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473Probability of Negative Portfolio Return via CLTA portfolio consists of n = 50 stocks with equal weight 1/n. The annual returns R 1, \ldots, R 50 are independent, each with mean = 0.08 (i.e., 8\%) and standard deviation = 0.20. The portfolio return is R = 1 50 \sum i=1 50 R i. **(a)** State what the LLN implies about R as n . **(b)** Using the CLT, approximate P( R < 0). You may use \Phi(2.83) \approx 0.9977.概率中等数值题未尝试免费475CLT with Estimated Variance via Slutsky's TheoremLet X 1, \ldots, X n be i.i.d.\ with mean and finite variance 2 > 0. Define the sample variance S n 2 = 1 n-1 \sum i=1 n (X i - X n) 2 and the studentized statistic T n = n \,( X n - ) S n . **(a)** Using the LLN and Slutsky's theorem, show that T n \xrightarrow d N(0,1). **(b)** In a study with n = 100 observations, you find X 100 = 12.5 and S 100 = 3.0. Assuming the true mean is \mu 0 = 12, approximate P( X 100 > 12.5) using T n. You may use \Phi(1.67) \approx 0.9525.概率困难derivation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528Hitting Time on the 3-Dimensional HypercubeA random walk moves on the 3-dimensional hypercube graph Q 3: the 8 vertices are binary strings of length 3, and two vertices are adjacent if they differ in exactly one coordinate. At each step, the walker picks one of the 3 coordinates uniformly at random and flips it. Starting at vertex 000, what is the expected number of steps to reach vertex 111 for the first time?概率中等数值题未尝试免费530Effective Resistance and Commute Time on the Hypercube Q₃Consider the 3-dimensional hypercube graph Q 3 (vertices are binary strings of length 3; edges connect strings differing in exactly one bit). Each edge has unit resistance. (a) Using the symmetry of Q 3, compute the effective resistance R eff (000, 111) between the two antipodal vertices. (b) The commute time C(u, v) of a random walk between vertices u and v on a graph satisfies C(u, v) = 2m R eff (u, v), where m is the number of edges. Use this to find the commute time between 000 and 111.概率困难derivation未尝试面试订阅532Hitting Time on the Petersen GraphThe Petersen graph has 10 vertices and 15 edges; it is 3-regular, vertex-transitive, and has diameter 2 (every pair of non-adjacent vertices has exactly one common neighbor, and the graph has girth 5). A random walk at each step moves to one of the 3 neighbors uniformly at random. Starting from a vertex u, what is the expected number of steps to reach a specified non-adjacent vertex v for the first time?概率中等数值题未尝试免费533Spectral Gap and Mixing Time of the Lazy Walk on a CycleConsider the lazy random walk on the cycle graph C n: at each step, the walker stays put with probability \tfrac 1 2 , and moves to each of the two neighbors with probability \tfrac 1 4 . The transition matrix has eigenvalues \lambda k = \tfrac 1 2 (1 + \cos(2 k / n)) for k = 0, 1, \ldots, n-1. (a) Find the spectral gap = 1 - \lambda 1 in terms of n. (b) Using the relation t mix \asymp 1 (up to logarithmic factors), determine the order of the mixing time for the lazy walk on C n as n .概率中等derivation未尝试免费539Cover Time of the Complete Graph K₄A simple random walk moves on the complete graph K 4. At each step, the walker moves to one of the 3 neighbors uniformly at random. (a) Compute the maximum hitting time t hit = \max u,v h(u v). (b) Using Matthews' theorem, which gives t hit H n-1 \ge E[\tau cov ] \ge t hit H n-1 n-1 n , where n = 4 and H k = \sum i=1 k 1/i, bound the expected cover time. (c) Compute E[\tau cov ] exactly by decomposing into phases: after visiting k distinct vertices, find the expected time to discover the (k+1)-th.概率困难derivation未尝试面试订阅545Mixing Time of the Lazy Random Walk on KₙConsider the lazy random walk on the complete graph K n: at each step, the walker stays put with probability \tfrac 1 2 and moves to a uniformly random neighbor with probability \tfrac 1 2 . (a) Show that the transition matrix has two distinct eigenvalues: \lambda 0 = 1 (with multiplicity 1) and \lambda 1 = \tfrac 1 2 \bigl(1 - 1 n-1 \bigr) = n-2 2(n-1) (with multiplicity n-1). (b) Find the spectral gap = 1 - \lambda 1 and determine the mixing time t mix up to leading order in n.概率困难derivation未尝试面试订阅550Expected Cover Time of the Cycle C₆A simple random walk moves on the cycle graph C 6 (vertices 0, 1, \ldots, 5). At each step, the walker moves clockwise or counterclockwise with equal probability. Starting at vertex 0, what is the expected number of steps to visit all 6 vertices (the expected cover time)?概率困难derivation未尝试面试订阅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1428Exponential Quadratic Remainder 3Compute lim (x->0) [exp(3x + -2x 2) - 1 - 3x] / x 2.数学中等数值题未尝试免费1430Fifth-Order Trigonometric ResidueCompute lim x->0 [sin(3x) - 3x + (9/2)x 3] / x 5.数学困难数值题未尝试免费1433Fractional Linear RemainderCompute lim x->0 [((1+x)/(1-x)) - 1 - 2x] / x 2.数学中等数值题未尝试免费