Poisonal
Martingales And Stopping Times Checkup
EXPLAINER
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Martingales And Stopping Times Checkup

Test your instinct on fair bets and exit thresholds.
In 1934, Paul Lévy introduced the martingale to describe a game where tomorrow’s fortune equals today’s balance, regardless of past history. A decade later, Joseph Leo Doob proved that no clever exit rule can turn a fair game profitable, provided you play under finite constraints. Most people accept this algebraically while secretly believing their personal stop-loss rule outsmarts Doob's optional stopping theorem. This checkup tests where that intuition breaks.
Multiple-Choice Quiz
01/03
Question 01

You flip a fair coin starting with $0, winning $1 on heads and losing $1 on tails ($$). You decide to stop playing the moment your net profit reaches +$10. Why does the Optional Stopping Theorem not guarantee that your expected terminal wealth is $10?

A
The coin tosses are not independent and identically distributed.
B
The stopping time is not almost-surely finite.
C
The stopping time is not bounded, and the process is not uniformly integrable across the infinite state space.
D
A symmetric random walk has non-zero drift after the tenth step.

You flip a fair coin starting with $0, winning $1 on heads and losing $1 on tails ($$). You decide to stop playing the moment your net profit reaches +$10. Why does the Optional Stopping Theorem not guarantee that your expected terminal wealth is $10?

Your Answer
Correct Answer
The stopping time is not bounded, and the process is not uniformly integrable across the infinite state space.
✗Incorrect
Explanation

While the symmetric random walk will hit +$10 with probability 1, the expected stopping time is infinite. Because the walk can drift arbitrarily far into negative territory before hitting +$10, uniform integrability fails, breaking the theorem's core hypothesis.

Score0/ 3
Question 02

Which of the following defines a valid stopping time with respect to a filtration ?

A
T is the first time the stock price hits its maximum over the next five trading days.
B
T is the first step n where cumulative wealth drops below -$50.
C
T is the step immediately preceding the largest drawdown of the year.
D
T is the round where the game achieves its eventual global peak.

Which of the following defines a valid stopping time with respect to a filtration ?

Your Answer
Correct Answer
T is the first step n where cumulative wealth drops below -$50.
✗Incorrect
Explanation

A stopping time requires that the event {T = n} depends strictly on information known up to time n (measurable with respect to ). Any rule requiring future knowledge—such as global peaks or upcoming drawdowns—is invalid.

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Question 03

A bettor uses the classic martingale doubling system on red at a roulette wheel with no zero (fair game). Starting with a $1 bet, they double after every loss and quit immediately after their first win. Why does this system yield an expected net gain of $0 under realistic conditions?

A
The probability of winning on red decays geometrically with consecutive losses.
B
Any finite bankroll or table limit bounds the stopping time, concentrating probability into a catastrophic tail loss.
C
The house edge increases with each doubled bet.
D
Doubling bets alters the underlying expectation of individual spins.

A bettor uses the classic martingale doubling system on red at a roulette wheel with no zero (fair game). Starting with a $1 bet, they double after every loss and quit immediately after their first win. Why does this system yield an expected net gain of $0 under realistic conditions?

Your Answer
Correct Answer
Any finite bankroll or table limit bounds the stopping time, concentrating probability into a catastrophic tail loss.
✗Incorrect
Explanation

With a bounded bankroll , you cannot absorb an arbitrarily long losing run. The rare event of exhausting carries a loss that exactly cancels the frequent $1 wins, preserving $\mathbb{E} = 0$.

Score0/ 3
The fatal error in applying stopping rules is confusing almost sure termination with finite expectation. In a one-dimensional symmetric random walk, the probability of eventually reaching a target of +$1 is exactly 1. However, the expected time $\mathbb{E}T$ to reach it is infinite. If you lack infinite time and infinite credit, the downside tail quietly absorbs every cent of apparent edge.
When you establish two finite boundaries—such as quitting at -$100 or +$10—the stopping time becomes bounded in expectation. Here, Wald's identity applies cleanly: the probability of hitting +$10 before -$100 is exactly 100/110, giving an expected terminal value of $(100/110)(10) + (10/110)(-100) = 0$. The threshold shifts your win rate, but it leaves the net expectation untouched.
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