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497Hitting Time with Harmonically Increasing Forward ProbabilityA Markov chain on \ 0, 1, 2, 3\ has transitions: from state i (0 \le i \le 2), move to i+1 with probability i+1 i+2 and stay at i with probability 1 i+2 . State 3 is absorbing. Compute E[T 3 \mid X 0 = 0].概率简单derivation未尝试免费498Hitting Time with Interior Reflecting BarrierA Markov chain on \ 0, 1, 2, 3, 4\ has the following transitions: - State 0 is absorbing. - From state 1: p(1, 0) = \tfrac 1 2 , p(1, 2) = \tfrac 1 2 . - From state 2: p(2, 1) = \tfrac 1 3 , p(2, 3) = \tfrac 2 3 . - From state 3: p(3, 2) = \tfrac 1 2 , p(3, 4) = \tfrac 1 2 . - From state 4: p(4, 3) = 1 (reflecting barrier). Compute E[T 0 \mid X 0 = 3].概率中等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500Harmonic Function Method for Splitting Probability and Hitting TimeA Markov chain on \ 0, 1, 2, 3, 4, 5\ has absorbing states 0 and 5. For interior states: p(1,0) = \tfrac 1 4 , \quad p(1,2) = \tfrac 3 4 , p(2,1) = \tfrac 1 3 , \quad p(2,3) = \tfrac 2 3 , p(3,2) = \tfrac 1 2 , \quad p(3,4) = \tfrac 1 2 , p(4,3) = \tfrac 2 3 , \quad p(4,5) = \tfrac 1 3 . Let T = \inf\ n \ge 0 : X n \in \ 0, 5\ \ . **(a)** A function f on \ 0,\ldots,5\ is *harmonic* on \ 1,2,3,4\ if f(i) = \sum j p(i,j) f(j) for i \in \ 1,2,3,4\ . Find the harmonic function with f(0) = 0, f(5) = 1, and use it to compute P(X T = 5 \mid X 0 = 2). **(b)** Compute E[T \mid X 0 = 2].概率困难derivation未尝试免费