# How to show expected number of false rejection $E[V]\leq \sum_{i,H_0 true} Pr(p_i\leq\frac{\alpha}{n})$

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

• 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. Commented Nov 15, 2021 at 17:51
• @MishaLavrov Since $E[V]=\sum_{v\leq n_0}vP(V=v)$. How is that bound transferred to each term here? Commented Nov 15, 2021 at 17:55
• Use linearity of expectation on a sum of indicator random variables. Commented Nov 15, 2021 at 17:55
• @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 $E[X]=\int P(X\geq x)dx$ Commented Nov 15, 2021 at 18:17

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 $$0\le x\le 1$$.

• 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. Commented Nov 15, 2021 at 22:59