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450Head Start in an Exponential RaceLet X \sim Exp ( ) and Y \sim Exp ( ) be independent. Player A finishes at time X and player B finishes at time Y + c where c > 0 is a head start for player A (player B starts c time units later). (a) Derive P(X < Y + c) — the probability that A finishes before B. (b) Show that as c 0, the result recovers the standard competing-exponentials formula. (c) Evaluate for = 3, = 2, c = 1 and interpret how the head start affects A's winning probability.概率困难multi part未尝试面试订阅477Mean Return Time and the Stationary DistributionA Markov chain on \ 1, 2, 3\ has transition matrix P = \begin pmatrix 0 & \tfrac 1 2 & \tfrac 1 2 \\ \tfrac 1 4 & \tfrac 1 2 & \tfrac 1 4 \\ \tfrac 1 3 & \tfrac 1 3 & \tfrac 1 3 \end pmatrix . **(a)** Find the stationary distribution . **(b)** Using the relationship between the stationary distribution and mean return times, compute the expected number of steps to return to state 1 starting from state 1.概率简单数值题未尝试免费479State-Dependent Drift and First-Passage to a BoundaryA particle moves on \ 0, 1, 2, 3, 4\ . From state i (0 < i < 4), it jumps to i+1 with probability p i = i/4 and to i-1 with probability q i = 1 - i/4. States 0 and 4 are absorbing. Starting from state 2: **(a)** Find the probability of being absorbed at state 4. **(b)** Find the expected number of steps until absorption (hitting either boundary).概率中等derivation未尝试免费480Hitting-Time Variance via a Compensated MartingaleConsider a Markov chain on \ 0, 1, 2, 3\ with transitions: from state i (0 < i < 3), the chain moves to i+1 with probability p = \tfrac 2 3 and to i-1 with probability q = \tfrac 1 3 . State 0 and state 3 are absorbing. Let T = \inf\ n \ge 0 : X n \in \ 0, 3\ \ . **(a)** Define M n = X n \wedge T - (p - q)(n \wedge T). Verify that M n is a martingale and use the Optional Stopping Theorem to find E[T \mid X 0 = 2]. **(b)** Find a second martingale of the form N n = (X n \wedge T - (p-q)(n \wedge T)) 2 - 4pq(n \wedge T) and use it to compute Var (T \mid X 0 = 2). **(c)** Verify your answer for E[T] by first-step analysis.概率困难derivation未尝试面试订阅482First Passage Through a Randomizing StateA Markov chain on \ 0, 1, 2\ has the transition matrix P = \begin pmatrix 1 & 0 & 0 \\ \tfrac 1 4 & 0 & \tfrac 3 4 \\ \tfrac 1 2 & \tfrac 1 2 & 0 \end pmatrix . State 0 is absorbing. Starting from state 1, let T = \inf\ n \ge 1 : X n = 0\ . **(a)** Compute E[T \mid X 0 = 1]. **(b)** Compute E[T \mid X 0 = 2].概率中等数值题未尝试免费484Hitting Time on a Product of Two-State ChainsLet (X n, Y n) be a Markov chain on \ 0,1\ 2 where X n and Y n evolve independently. At each step, X n flips (i.e., X n+1 = 1 - X n) with probability \tfrac 1 3 and stays with probability \tfrac 2 3 . Similarly, Y n flips with probability \tfrac 1 2 and stays with probability \tfrac 1 2 . Starting from (X 0, Y 0) = (1, 1), compute the expected number of steps to first reach the state (0, 0).概率中等derivation未尝试免费488First Passage with Parity-Dependent TransitionsConsider a Markov chain on \ 0, 1, 2, 3\ where states 0 and 3 are absorbing. The transition probabilities for transient states depend on the parity of the state: - From any **odd** state i: p(i, i-1) = \tfrac 3 4 , p(i, i+1) = \tfrac 1 4 . - From any **even** transient state i (i.e., i = 2): p(i, i-1) = \tfrac 1 4 , p(i, i+1) = \tfrac 3 4 . Compute E[T \mid X 0 = 1] where T = \inf\ n \ge 0 : X n \in \ 0, 3\ \ .概率中等derivation未尝试免费492Hitting Time on a Six-State Chain with Paired CoordinatesA Markov chain lives on S = \ 0, 1, 2\ \ 0, 1\ . The state (0, y) for any y is absorbing. From state (i, j) with i \ge 1, the transitions are: - Move to (i-1, j) with probability \tfrac 1 3 . - Move to (i, 1-j) with probability \tfrac 1 3 (flip the second coordinate). - Move to (i+1, j) with probability \tfrac 1 3 , provided i+1 \le 2; if i = 2, this probability is redistributed equally to the other two moves. Let A = \ (0, 0), (0, 1)\ and T = \inf\ n \ge 0 : X n \in A\ . Compute E[T \mid X 0 = (1, 1)].概率中等derivation未尝试免费494Expected Time to Visit All States in an Ergodic ChainA Markov chain on \ 1, 2, 3, 4\ has transition matrix P = \begin pmatrix 0 & 1 & 0 & 0 \\ \tfrac 1 3 & 0 & \tfrac 2 3 & 0 \\ 0 & \tfrac 1 2 & 0 & \tfrac 1 2 \\ 0 & 0 & \tfrac 1 2 & \tfrac 1 2 \end pmatrix . Starting from state 1, let T cover = \inf\ n \ge 0 : \ 1,2,3,4\ \subseteq \ X 0, X 1, \ldots, X n\ \ be the first time all four states have been visited. Compute E[T cover \mid X 0 = 1]. *Hint:* Decompose the cover time by tracking which states remain unvisited. Since the chain starts at state 1 and must go to 2, then eventually reach 3 and 4, compute the expected time to reach each new state sequentially.概率困难derivation未尝试免费495Finite Hitting Time Between Communicating ComponentsConsider a Markov chain on \ 1, 2, 3, 4, 5\ with transition matrix P = \begin pmatrix \tfrac 1 2 & \tfrac 1 2 & 0 & 0 & 0 \\ \tfrac 1 3 & 0 & \tfrac 2 3 & 0 & 0 \\ 0 & \tfrac 1 4 & 0 & \tfrac 1 2 & \tfrac 1 4 \\ 0 & 0 & 0 & \tfrac 1 2 & \tfrac 1 2 \\ 0 & 0 & 0 & \tfrac 1 3 & \tfrac 2 3 \end pmatrix . Let B = \ 4, 5\ and T B = \inf\ n \ge 0 : X n \in B\ . **(a)** Show that E[T B \mid X 0 = i] < for all i \in \ 1, 2, 3\ . **(b)** Compute E[T B \mid X 0 = 1].概率困难derivation未尝试免费499First Passage with Periodically Varying DriftA Markov chain on \ 0, 1, 2, 3, 4, 5, 6\ has absorbing state 0 and reflecting state 6 (i.e., p(6, 5) = 1). For transient states 1 \le i \le 5, the transition probabilities depend on i \bmod 3: - If i \equiv 0 \pmod 3 (i.e., i = 3): p(i, i-1) = \tfrac 2 3 , p(i, i+1) = \tfrac 1 3 . - If i \equiv 1 \pmod 3 (i.e., i \in \ 1, 4\ ): p(i, i-1) = \tfrac 1 2 , p(i, i+1) = \tfrac 1 2 . - If i \equiv 2 \pmod 3 (i.e., i \in \ 2, 5\ ): p(i, i-1) = \tfrac 1 3 , p(i, i+1) = \tfrac 2 3 . Compute E[T 0 \mid X 0 = 3].概率中等derivation未尝试免费590Weighted Offer Stop Rule 5You may inspect up to 3 independent offers. Each offer takes values 0 with probability 1/4, 4 with probability 1/4, 7 with probability 1/4, 12 with probability 1/4. Rejecting an offer and continuing costs 1 point(s), and if you reach the last draw you must accept it. What first-round acceptance threshold is optimal, and what is the resulting expected net payoff?概率困难derivation未尝试免费606Target-Hitting Stake Choice 6You start with wealth 5. In each of at most 3 rounds, you may bet any integer stake between 0 and your current wealth on an even-money coin that wins with probability 3/5. If you win, your wealth increases by the stake; if you lose, it decreases by the stake. What first-round stake maximizes the probability of finishing with wealth at least 9 after 3 rounds, and what is that maximal probability?概率简单数值题未尝试免费626Weighted Centered Sum 1Let X 1, X 2, ... be iid Bernoulli(2/5) random variables, and let F n = sigma(X 1,...,X n). Define M n = sum (k=1) n k*(X k-2/5). Is (M n) a martingale with respect to (F n)?概率简单derivation未尝试免费628Weighted Centered Sum 3Let X 1, X 2, ... be iid Bernoulli(3/7) random variables, and let F n = sigma(X 1,...,X n). Define M n = sum (k=1) n k 2*(X k-3/7). Is (M n) a martingale with respect to (F n)?概率中等derivation未尝试免费630Weighted Centered Sum 5Let X 1, X 2, ... be iid Bernoulli(3/5) random variables, and let F n = sigma(X 1,...,X n). Define M n = sum (k=1) n 3k*(X k-3/5). Is (M n) a martingale with respect to (F n)?概率困难derivation未尝试免费631Terminal-Variable Projection 1Let X 1, X 2, X 3, X 4 be iid symmetric ±1 variables with natural filtration F n. Define Y = 1 X 1+X 2+X 3 >= 2 and M n = E[Y | F n]. Is (M n) a martingale?概率简单derivation未尝试免费632Terminal-Variable Projection 2Let X 1, X 2, X 3, X 4 be iid symmetric ±1 variables with natural filtration F n. Define Y = 1 X 1+X 2+X 3+X 4 = 0 and M n = E[Y | F n]. Is (M n) a martingale?概率中等derivation未尝试免费633Terminal-Variable Projection 3Let X 1, X 2, X 3, X 4 be iid symmetric ±1 variables with natural filtration F n. Define Y = 1 max(X 1,X 2,X 3) = 1 and M n = E[Y | F n]. Is (M n) a martingale?概率中等derivation未尝试免费634Terminal-Variable Projection 4Let X 1, X 2, X 3, X 4 be iid symmetric ±1 variables with natural filtration F n. Define Y = X 1+X 2+X 3+X 4 and M n = E[Y | F n]. Is (M n) a martingale?概率中等derivation未尝试免费