Sophie

Why is the Bonferroni correction the value that it is, intuitively? I understand the proof but consider the following argument:

Imagine I have $n=10$ hypotheses and I a desired $p=0.05$. Then if the null hypothesis holds then the probability that at least one of them will appear true is $1-(1-p)^n=0.4$. We wish this was $p$ instead so we have to divide $p$ by $(1-(1-p)^n)p^{-1}$ so our new threshold is $p^2(1-(1-p)^n)^{-1}=0.006$. This is more than the true value $p/n=0.005$, so the argument fails. Why?

Consider the group generated by the two functions $x\mapsto \frac 1x$ and $x\mapsto -1-x$. It is isomorphic to $C_3$ and contains the following elements:

$$G=\left\{x,\frac 1x, -1-x,-\frac{1}{1+x},-\frac{x}{1+x},-\frac{1+x}{x}\right\}$$

I want to prove that for a prime $p$, when this group acts on an element $a$ of $\mathbb{Z}/p\mathbb{Z}$ such that $a^3\not\equiv 1\pmod p$ no pair of the group's actions maps $a$ to the same number. Assuming that $a$ and $a+1$ are invertible, I found by equating the identity to the other 5 transformations that there are collisions when $a\equiv -2^{-1}, -…