Oct 26, 2026
Joe Doob’s upcrossing Lemma (Doob 1953) for a martingale (M_t) states the number of upcrossings of the interval [a,b] over [0,t] times (b - a) is less than {E[\max\{M_t - a, 0\}] - \max\{M_0 - a, 0\}}. It is just a lemma because Doob needed it to prove martingale convergence theorems – generalizations to stochastic processes of the fact bounded, monotone sequences converge.
His proof has a simple financial interpretation: the number of upcrossings can be used to find an option premium, its value minus the intrinsic, using historical data. This is mentioned in mathematical finance literature but there does not seem to be any detailed analysis. This short note begins to rectifying that.
Suppose a money-market account R_t is available for hedging and S_t is the price of a stock at time t. To hedge a call option with strike K and expiration T you need to find a self-financing trading strategy that pays \max\{S_T - K,0\} at time T.
For now, lets assume zero interest rates so R_t = 1, S_0 < K, and the stock pays no dividends. A naive approach is to borrow K and buy one share of stock when it hits level K. If the stock remains above K until expiration T then sell the stock for S_T, use that to pay back the K you borrowed, and the remaining S_T - K to cover the option obligation If the stock goes back down to level K then sell it at K to preserve self-financing. Rinse and repeat.
Exercise. If S_0 > K immediately buy one share. Show the P&L is \max\{S_T - K,0\} - \max\{S_0 - K,0\}.
Hint: S_T - S_0 = S_T - K - (S_0 - K).
Note this is true for every stock trajectory and does not involve probability.
When I showed this argument to Peter Carr he laughed and said, “That can’t be right!” It isn’t. He eventually turned this into the paper (Carr and Mahayni 2010). If you try to implement this strategy you will find out that local time is a mathematical fiction.
In reality stocks trade at integer multiples of their minimum tick size. To hedge a call option with strike K and expiration T specify possible levels a < K and b > K, buy one share when S_t > b, then sell it when S_t < a. If limit orders exist they can be used to execute at exact levels with no slippage. What is not clear is how small b - a can be and still result in a practical hedging strategy.
Define trading times \begin{aligned} \tau_0 &= \inf\{t > 0\mid S_t > b\} \\ \tau_1 &= \inf\{t > \tau_1\mid S_t < a\} \\ &\cdots \\ \tau_{2n} &= \inf\{t > \tau_{2n-1}\mid S_t > b\} \\ \tau_{2n+1} &= \inf\{t > \tau_{2n}\mid S_t < a\} \\ \end{aligned} for n > 0. The trading strategy is to buy 1 share at \tau_{2n} and sell it at \tau_{2n+1}.
For simplicity assume a = K. Assuming S_0 < b this results in cash flows -b at \tau_0, K at \tau_1, \ldots, that alternate until expiration T. We pay b - K at each down crossing. This costs (b - a)N_T(a,b) where N_t(a,b) = \sum_j 1(\tau_j \le t), the number of crossings of [a,b] at time t.