# Using Leave-one-out cross-validation for large data

#### Mans Magnusson, Paul Bürkner, Aki Vehtari and Jonah Gabry

#### 2019-12-18

# 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 approximatations to posterior distributions. Some sections from this vignette are excerpted from the papers

- Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2019). Leave-One-Out Cross-Validation for Model Comparison in Large Data.

- Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2019). Leave-One-Out Cross-Validation for Large Data. In International Conference on Machine Learning

- 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

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:

    llfun_logistic <- function(data_i, draws) {
      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 = TRUE)
    }

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:

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

    # 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.

    parameter_draws_1 <- extract(fit_1)$beta
    stan_df_1 <- as.data.frame(standata)
    loo_i(1, llfun_logistic, data = stan_df_1, draws = parameter_draws_1)

    $pointwise
        elpd_loo mcse_elpd_loo        p_loo     looic
    1 -0.3310342  0.0002908997 0.0003487243 0.6620683
    ...

# 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:

    library("loo")
    parameter_draws_1 <- extract(fit_1)$beta
    stan_df_1 <- as.data.frame(standata)

    set.seed(4711)
    loo_ss_1 <-
      loo_subsample(
        llfun_logistic,
        draws = parameter_draws_1,
        data = stan_df_1,
        observations = 100 # take a subsample of size 100
      )
    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, Gelman, and Gabry (2017)). 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, draws = parameter_draws_1, data = stan_df_1,
                        observations = 200) # subsample 200 instead of 100
    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,
        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
      )
    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
      )
    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
      )
    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
    set.seed(4711)
    loo_ss_1 <-
      loo_subsample(
        x = llfun_logistic,
        draws = parameter_draws_1,
        data = stan_df_1,
        observations = 200
      )

    # compute subsampling loo for second model (with log-arsenic)
    loo_ss_2 <-
      loo_subsample(
        x = llfun_logistic,
        draws = parameter_draws_2,
        data = stan_df_2,
        observations = 200
      )

    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,
        draws = parameter_draws_2,
        data = stan_df_2,
        observations = loo_ss_1
      )

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,
        draws = parameter_draws_2,
        data = stan_df_2,
        observations = idx
      )

    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 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, draws = parameter_draws_2, data = stan_df_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. http://mc-stan.org

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

Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2019). Leave-One-Out Cross-Validation for Model Comparison in Large Data.

Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2019). Leave-One-Out Cross-Validation for Large Data. In International Conference on Machine Learning

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.