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449Memoryless Message Relay ChainA message must traverse a chain of relay nodes to reach its destination. Each node independently takes Geom (1/3) attempts to successfully forward the message to the next node. However, on each attempt, there is an independent probability 1/5 that the node permanently fails, destroying the message. If the node fails, the message is lost. If the chain has 2 nodes, find: (i) The probability the message reaches the destination (traverses both nodes). (ii) The expected total number of attempts across both nodes, given the message reaches the destination.概率中等数值题未尝试免费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