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5967Doubling Strategy and Optional Stopping FailureA gambler starts with 0 net and makes fair 1-doubling bets (bet 1, then 2, then 4, ...) on a sequence of fair coin flips, stopping the first time they win a single flip (guaranteeing +1 net). Let T be that stopping time. Compute E[net wealth at T], and explain whether E[net at T] equals net at time 0 as naive optional stopping would suggest.概率中等数值题未尝试免费5968Waiting Time for the Pattern HTHHA fair coin is flipped repeatedly. Using a martingale (gambling-team) argument, find the expected number of flips until the pattern H, T, H, H first appears.概率中等数值题未尝试免费5970Ballot Problem via MartingaleIn an election candidate A receives 7 votes and candidate B receives 3 votes; the 10 votes are counted in a uniformly random order. Using a martingale / optional stopping argument, find the probability that A is strictly ahead of B throughout the entire count.概率困难数值题未尝试面试订阅5971Expected Duration of Biased Gambler's RuinA walk starts at 2, moves +1 with probability 2/3 and -1 with probability 1/3, and stops on hitting 0 or 5. First find the probability it exits at 5, then use the linear-drift martingale to find the expected duration E[T].概率困难数值题未尝试面试订阅5973Ruin Duration With Holding TiesA score starts at 3 and each round goes +1 with probability 0.3, -1 with probability 0.3, and stays the same (a tie) with probability 0.4. The game ends when the score first reaches 0 or 8. Find the expected number of rounds until the game ends.概率中等数值题未尝试免费5976Expected Maximum Before RuinA fair simple random walk starts at 2 and is absorbed when it first hits 0 or 5. Let H be the highest level the walk reaches over its whole path (the running maximum at absorption). Find E[H].概率困难数值题未尝试面试订阅5977Optional Stopping on a Product MartingaleLet X 1, X 2, ... be i.i.d. fair +-1 steps and define the product P n = prod i=1 n (1 + (1/2) X i), with P 0 = 1. Let N be any almost-surely finite stopping time. Treating P n as a martingale, what is E[P N]?概率中等数值题未尝试免费5978First Passage Probability for a Lazy Biased WalkA lazy walk starts at 1. Each step it moves +1 with probability 0.3, -1 with probability 0.2, and stays put with probability 0.5. It stops on first reaching +4 or -4. Find the probability it exits at +4.概率中等数值题未尝试免费5979Three-Level Exit With a Non-Absorbing BarrierA fair simple random walk starts at 0. Three levels are marked: -3, +2, and +6. Only the two extreme levels, -3 and +6, are absorbing; the walk passes freely through +2. Find the probability the walk is absorbed at -3, and the expected absorption value E[S T].概率困难数值题未尝试面试订阅5980Random Walk With Rest PeriodsA process evolves in rounds. Each round, independently, with probability 1/2 the walker rests (position unchanged) and with probability 1/2 it takes a step that is +1 or -1 with equal odds. The walker stops at the first round in which it completes its 8th actual (non-rest) step. Let S be the position at that stopping time. Find E[S 2].概率中等数值题未尝试免费5981Total Stake Until the First SixYou roll a fair die repeatedly until the first time a 6 appears; let N be the number of rolls (including the winning roll). On each roll you independently collect a payoff X i with E[X i]=1.5, where the X i are i.i.d. and independent of the roll values. Compute E\! [\sum i=1 N X i ].概率简单derivation未尝试免费5982Variance of Fills Over a Binomial Number of QuotesOut of n=10 resting quotes, each fills independently with probability 0.3, so the number of fills N is Binomial(10,0.3). Each fill produces an i.i.d. PnL X i with E[X i]=2 and Var (X i)=9, independent of which quotes fill. For the stopped sum S N=\sum i=1 N X i, compute Var (S N).概率中等derivation未尝试免费5983Number of Trades to Cross a Profit TargetA strategy books i.i.d. positive profits X 1,X 2,\dots with E[X i]=2.5. Let N be the first time the running total S n=\sum i=1 n X i strictly exceeds 10; that is, N=\min\ n:S n>10\ . Assuming E[N]< , the expected overshoot is known to satisfy E[S N]=14. Use a Wald-style identity to compute E[N].概率中等derivation未尝试免费5984Expected Inspection Cost Until the First DefectA quality line inspects items one at a time; each item is defective independently with probability 0.05. Inspection stops at the first defective item. Each inspection (defective or not) costs an i.i.d. amount C i with E[C i]=\8, independent of the defect outcomes. Let N be the number of items inspected. Find the expected total inspection cost E\! [\sum i=1 N C i ].概率简单数值题未尝试免费5985Expected Total Slippage With Negative DriftA market-making desk incurs i.i.d. per-trade adverse-selection costs X 1,X 2,\dots with E[X i]=-0.4 (a net loss per trade). The number of trades in a session, N, is independent of the costs and is Poisson with mean 15. Compute the expected cumulative cost E\! [\sum i=1 N X i ].概率简单derivation未尝试免费5986Expected Winnings Over a Random Number of BetsA gambler places bets until a random stopping rule halts play; the number of bets N is a stopping time for the i.i.d. bet outcomes with E[N]=8. Each bet has an i.i.d. net result X i with E[X i]=-0.05 (a 5\% house edge per unit staked, with unit stakes), and the decision to stop after bet n depends only on outcomes up to bet n. Compute the gambler's expected total winnings E\! [\sum i=1 N X i ], and state whether any stopping rule with E[N]=8 can make this positive.概率中等derivation未尝试免费5987When the Stopping Rule Looks at the Last DrawDraw i.i.d. values X 1,X 2,\dots uniform on \ 1,2,3\ (so E[X i]=2). Define N as follows: keep drawing and stop the first time you draw a 3; let N be the number of draws. Let S N=\sum i=1 N X i. A candidate computes E[N]E[X 1]=3 2=6 and claims E[S N]=6. Compute the correct value of E[S N] and explain in one sentence why E[N]E[X 1] is the wrong formula here.概率困难essay未尝试面试订阅5997Expected Count Under an Unknown RegimeOn any given day the market is in a 'calm' regime with probability 0.5, where news events arrive as a Poisson process with rate 6 per hour, or a 'busy' regime with probability 0.5, with rate 14 per hour. Before observing the day you do not know the regime. What is the expected number of news events in a 1-hour window?概率中等数值题未尝试免费6009Expected Large Trades in Two HoursTrades print as a Poisson process at rate =30 per hour. Independently, each trade is a block (large) trade with probability 0.15. What is the expected number of block trades over the next 2 hours?概率简单derivation未尝试面试订阅6010No Orders Routed to Venue COrders arrive as a Poisson process at rate =20 per minute and are independently routed to venue A, B, or C with probabilities 0.5, 0.25, and 0.25. What is the probability that venue C receives no orders during the next 6 seconds?概率中等derivation未尝试面试订阅