Since the subsampling uncertainty is so small in this case it can effectively be ignored. If we need larger subsamples we can simply add samples using the update() method demonstrated earlier.

It is also possible to compare a subsampled loo computation with a full loo object.

# use loo() instead of loo_subsample() to compute full PSIS-LOO for model 2
loo_full_2 <- loo(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
cores = 2
)
loo_compare(loo_ss_1, loo_full_2)
Estimated elpd_diff using observations included in loo calculations for all models.

Because we are comparing a non-subsampled loo calculation to a subsampled calculation we get the message that only the observations that are included in the loo calculations for both model1 and model2 are included in the computations for the comparison.

       elpd_diff se_diff subsampling_se_diff
model2  0.0       0.0     0.0
model1 16.3       4.4     0.3   

Here we actually see an increase in subsampling_se_diff, but this is due to a technical detail not elaborated here. In general, the difference should be better or negligible.

# 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. https://mc-stan.org/

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

Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2020). Leave-One-Out Cross-Validation for Model Comparison in Large Data. Proceedings of the 23rd International Conference on Artificial Intelligence and Statistics (AISTATS), in PMLR 108. arXiv preprint arXiv:2001.00980.

Magnusson, M., Andersen, M., Jonasson, J. & Vehtari, A. (2019). Bayesian leave-one-out cross-validation for large data. Proceedings of the 36th International Conference on Machine Learning, in PMLR 97:4244-4253 online, arXiv preprint arXiv:1904.10679.

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., Simpson, D., Gelman, A., Yao, Y., and Gabry, J. (2019). Pareto smoothed importance sampling. arXiv preprint arXiv:1507.02646.