## 26.4 Testing uniformity

A simple, though not very highly powered, \(\chi^2\)-squared test for uniformity can be formulated by binning the ranks \(0:M\) into \(J\) bins and testing that the bins all have roughly the expected number of draws in them. Many other tests for uniformity are possible. For example, Cook, Gelman, and Rubin (2006) transform the ranks using the inverse cumulative distribution function for the standard normal and then perform a test for normality. Talts et al. (2018) recommend visual inspection of the binned plots.

The bins don’t need to be exactly the same size. In general, if \(b_j\) is the number of ranks that fall into bin \(j\) and \(e_j\) is the number of ranks expected to fall into bin \(j\) (which will be proportional to its size under uniformity), the test statistic is \[ X^2 = \sum_{j = 1}^J \frac{(b_j - e_j)^2}{e_j}. \] The terms are approximately square standard normal, so that under the null hypothesis of uniformity, \[ X^2 \sim \textrm{chiSquared}(J - 1), \] with the corresponding \(p\)-value given by the complementary cumulative distribution function (CCDF) of \(\textrm{chiSquared}(J - 1)\) applied to \(X^2\). Because this test relies on the binomial being approximately normal, the traditional advice is to make sure the expected count in each bin is at least five, i.e., \(e_j \geq 5.\)

### 26.4.1 Indexing to simplify arithmetic

Because there are \(M + 1\) possible ranks, with \(J\) bins, it is easiest to have \(M + 1\) be divisible by \(J\). For instance, if \(J = 20\) and \(M = 999\), then there are \(1000\) possible ranks and an expected count in each bin of \(\frac{M + 1}{J} = 50.\)

Distributing the ranks into bins is another fiddly operation that can be done with integer arithmetic or the floor operation. Using floor, the following function determines the bin for a rank, \[ \textrm{bin}(r_{n, m}, M, J) = 1 + \left\lfloor \frac{r_{n, m}}{(M + 1) / J} \right\rfloor. \] For example, with \(M = 999\) and \(J = 20\), \((M + 1) / J = 50\). The lowest rank checks out, \[ \textrm{bin}(0, 999, 20) = 1 + \lfloor 0 / 50 \rfloor = 1, \] as does the 50th rank, \[ \textrm{bin}(49, 999, 20) = 1 + \lfloor 49 / 50 \rfloor = 1, \] and the 51st is appropriately put in the second bin, \[ \textrm{bin}(50, 999, 20) = 1 + \lfloor 50 / 50 \rfloor = 2. \] The highest rank also checks out, with \(\textrm{bin}(1000, 999, 20) = 50.\)

To summarize, the following pseudocode computes the \(b_j\) values for the \(\chi^2\) test or for visualization in a histogram.

```
Inputs: M draws, J bins, N parameters, ranks r[n, m]
b[1:J] = 0
for (m in 1:M)
++b[1 + floor(r[n, m] * J / (M + 1))]
```

where the `++b[n]`

notation is a common form of syntactic sugar
for `b[n] = b[n] + 1.`

In general, a great deal of care must be taken in visualizing discrete data because it’s easy to introduce off-by-one errors and artifacts at the edges because of the way boundaries are computed by default. That’s why so much attention must be devoted to indexing and binning.

*References*

*Journal of Computational and Graphical Statistics*15 (3): 675–92. https://doi.org/10.1198/106186006X136976.

*arXiv*, no. 1804.06788.