Martingale Mutiny - optional stopping practice
← Stochastic ProcessesA positive edge and a proportional storm. When do you drop anchor?
Take the first leg.
The path draws itself as you sail.
Deliberately no live E[Xn] here. Dropping anchor asks you for it, and a readout would answer the question for you.
About Martingale Mutiny
Sail a wealth process with a flat positive edge against a proportional storm that halves your hold - and get graded on predicting the recursion's expected value when you choose to stop.
Your boat starts with 100 cargo and sails up to 30 legs toward an island at 250 cargo, with a shipwreck line at 15. Each leg one of two things happens. With probability 1 - e^(-λ) (λ = 0.1, about 10% per leg) a storm hits and your cargo is halved. Otherwise it is a calm-water leg: cargo moves by a fixed drift of +8 plus a random term σₙ·z, where z is a standard Normal draw and the spread grows every leg as σₙ = 4·(1.04)ⁿ - later legs are wilder than early ones. Cargo is floored at zero on a calm leg.
You have three moves each leg: sail on, drop anchor, or (for the first 3 legs only) ask to be walked through it. The guided option rolls the real outcome, shows you the exact formula with the actual σₙ and z draw substituted in, and asks you to compute the resulting cargo yourself before it is committed - graded to within 0.5 cargo units against the same rounded numbers you were shown.
The run ends three ways: cargo at or below 15 is a shipwreck, cargo at or above 250 reaches the island, and 30 legs completed means the voyage stops. Dropping anchor is the deliberate ending - but before your haul is tallied you must first predict the theoretical expected cargo E[Xₙ] at your current leg count, with no live readout to lean on. Only then do you see how your prediction, your actual haul, and the recursion compare.
Why quant interviews test this
One-step conditioning into a linear recursion E[Xₙ] = a·E[Xₙ₋₁] + b, then solving for the fixed point b/(1-a), is a staple of quant interviews - it is the same machinery behind gambler's ruin expectations, expected hitting times, and AR(1) processes, where the fixed point is the stationary mean.
Interviewers also like the mix of a proportional loss and a flat gain. It is behind questions on geometric versus arithmetic growth and Kelly sizing, and on why a strategy with positive expected value per step can still be capped or ruinous. Say that the shock is proportional and the edge is flat, so expectation converges to (1-p)μ/(p/2). Then defend dropping anchor as a stopping time that uses no future information. That answers both the computation and the concept.
The Martingale Mutiny guide covers how scoring works, the strategy that wins, a worked example and the mistakes most players make.