The Riskiest Coin

Question

Maximize f(p) = p(1-p) over p in [0,1], value and location, with a proof that doesn't need calculus. Why does every statistician's worst case live at this point?

Accepted Answer

Answer: maximum value frac14, at p = frac12. No-calculus proof. Complete the square: p(1-p) = tfrac14 - left(p - tfrac12right)^2 le tfrac14, with equality iff p = frac12. (AM-GM gives the same: sqrtp(1-p) le fracp + (1-p)2 = frac12, equality when the factors are equal.) Calculus check if you must: f'(p) = 1 - 2p = 0 at p = frac12, f'' < 0. Why this tiny fact carries weight. p(1-p) is the variance of a Bernoulli(p) outcome, a binary bet, a fill-or-no-fill, a win-or-lose trade. The parabola says: - Uncertainty peaks at 50–50. A contract priced near 0.50 (an even-odds prediction market, an at-the-money binary option) has the largest possible outcome variance, frac14; one priced at 0.95 has variance 0.0475, more than five times smaller. Quoting width and position limits on event contracts should scale with sqrtp(1-p), tightest conviction, widest risk, both at the middle. - Sample-size planning. The standard error of an estimated hit rate is sqrtfracp(1-p)n le frac12sqrt n. The bound frac12sqrt n is why pollsters can quote "pm 3%" without knowing p: n approx 1,100 makes frac12sqrt n approx 0.015, i.e. a 95% interval of about pm 3% whatever the true p. Same logic sizes the backtest you n

The Riskiest Coin

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