Doob's Maximal Inequality

Markov's inequality bounds how large a single random variable is likely to be, but a strategy's cumulative P&L isn't just one number — it's a whole path over time, and what actually matters for risk is the largest value that path ever reaches along the way, not just where it ends up. Doob's maximal inequality extends the Markov-style bound to cover this entire running maximum at once, for any process that is a martingale (fair-game, no built-in drift).

Precisely: if $M_t$ is a non-negative martingale (or the absolute value of any martingale) over times $0$ to $T$, then for any threshold $a > 0$,

$$\mathbb{P}\left( \max_{0 \le t \le T} M_t \ge a \right) \le \frac{\mathbb{E}[M_T]}{a} .$$

In words: the chance the process ever exceeds level $a$ at any point during the whole window is bounded using only the expected value at the final time — you don't need to know the process's behavior at every intermediate time separately, one inequality covers the entire path's peak.

Suppose a market-neutral strategy's cumulative absolute P&L swing behaves like a martingale with \mathbb{E}[M_T] = \10{,}000 over a quarter. Doob's inequality says the probability that swing ever exceeds \100,000 at any point during the quarter is at most $10{,}000 / 100{,}000 = 10\%$ — a single bound on the worst point of the whole path, obtained from only the average final value.