Figure 1: A realization of a Brownian bridge together with the optimal stopping boundary. Numerical calculations show that B ≈ 0.8399 and V(0,0) = √(π⁄2) × (1 − B2) ≈ 0.3691.
…Remark: ..X pins at a at time T…[and] B is as in Theorem 2.1. Moreover,
τ✱ = inf{s ≥ t: Xs ≥ a + B√(T − s)}
is an optimal stopping time.
…The optimal stopping Problem (1.1) is a continuous version of the following classical urn problem. Suppose that an urn contains n red balls and n black balls which are drawn without replacement. Moreover, every red ball pays you a dollar and every black ball fines you a dollar. If you may stop the game at any time, what is your maximal expected profit and what strategy should you use? (In the guise of a card game, the same problem appears as Question 1.42 of Heard on the Street: Quantitative Questions from Wall Street Job Interviews, Crack2007 [cf. Problem 14].)
One potential application of Problem (1.1) is in financial theory. Indeed, it has recently been reported that stock prices tend to end up at the strikes of heavily traded options written on the stocks when they mature; see [1] and the references therein. A possible explanation for this pinning phenomenon is that hedgers with a long position in vanilla options are advised by standard Black-Scholes theory to buy stocks if the price falls and to sell stocks if the price rises. Moreover, this trading is more important at a strike and close to maturity since the option delta changes rapidly there [see delta neutrality]. By supply and demand arguments, the stock price therefore pins at the strike. Problem (1.1) serves as a first approximation to the problem of when to sell a stock in the presence of stock pinning.