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Fair games and when you may stop

Mathematics · Probability · ages 23-24

Name ______________________ Date ____________

M_n is F_n-measurable and E[|M_n|] is finite for every n. If E[M_(n+1) | F_n] = M_n almost surely for every n, then M_n is a martingale.
Circle one: True False
B_n starts at 0 with independent increments +1 of probability 3/4 and -1 of probability 1/4. Which adjustment makes it a martingale for its observed past?
- Subtract n from B_n at time n.
- Add n/2 to B_n at time n.
- Subtract n/2 from B_n at time n.
- Subtract 3n/2 from B_n at time n.
Z equals 1 or -1 with equal probability. Let X_n = (-1)^n Z, with the past revealing X_0. Since E[X_n] = 0 at every time, X_n is a martingale.
Circle one: True False
A simple symmetric walk starts at 0 and takes three independent steps, each +1 or -1 with equal probability. What is its expected position?
Answer: ______________
A fair unit walk starts at 4 and stops on first reaching 0 or 9. Because its stopped position lies in [0,9], its stopping time must be at most 9.
Circle one: True False
A fair unit walk starts at 0. T is its first hit of 1, which occurs almost surely. What is the expected position when stopped at T?
- E[S_T]=1; the mean has changed.
- E[S_T]=0; the initial mean is preserved.
- E[S_T] is undefined because T is infinite.
- E[S_T]=-1 because the first hit is negative.
A gambler starts with 3 coins and gains or loses 1 on each independent fair flip. Play stops on first reaching 0 or 10. What is the probability of reaching 10? Give a decimal.
Answer: ______________
A fair unit walk starts at 0, and S_n is its position at time n. Which rule is not a stopping time for its observed past?
- At time 2, set T=2 if S_2>0; otherwise set T=4.
- At time 2, set T=2 if S_2<0; otherwise set T=5.
- At time 3, set T=2 if S_3>0; otherwise set T=3.
With unlimited funds, a player doubles a fair bet after each loss and stops at the first win. Capped profits have mean 0; terminal profit is 1 almost surely. Why does the bounded-stopped-values theorem not justify passing to this limit?
- The capped profits have no common absolute bound.
- The capped profits have strictly positive means.
- The first-win rule uses unobserved future outcomes.
A fair unit walk starts at 0. Stop at time 1 if the first step is +1; otherwise stop at time 3. Conditional on the first step being -1, the expected stopped position is 0.
Circle one: True False

Answer key

For grown-ups. Fold this page away before handing over the rest.

Fair games and when you may stop W1-mt_198uXX3jRJ-s1

True · The statement includes adaptedness, integrability and the conditional-mean identity.
Subtract n/2 from B_n at time n. · The conditional mean increment is 1/2. Subtracting n/2 cancels that drift at each step.
False · The past determines the next value as -X_n. Its conditional mean is -X_n rather than X_n.
0 · Each increment has mean 0. Adding three increments leaves the initial mean at 0.
False · A position bound is not a time bound. Nine steps alternating +1 and -1 can keep the walk at 4 or 5 without exiting.
E[S_T]=1; the mean has changed. · The terminal value is 1 almost surely, so its expectation is 1. This stopping time does not preserve the initial mean 0.
0.3 · The terminal mean is 10p, where p is the upper-exit probability. Optional stopping gives 10p=3, so p=0.3.
At time 3, set T=2 if S_3>0; otherwise set T=3. · The retrospective rule uses S_3 to decide whether stopping occurred at time 2. When S_2=0, that decision depends on the still-unknown third step. The other rules decide at time 2 using S_2.
The capped profits have no common absolute bound. · After r losses, profit is -(2^r-1). No fixed absolute bound controls the capped stopped profits, so the theorem's bounded-convergence argument is unavailable.
False · On that branch the position is -1 at time 1. The next two independent fair increments have mean 0, giving conditional stopped mean -1.