# Bonferroni Correction: Independence vs Disjoint Events

Let $$T_i$$ and $$c_i$$ be a test statistic and critical value, respectively, for testing hypothesis $$H_i$$, $$i = 1,2,\dots, n$$. Assume a Type-1 error rate of $$\alpha$$ for each test. The family-wise error rate is given by

$$\textit{FWER} :=P\left(\bigcup_{i=1}^n\{T_i>c_i\}\right).$$

If the events $$\{T_i\leq c_i\}$$, $$i=1,2,\dots,n$$ are independent, the $$\textit{FWER}$$ can be rewritten as $$P\left(\bigcup_{i=1}^n\{T_i>c_i\}\right) = 1-P\left(\bigcap_{i=1}^n\{T_i\leq c_i\}\right) = 1 - \prod_{i=1}^nP(\{T_i\leq c_i\} = 1 - (1-\alpha)^n.$$

The Bonferroni method replaces $$\alpha$$ with $$\alpha^* := \frac\alpha n$$ to ensure that $$\textit{FWER} \leq \alpha$$. In fact, using the inequality $$1-\exp(-x)\leq x$$ for all $$x\in\mathbb R$$, we can show that $$\textit{FWER}\leq\alpha$$ if we chose a Type-1 error rate of $$\alpha^*$$ for each test.

The Bonferroni correction is a consequence of the sub-additivity property of probability measures:

$$P\left(\bigcup_{i=1}^n\{T_i>c\}\right) \leq \sum_{i=1}^n P(\{T_i>c_i\} = n\alpha.$$

Now I was wondering when is the bound attained? By the additivity property of probability measures for disjoint sets, this is the case if the events $$\{T_i>c_i\}$$, $$i=1,2,\dots,n$$ are mutually disjoint.

Since the events $$\{T_i\leq c_i\}$$ cannot be independent while the events $$\{T_i>c_i\}$$ are disjoint, and vice versa (see my other question on MSE https://math.stackexchange.com/questions/4723024/subadditivity-of-probability-measure-and-independence), it seems to be true that independence is not as worse as disjointness of events with regard to Type-1 error rate cumulation. Is this conclusion correct? Because I have often read that independence of hypotheses is the worst case.

I suppose that hypotheses concerning different data sets are independent. But what would be an example of disjoint events?

A very simple example for the worst case is standard testing of a parametric hypothesis $$H_0:\ \mu=0$$ on the same data separately against both $$\mu<0$$ and $$\mu>0$$. The rejection events will be disjoint in most cases (think of a t-test of a mean). They're obviously not independent, because if you reject one, you will not reject the other. If you run both of these tests at level $$\alpha/2$$ as prescribed by Bonferroni, putting the tests together gives the usual two-sided test against $$\mu\neq 0$$ at level $$\alpha$$.