Chapter 8.1, Exercise 7, Page 312

Find the maximum possible value for \(p(1 - p)\) if \(0 < p < 1\).

Using this result and Exercise 6, show that the estimate

\(\quad \quad \quad P \left( \left| \frac{S_n}{n} -p \right| \ge \epsilon \right) \le \frac{1}{4n \epsilon ^2}\)

is valid for any p. 

We note that the following conditions are satisfied:

  1. \(f(p)=p(1-p)=p-p^2\) is continuous on the closed interval \([0,1]\) .
  2. \(f(p)=p(1-p)=p-p^2\) is differentiable on the open interval \((0,1)\) .
  3. \(f(0)=f(1)=0\) .

By Rolle’s Theorem there must be some number \(c\) such that \(0<c<1\) and \(f'(c)=0\),
which means that f(p) must have a critical point in \((0,1)\) .

We note that setting \(\frac{df}{dp}= 1-2p = 0\) gives such critical point, \(p=\frac{1}{2}\) , and \(f\left( \frac{1}{2}\right)=\frac{1}{2} \left(1-\frac{1}{2} \right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) .

We also note that \(\frac{d^2f}{dp^2}= -2 < 0\) , which confirms that this critical point is a maximum.

Therefore, \[\max\limits_{0<p<1} f(p)=p(1-p)=\frac{1}{4}\] .
From Exercise 6, we have

\[\begin{aligned} P \left( \left| \frac{S_n}{n} -p \right| \ge \epsilon \right) &\le \frac{p(1-p)}{n \epsilon ^2} \\ &=p(1-p) \cdot \frac{1}{n \epsilon ^2} \\ &\le \frac{1}{4} \cdot \frac{1}{n \epsilon ^2} \end{aligned}\] .

QED.