# Bayesian One-Sample T-Test (Stan)

#### Quentin F. Gronau

#### 2025-11-18

In this vignette, we explain how we can compute the (log) marginal likelihood and the Bayes factor for models fitted in `Stan`. This approach has the advantage that the user only needs to pass the fitted `stanfit` object which contains all information that is necessary to compute the (log) marginal likelihood. Here we show how one can conduct a Bayesian one-sample t-test as implemented in the `BayesFactor` package (Morey & Rouder, 2015).

## Model

The Bayesian one-sample t-test makes the assumption that the observations are normally distributed with mean \(\mu\) and variance \(\sigma^2\). The model is then reparametrized in terms of the standardized effect size \(\delta = \mu/\sigma\). For the standardized effect size, a Cauchy prior with location zero and scale \(r = 1/\sqrt{2}\) is used. For the variance \(\sigma^2\), Jeffreys’s prior is used: \(p(\sigma^2) \propto 1/\sigma^2\).

In this example, we are interested in comparing the null model \(\mathcal{H}_0\), which posits that the effect size \(\delta\) is zero, to the alternative hypothesis \(\mathcal{H}_1\), which assigns \(\delta\) the above described Cauchy prior.

## Data

In this example, we will analyze the `sleep` data set from the `t.test` example. This data set shows the effect of two soporific drugs (increase in hours of sleep compared to control) on 10 patients. These data can be analyzed via a one-sample t-test by first computing the difference scores and then conducting the t-test using these difference scores as data. The difference scores are calculated as follows:

    library(bridgesampling)

    set.seed(12345)

    # Sleep data from t.test example
    data(sleep)

    # compute difference scores
    y <- sleep$extra[sleep$group == 2] - sleep$extra[sleep$group == 1]
    n <- length(y)

## Specifying the Models

Next, we implement the models in `Stan`. Note that to compute the (log) marginal likelihood for a `Stan` model, we need to specify the model in a certain way. Instad of using `"~"` signs for specifying distributions, we need to directly use the (log) density functions. The reason for this is that when using the `"~"` sign, constant terms are dropped which are not needed for sampling from the posterior. However, for computing the marginal likelihood, these constants need to be retained. For instance, instead of writing `y ~ normal(mu, sigma)` we would need to write `target += normal_lpdf(y | mu, sigma)`. The models can then be specified and compiled as follows (note that it is necessary to install `rstan` for this):

    library(rstan)

    # models
    stancodeH0 <- '
    data {
      int<lower=1> n; // number of observations
      vector[n] y; // observations
    }
    parameters {
      real<lower=0> sigma2; // variance parameter
    }
    model {
      target += log(1/sigma2); // Jeffreys prior on sigma2
      target += normal_lpdf(y | 0, sqrt(sigma2)); // likelihood
    }
    '
    stancodeH1 <- '
    data {
      int<lower=1> n; // number of observations
      vector[n] y; // observations
      real<lower=0> r; // Cauchy prior scale
    }
    parameters {
      real delta;
      real<lower=0> sigma2;// variance parameter
    }
    model {
      target += cauchy_lpdf(delta | 0, r); // Cauchy prior on delta
      target += log(1/sigma2); // Jeffreys prior on sigma2
      target += normal_lpdf(y | delta*sqrt(sigma2), sqrt(sigma2));  // likelihood
    }
    '
    # compile models
    stanmodelH0 <- stan_model(model_code = stancodeH0, model_name="stanmodel")
    stanmodelH1 <- stan_model(model_code = stancodeH1, model_name="stanmodel")

## Fitting the Models

Now we can fit the null and the alternative model in `Stan`. One usually requires a larger number of posterior samples for estimating the marginal likelihood than for simply estimating the model parameters. This is the reason for using a comparatively large number of samples for these simple models.

    # fit models
    stanfitH0 <- sampling(stanmodelH0, data = list(y = y, n = n),
                          iter = 20000, warmup = 1000, chains = 4, cores = 1,
                          control = list(adapt_delta = .99))
    stanfitH1 <- sampling(stanmodelH1, data = list(y = y, n = n, r = 1/sqrt(2)),
                          iter = 20000, warmup = 1000, chains = 4, cores = 1,
                          control = list(adapt_delta = .99))

## Computing the (Log) Marginal Likelihoods

Computing the (log) marginal likelihoods via the `bridge_sampler` function is now easy: we only need to pass the `stanfit` objects which contain all information necessary. We use `silent = TRUE` to suppress printing the number of iterations to the console:

    H0 <- bridge_sampler(stanfitH0, silent = TRUE)
    H1 <- bridge_sampler(stanfitH1, silent = TRUE)

We obtain:

    print(H0)

    ## Bridge sampling estimate of the log marginal likelihood: -20.80934
    ## Estimate obtained in 4 iteration(s) via method "normal".

    print(H1)

    ## Bridge sampling estimate of the log marginal likelihood: -17.96114
    ## Estimate obtained in 4 iteration(s) via method "normal".

We can use the `error_measures` function to compute an approximate percentage error of the estimates:

    # compute percentage errors
    H0.error <- error_measures(H0)$percentage
    H1.error <- error_measures(H1)$percentage

We obtain:

    print(H0.error)

    ## [1] "0.049%"

    print(H1.error)

    ## [1] "0.0655%"

## Computing the Bayes Factor

To compare the null model and the alternative model, we can compute the Bayes factor by using the `bf` function. In our case, we compute \(\text{BF}_{10}\), that is, the Bayes factor which quantifies how much more likely the data are under the alternative versus the null hypothesis:

    # compute Bayes factor
    BF10 <- bf(H1, H0)
    print(BF10)

    ## Estimated Bayes factor in favor of H1 over H0: 17.25678

We can compare the bridge sampling result to the `BayesFactor` package result:

    library(BayesFactor)
    BF10.BayesFactor <- extractBF(ttestBF(y), onlybf = TRUE)

We obtain:

    print(BF10.BayesFactor)

    ## [1] 17.25888

## One-sided Test

We can also conduct one-sided tests. For instance, we could test the hypothesis that the effect size is positive versus the null hypothesis. Since we already fitted the null model and computed its marginal likelihood, we only need to slightly adjust the alternative model to reflect the directed hypothesis. To achieve this, we need to truncate the Cauchy prior distribution for \(\delta\) at zero and then renormalize the (log) density. This is easily achieved via the `Stan` function `cauchy_lccdf` which corresponds to the log of the complementary cumulative distribution function of the Cauchy distribution. Thus, `cauchy_lccdf(0 | 0, r)` gives us the log of the area greater than zero which is required for renormalizing the truncated Cauchy prior. The model can then be specified and fitted as follows:

    stancodeHplus <- '
    data {
      int<lower=1> n; // number of observations
      vector[n] y; // observations
      real<lower=0> r; // Cauchy prior scale
    }
    parameters {
      real<lower=0> delta; // constrained to be positive
      real<lower=0> sigma2;// variance parameter
    }
    model {
      target += cauchy_lpdf(delta | 0, r) - cauchy_lccdf(0 | 0, r); // Cauchy prior on delta
      target += log(1/sigma2); // Jeffreys prior on sigma2
      target += normal_lpdf(y | delta*sqrt(sigma2), sqrt(sigma2));  // likelihood
    }
    '
    # compile and fit model
    stanmodelHplus <- stan_model(model_code = stancodeHplus, model_name="stanmodel")
    stanfitHplus <- sampling(stanmodelHplus, data = list(y = y, n = n, r = 1/sqrt(2)),
                             iter = 30000, warmup = 1000, chains = 4,
                             control = list(adapt_delta = .99))

The (log) marginal likelihood is then computed as follows:

    Hplus <- bridge_sampler(stanfitHplus, silent = TRUE)

We obtain:

    print(Hplus)

    ## Bridge sampling estimate of the log marginal likelihood: -17.27045
    ## Estimate obtained in 4 iteration(s) via method "normal".

We can again use the `error_measures` function to compute an approximate percentage error of the estimate:

    Hplus.error <- error_measures(Hplus)$percentage

We obtain:

    print(Hplus.error)

    ## [1] "0.136%"

The one-sided Bayes factor in favor of a positive effect versus the null hypothesis can be computed as follows:

    # compute Bayes factor
    BFplus0 <- bf(Hplus, H0)
    print(BFplus0)

    ## Estimated Bayes factor in favor of Hplus over H0: 34.42872

We can compare the bridge sampling result to the `BayesFactor` package result:

    BFplus0.BayesFactor <- extractBF(ttestBF(y, nullInterval = c(0, Inf)), onlybf = TRUE)[1]

We obtain:

    print(BFplus0.BayesFactor)

    ## [1] 34.41694

## References

Richard D. Morey and Jeffrey N. Rouder (2015). BayesFactor: Computation of Bayes Factors for Common Designs. R package version 0.9.12-2.