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Consider $n$ hypothesis testing with size $\alpha/n$. Suppose there are $n_0$ of them are true. Let $V$ be the random number of falsely rejected hypothesis of those true ones.

$\textbf{Q:}$ How to show expected number of false rejection $E[V]\leq \sum_{i,H_0 true} Pr(p_i\leq\frac{\alpha}{n})$ where sum is over those true hypothesis, $Pr(-)$ is the probability $p_i$ is the p-value. One can see that $E[V]=\sum_{v\leq n_0}P(V\geq v)$. This is shown in the following document's 4.4.1. https://statweb.stanford.edu/~candes/teaching/stats300c/Lectures/Lecture04.pdf

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  • $\begingroup$ You don't need the fancy expected value theorem. This is just the sum over all $n_0$ true hypotheses of the probability of being falsely rejected. $\endgroup$
    – Misha Lavrov
    Commented Nov 15, 2021 at 17:51
  • $\begingroup$ @MishaLavrov Since $E[V]=\sum_{v\leq n_0}vP(V=v)$. How is that bound transferred to each term here? $\endgroup$
    – user45765
    Commented Nov 15, 2021 at 17:55
  • $\begingroup$ Use linearity of expectation on a sum of indicator random variables. $\endgroup$
    – Misha Lavrov
    Commented Nov 15, 2021 at 17:55
  • $\begingroup$ @MishaLavrov What are the sets here? I do not see how to connect $E[V]$ and a set of indictor random variables here. I could say $E[V]=E[\sum_{v\leq n_0}v I(V=v)]$. I do not see this will tell me anything useful here. If I use $P(V\geq v)\leq P(V\geq 1)$, this will produce me an extra factor of $n_0$ after the sum when I use $E[X]=\int P(X\geq x)dx$ $\endgroup$
    – user45765
    Commented Nov 15, 2021 at 18:17

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I don't think this is quite right.

Let $I$ be the set of indices corresponding to the true null hypotheses, so there are $n_0$ elements in $I$.

And suppose $\hat p_i$ is the p-value corresponding to the $i$th null hypothesis $H_{0i}$, $i=1,2,\ldots,n$.

Then, $$V=\sum_{i\in I}I\left(\hat p_i\le \frac{\alpha}{n}\right)$$

Therefore,

$$E\left[V\right]=\sum_{i \in I}E\left[I\left(\hat p_i\le \frac{\alpha}{n}\right)\right]=\sum_{i \in I} P\left(\hat p_i\le \frac{\alpha}{n}\right)\le \sum_{i \in I} \frac{\alpha}{n}=\frac{n_0\alpha}{n}$$

The only inequality used here is $P(\hat p_i\le x)\le x$ for $0\le x\le 1$.

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    $\begingroup$ I think the crucial step is $V=\sum_{i\in H_0}I(p_i\leq\frac{\alpha}{n})$. However, I do not think inequality shows up at the correct step as you have shown equality here. $\endgroup$
    – user45765
    Commented Nov 15, 2021 at 22:59

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