# Proof that gatekeeping controls alpha?

Consider the following gatekeeping strategy (for the sake of simplicity, let's assume that family $$F_i$$ has only 1 hypothesis, $$H_i$$),

• Hypothesis $$H_1$$ is tested at the 5% significance level,
• If $$H_1$$ cannot be reject, then stop
• Otherwise, $$H_2$$ is tested at the 5% significance level,
• And so forth.

Why/How does this procedure maintain the type 1 error rate at 5%? I.e.,

$$\Pr(\text{reject } H_1 \text{ or reject } H_2 \,|\, H_1 \text{ is true and } H_2 \text{ is true}) = 0.05$$

• It doesn't (look like). Can you give us some context? What are these hypotheses? What data are they being tested on? Commented Sep 5, 2023 at 8:50
• That's my understanding of statmodeling.stat.columbia.edu/wp-content/uploads/2017/11/…. Post edited accordingly Commented Sep 5, 2023 at 8:53
• I don't buy this argument. One of the key steps of this procedure is to order the hypotheses accordingly (from best to worst in a sense), which is just a different way of doing multiple comparisons before doing multiple comparisons. Commented Sep 5, 2023 at 9:01
• @user2974951: Could you please make it an answer with the more detailed procedure and the calculation of the type I error? Commented Sep 5, 2023 at 10:06

• the intersection of the two null hypotheses is rejected iff $$p_1 \leq \alpha$$,
• the first elementary null hypothesis is rejected iff $$p_1 \leq \alpha$$, and
• the second elementary null hypothesis is rejected iff $$p_2 \leq \alpha$$.
If such a procedure has total initially assigned $$\alpha$$ smaller than the intended level of the test and all outgoing vertices having weights than sum to less than 1, then you know that this translates into a closed testing procedure that controls the familywise type I error rate at level $$\alpha$$ (see the linked paper).