# How does Stigler derive this result from Bernoulli's weak law of large numbers?

In Stephen Stigler's History of Statistics, there's a section on Jacob Bernoulli and his attempt to formalize uncertainty about an unknown proportion given an accumulation of evidence, leading ultimately to the weak law of large numbers.

I'm having trouble understanding Stigler's modern derivation of Bernoulli's law. Stigler writes that

a modern statement of Bernoulli's solution is that for any given small positive number $\epsilon$ and any given large positive number $c$ (say, $c$ = 10, 100, or 1,000), $N$ may be specified so that

$$P \bigg( \left| \frac{X}{N} - p \right| \leq \epsilon \bigg) > cP \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg) .$$

This statement can be easily converted into what is now known as Bernoulli's weak law of large numbers. By simple algebra this becomes

$$(1) \quad P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg) < \frac{1}{(c+1)} .$$

Thus, since we recognize that $c$ is arbitrary, we have that given any $\epsilon > 0$ and any $c$ (however large) $N$ can be specified large enough that (1) holds — and Bernoulli's law is proved.

I'm familiar with modern proofs that make use of Chebyshev's inequality. But I'm having trouble understanding the "simple algebra" that Stigler uses to derive $\frac{1}{(c+1)}$. Can someone help me with this or suggest some reading to make it clearer?

It is indeed very simple algebra. Clearly you need to end up with no term in $P \big( \left| \frac{X}{N} - p \right| \leq \epsilon \big)$; since there's already a term in the complementary event, you have an obvious substitution you can perform.
$$P \bigg( \left| \frac{X}{N} - p \right| \leq \epsilon \bigg) > c\,P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg)$$
$$1-P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg)> c\,P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg)$$
$$1>(1+c)\,P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg)$$
$$\frac{1}{1+c}>\,P \bigg( \left| \frac{X}{N} - p \right| > \epsilon \bigg)$$
etc. (you need $c>-1$ for this last step to work like that but it's said to be "large positive" at the top so we should be fine)