# Why applying the same test to several hypotheses reduces its rejection power?

UPDATE: after more study on the subject, I noticed that I didn't have a good grasp on the difference between $p$-value and significance. In particular in the last portion of the question I use the term $p$-value instead of significance. I'll leave the question as is for future reference, just keep in mind that it is a bit imprecise :)

Reading "Kernel Methods for Pattern Analysis", by John Shawe-Taylor, I've got stuck into this sentence:

[...] if we applied the same test for $n$ hypotheses $P_1, \dots, P_n$, and found that for one of the hypotheses, say $P^*$, a significance of $p$ is measured, we can only assert the hypothesis with significance $np$. This is because the data could have misled us about any one of the hypotheses, so that even if none were true there is still a probability $p$ for each hypothesis that it could have appeared significant, giving in the worst case a probability $np$ that one of the hypotheses appears significant at level $p$.

This sentence doesn't make any sense to me, since each test is independent of the other $n-1$. Can you provide me with a clear example of the contrary? Can you provide me a mathematical proof, or at least an insight?

EDIT: Thanks for the link in the comment (for future reference: xkcd: Significant). Let me elaborate further.

Let's suppose I have a training algorithm $f(\alpha)$, that is, a function $f$ that returns the parameters of a classifier. $\alpha$ is a custom parameter that can be set by the experimenter. If I vary $\alpha$, let's say, between $0$ and $1$ with step $0.1$, I'll get $11$ different results for the classifier parameters. Let's suppose that only in one case this trained classifier (e.g., for $\alpha = 0.6$) is better than a "reference", with $p$-value $0.01$. I still have to say that the overall $p$-value for the null hypothesis is $11 \times 0.11 = 0.121$?

• I believe this question may be answered at stats.stackexchange.com/questions/88065. Certainly the "clear example" portion of it is! – whuber Aug 25 '15 at 12:45
• I'm having trouble following the new paragraph because classification and hypothesis testing differ in aims and approach. Could you explain what "null hypothesis" is being evaluated in this procedure? – whuber Aug 25 '15 at 13:42
• @whuber the null hypothesis is "there is no difference in classification performance (error rate) between the reference classifier and the classifier trained using $f(\alpha)$" – etamponi Aug 25 '15 at 15:19
• Thank you--that makes the situation clearer. In order to test such a hypothesis, we need a probability model for the distribution under the null. That model will depend on whether you are using the same data to test each classifier. This is an important issue, because any overlap in such datasets will introduce statistical dependencies among the results that ought to be accommodated in any p-value calculation. – whuber Aug 25 '15 at 17:21
• Thank you for the insight! I'm using the "standard" paired t-test, using a large number of cross-validation runs to build the sample. What I didn't consider is the dependence between two experiments with different $\alpha$ values! I guess it would be easier to make each experiment as independent as possible from the others, by using a different seed each time. – etamponi Aug 25 '15 at 17:51

Look at this from a different perspective with just P1 and P2. lets say you're looking for p significance level, that is, the chance to regect P1 if P1 is true is less or equal to p.
lets assume that P1 and P2 are indepentent.
what is the chance to not reject P1 and P2 if both are true: (1-p)^2. so the chance to reject atleast one: 1-(1-p)^2 = 2*p - p^2 --> 2*p for small p. The same methodology applies for n>2. When we use instead of P, P/n than we will have the desired significance level of p that atleast one of the hypothesis will be rejected when all the null are trues.
• Isn't $p$ the chance of accepting $P_1$ even if $P_1$ is false? – etamponi Aug 25 '15 at 12:58