# On the exactness of permutation tests

In this article, we discuss the exactness property of permutation tests, which is closely related to how \(p\)-values are computed from the permutations. This article is a summary of the paper by Phipson and Smyth (2010).

## Traditional formulation of a statistical test

A statistical test aims at determining whether some observed data can be considered to be strong evidence in favor of a so-called alternative hypothesis \(H_a\) compared to a so-called null hypothesis \(H_0\). To do that, a test statistic \(T\) is defined such that:

- its observed value under \(H_0\) can be computed once you observed some data;
- large values of the statistic are evidence in favor of \(H_a\).
- you know (an approximation of) the distribution of \(T\) under the null hypothesis, also known as the null distribution.

Once such a test statistic is available and we observe some data, we can denote by \(t_\mathrm{obs}\) the value of the test statistic computed from the observed data and define the so-called p-value as the null hypothesis tail probability: \[ p_\infty = \mathbb{P}_{H_0} \left( T \ge t_\mathrm{obs} \right). \]The p-value \(p_\infty\) is by definition uniformly distributed on \((0,1)\) under the null hypothesis. Hence, we can define the so-called significance level \(\alpha \in (0,1)\) and decide to reject \(H_0\) in favor of \(H_1\) when \(p_\infty \le \alpha\). By doing this, the probability of wrongly rejecting \(H_0\), also known as the probability of type I errors, is simply: \[ \mathbb{P}_{H_0} \left( p_\infty \le \alpha \right) = \alpha. \]The significance level \(\alpha\) therefore matches by design the probability of type I errors, which means that choosing \(\alpha\) allows to control the probability of type I errors. We say that the test is exact.

When the null distribution of \(T\) is not known, you do not have access to \(p_\infty\). A possible solution is then to resort to resampling techniques to approach the null hypothesis. Once you get an approximate null distribution, the question is how do you compute a p-value that provides an exact statistical test. There are two approaches to this problem: the first estimates the true p-value \(p_\infty\) from the approximate null distribution while the second proposes an alternative definition of the p-value that can be straightforwardly computed. Let us expand on both approaches.

## The two-sample problem in a permutational framework

We start with two samples \(X_1, \dots, X_{n_x} \stackrel{iid}{\sim} \mathcal{D}(\theta_x)\) and \(Y_1, \dots, Y_{n_y} \stackrel{iid}{\sim} \mathcal{D}(\theta_y)\). We want to know whether the two distributions are the same or not on the basis of the two samples we collected. In this parametric setting, it boils down to testing the following hypotheses: \[ H_0: \theta_x = \theta_y \quad \mbox{vs} \quad \theta_x \neq \theta_y. \]Let \(T\) be a statistic that depends on the two samples which is suited for elucidating this test, i.e.:

- you can compute its observed value under the null hypothesis once you observed some data;
- large values of the statistic are evidence in favor of the alternative hypothesis.

Now, for performing the test, one should also know (an approximation of) the distribution of \(T\) under the null hypothesis, also known as the null distribution, from which the p-value associated to this test can be computed for instance.

If one knows the exact null distribution, then there is no need to resort to permutations. However, if the null distribution is not known, permutations come in handy for approaching it.

The idea is that, under the null hypothesis, the two samples come from the same distribution. Hence, we can pull them together as one big sample of size \(n = n_x + n_y\) generated by that common distribution. At this point, we can split the pooled sample into two random subsets of size \(n_x\) and \(n_y\) respectively, and use them to compute a value of the statistic \(T\). If we repeat many times this splitting strategy, say \(m\) times, we end up with \(m\) values of the statistic from which we can compute the empirical distribution, known as the permutation distribution, which approaches the null distribution.

## Permutation p-value as an unbiased estimator of \(p_\infty\)

Let \(t_\mathrm{obs}\) be the value of the statistic computed from the original two samples, \(m\) be the number of permutations used to approach the null distribution and \(B\) be a random variable that counts the number of test statistic values greater than or equal to \(t_\mathrm{obs}\).

By definition, the random variable \(B\) follows a binomial distribution of size \(m\) and rate of success \(p_\infty\). Hence, we can define the following unbiased estimator of \(p_\infty\): \[ \widehat{p_\infty} = \frac{B}{m}. \]

However, when one uses this estimator of the p-value for the purpose of hypothesis testing, the resulting test is not exact. Let us investigate why.

First, remember that the true p-value \(p_\infty\) is a random variable itself, in the sense that its value changes as soon as \(t_\mathrm{obs}\) changes i.e. each time the whole experiment is reconducted. Hence, the probability of wrongly rejecting the null hypothesis using \(\widehat{p_\infty}\) reads: \[ \mathbb{P} \left( \widehat{p_\infty} \le \alpha \right) = \int_\mathbb{R} \mathbb{P} \left( \widehat{p_\infty} \le \alpha | p \right) f_{p_\infty}(p) dp = \int_0^1 \mathbb{P} \left( \widehat{p_\infty} \le \alpha | p \right) dp, \]because \(p_\infty\) is uniformly distributed on \((0,1)\) under the null hypothesis.

Next, notice that \(\widehat{p_\infty}\) can only take on a finite set of values \(\left\{ 0, \frac{1}{m}, \frac{2}{m}, \dots, \frac{m-1}{m}, 1 \right\}\). Hence, we have for any \(b \in 0, 1, \dots, m\): \[ \mathbb{P} \left( \widehat{p_\infty} = \frac{b}{m} \right) = \int_0^1 \mathbb{P} \left( \widehat{p_\infty} = \left. \frac{b}{m} \right| p \right) dp = \int_0^1 \binom{m}{b} p^b (1-p)^{m-b} dp = \frac{1}{m + 1}. \]

We can therefore deduce that: \[ \mathbb{P} \left( \widehat{p_\infty} \le \alpha \right) = \frac{\lfloor m \alpha \rfloor + 1}{m + 1} \neq \alpha. \]

The following R code shows graphically that using \(\widehat{p_\infty}\) as p-value does not provide an exact test:

    alpha <- seq(0.01, 0.1, by = 0.01)
    m <- c(10, 100, 1000)
    p1 <- crossing(alpha, m) %>%
      mutate(
        p = (floor(m * alpha) + 1) / (m + 1),
        mf = paste("m =", m)
      ) %>%
      ggplot(aes(alpha, p, color = mf)) +
      geom_point() +
      geom_abline(aes(intercept = 0, slope = 1)) +
      labs(
        x = "Significance level",
        y = "Probability of wrongly rejecting H0"
      ) +
      facet_wrap(vars(mf)) +
      scale_color_viridis_d() +
      scale_y_continuous(limits = c(0, 0.1)) +
      coord_equal() +
      theme_bw()
    fig <- p1 %>%
      plotly::ggplotly() %>%
      plotly::hide_legend()
    htmlwidgets::saveWidget(
      widget = fig,
      file = "exactness-fig1.html",
      selfcontained = rmarkdown::pandoc_available("1.12.3")
    )
    htmltools::tags$iframe(
      src = "exactness-fig1.html",
      scrolling = "no",
      seamless = "seamless",
      frameBorder = "0",
      width = "100%",
      height = 400
    )

## Permutation p-value as the tail probability of a resampling distribution

An alternative strategy is to define the p-value by looking at the random variable \(B\) instead of the test statistic \(T\). While the p-value is the tail probability of the null distribution, in the context of randomization tests, we can define it as the tail probability of the distribution of \(B\). Given a fixed number \(m\) of sampled permutations, recall that the random variable \(B\) counts the number of test statistic values larger than or equal to \(t_\mathrm{obs}\). Hence, an alternative equivalent definition of the p-value is given by the so-called exact permutation p-value: \[ p_e = \mathbb{P}_{H_0} \left( B \le b \right), \]where \(b\) is the observed number of test statistics larger than or equal to \(t_\mathrm{obs}\) (using the observed sample of permutations that was drawn).

Let \(B_t\) be a random variable that counts the total number of possible distinct test statistic values exceeding \(t_\mathrm{obs}\) and recall that \(m_t\) is the total number of possible distinct permutations. We denote by \[ p_t = \frac{B_t + 1}{m_t + 1}, \]the permutation p-value when the exhaustive list of all permutations is used.

As we have seen before, it is straightforward to show that \(B_t\) follows a discrete uniform distribution on the integers \(0, \dots, m_t\) and that, conditional on \(B_t = b_t\), the random variable \(B\) follows a binomial distribution of size \(m\) and rate of success \(p_t\). We can thus write: \[ p_e = \sum_{b_t=0}^{B_t} \mathbb{P}_{H_0} \left( B \le b | B_t = b_t \right) \mathbb{P}_{H_0} \left( B_t = b_t \right) = \frac{1}{m_t + 1} \sum_{b_t=0}^{B_t} F_B \left( b; m, \frac{b_t + 1}{m_t + 1} \right), \] where \(F_B \left( \cdot; m, \frac{b_t + 1}{m_t + 1} \right)\) is the cumulative probability function of the binomial distribution of size \(m\) and probability of success \(\frac{b_t + 1}{m_t + 1}\).

This can be computationally intense to compute for large values of \(m_t\), in which case one might use the following integral approximation: \[ p_e \approx \frac{b+1}{m+1} - \int_0^{\frac{0.5}{m_t+1}} F_B (b; m, p_t) dp_t. \]

This approximation shows that the exact p-value \(p_e\) is upper bounded by \[ p_u = \frac{b+1}{m+1}, \] which happens to be the exact p-value in the case of sampling permutations without replacement.

## Comparison by empirical evidence

In flipr, you perform a permutation test using the estimator \(\widehat{p_\infty}\) of the p-value \(p_\infty\) by setting `type == "estimate"`. This provides a non-exact test but an unbiased estimate of \(p_\infty\). You perform a permutation test using the permutation p-value \(p_e\) by setting `type == "exact"`. This provides an exact test. You also have the possibility of using the upper bound p-value \(p_u\) using `type == "upper_bound"`.

The following R code runs simulations to empirically estimate the probability of wrongly rejecting the null hypothesis. The generative model for both samples is the standard normal distribution. Sample sizes are set to \(n_1 = n_2 = 5\). We draw \(20\) permutations for each test. We used a significance level of \(5\%\).

    # General setup
    nreps <- 1e4
    n1 <- 5
    n2 <- 5
    set.seed(12345)
    sim <- map(sample.int(.Machine$integer.max, nreps, replace = TRUE), ~ {
        list(
          x = rnorm(n = n1, mean = 0, sd = 1),
          y = rnorm(n = n2, mean = 0, sd = 1),
          s = .x
        )
      })

    # Cluster setup
    cl <- makeCluster(detectCores(logical = FALSE))
    clusterEvalQ(cl, {
      library(tidyverse)
      library(flipr)
      null_spec <- function(y, parameters) {
        map(y, ~ .x - parameters)
      }
      stat_functions <- list(stat_t)
      stat_assignments <- list(delta = 1)
      nperms <- 20
      alpha <- 0.05
    })

    alpha_estimates <- pbapply::pblapply(sim, function(.l) {
        pf <- PlausibilityFunction$new(
          null_spec = null_spec,
          stat_functions = stat_functions,
          stat_assignments = stat_assignments,
          .l$x, .l$y,
          seed = .l$s
        )
        pf$set_nperms(nperms)
        pf$set_alternative("left_tail")
        pf$set_pvalue_formula("exact")
        pv_exact <- pf$get_value(0)
        pf$set_pvalue_formula("upper_bound")
        pv_upper_bound <- pf$get_value(0)
        pf$set_pvalue_formula("estimate")
        pv_estimate <- pf$get_value(0)
        c(
          exact       = pv_exact       <= alpha,
          upper_bound = pv_upper_bound <= alpha,
          estimate    = pv_estimate    <= alpha
        )
      }, cl = cl) %>%
      transpose() %>%
      simplify_all() %>%
      map(mean)
    stopCluster(cl)

    as_tibble(alpha_estimates)
    #> # A tibble: 1 × 3
    #>    exact upper_bound estimate
    #>    <dbl>       <dbl>    <dbl>
    #> 1 0.0507      0.0507   0.0978

This simulation provides numerical evidence that using the permutation p-value \(p_e\) yields an exact test. This is by design since \(p_e\) is a genuine probability hence it is uniformly distributed on \((0,1)\) under the null hypothesis. We also observe that the upper bound \(p_u\) is very close to the exact p-value \(p_e\).

Of course, one might criticize this simulation in which the number of permutations \(m\) has been chosen small on purpose to prove our point. In effect, when \(m\) is larger, all formulae to compute the p-value yield an asymptotically exact test. Quoting Phipson and Smyth (2010), while this is true, the urgent need to avoid small \(m\) disappears when the exact p-value \(p_e\) is used in place of \(\widehat{p_\infty}\), because \(p_e\) ensures a valid statistical test regardless of the sample sizes or the number of permutations. When exact p-values are used, the only penalty for choosing \(m\) small is a loss of statistical power to reject the null hypothesis. This is as it should be: more permutations should generally provide greater power.

## References

Phipson, Belinda, and Gordon K Smyth. 2010. “Permutation p-Values Should Never Be Zero: Calculating Exact p-Values When Permutations Are Randomly Drawn.” Statistical Applications in Genetics and Molecular Biology 9 (1). https://doi.org/10.2202/1544-6115.1585.