Lesson 2

Problem. Let \(\alpha\) be the proportion of heads in \(n\) tosses of a fair coin: \(\alpha=\frac{\beta}{n}\), where \(\beta \sim \operatorname{Binomial}(n,1/2)\). We know that \(\operatorname{E}[\alpha]=\frac{1}{2}\). We want to estimate the probability that \(\alpha\) deviates from its mean by at least \(\frac{1}{4}\): \[\Pr\left[\left|\alpha-\frac{1}{2}\right| \ge \frac{1}{4} \right] \ .\] What upper bound on this probability does Chebyshev's inequality give?