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544Hitting Time on the Diamond GraphTake the complete graph K 4 on vertices \ A, B, C, D\ and remove edge A - D, leaving 5 edges (the "diamond" or "kite" graph). The resulting degrees are d(A) = d(D) = 2 and d(B) = d(C) = 3. A simple random walk moves at each step to a uniformly random neighbor. (a) Starting from vertex A, find the expected hitting time h(A D). (b) Starting from vertex D, find h(D A). (c) Compute the commute time C(A, D) and verify it using the effective-resistance formula.概率困难数值题未尝试面试订阅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546Hitting Time on the Path Graph P₅A simple random walk moves on the path graph P 5 with vertices \ 0,1,2,3,4\ and edges connecting consecutive vertices. At the interior vertices (1, 2, 3), the walker moves left or right with equal probability \tfrac 1 2 . At the endpoints (0 and 4), the walker moves to the unique neighbor with probability 1. Starting at vertex 0, what is the expected number of steps to reach vertex 4 for the first time?概率简单数值题未尝试免费547Hitting Time on K₄ Minus One EdgeTake the complete graph K 4 on vertices \ 1,2,3,4\ and remove edge \ 1,4\ . The resulting graph has 5 edges, with d(1)=d(4)=2 and d(2)=d(3)=3. A simple random walk moves at each step to a uniformly random neighbor. Starting from vertex 2, what is the expected number of steps to reach vertex 4?概率简单数值题未尝试免费548Commute Time Between Endpoints of a PathConsider the path graph P n with vertices \ 0, 1, \ldots, n-1\ and n-1 edges (each of unit resistance), where the walk at interior vertices moves left or right with equal probability, and at the endpoints moves to the unique neighbor. (a) Compute the effective resistance R eff (0, n-1) between the two endpoints. (b) Using C(u,v) = 2m R eff (u,v), find the commute time between the two endpoints. (c) Verify for n = 4 by computing h(0 3) and h(3 0) directly.概率中等derivation未尝试免费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2757Reserve Price as a Participation FilterIn a second-price auction with reserve r, a bidder with value v faces a strategic question only when v is near r. What simple cutoff rule determines whether the bidder should participate at all?脑筋急转弯中等derivation未尝试面试订阅2761Winner's-Curse Adjustment From a Symmetric Signal BandA common-value asset has expected resale value equal to the average of all bidders' signals. Your private signal is s, and you know that winning means your signal was the highest among a symmetric signal band around the truth. Why should your bid be below s even if E[value|signal=s]=s before conditioning on winning?脑筋急转弯简单derivation未尝试面试订阅2891Two-State Regime SwitchingA market regime is either Calm or Volatile. From Calm, the chain moves to Volatile with probability ; from Volatile, it moves to Calm with probability . Find the stationary distribution.概率简单derivation未尝试面试订阅2892A Three-State Birth-Death Regime ChainConsider states Bull, Neutral, and Bear with transition matrix \[ P= \begin pmatrix 0.8 & 0.2 & 0\\ 0.3 & 0.4 & 0.3\\ 0 & 0.2 & 0.8 \end pmatrix . \] Find the stationary distribution.概率中等derivation未尝试面试订阅2901A Tiny Queue with Capacity TwoA queue length process lives on \ 0,1,2\ . From state 0 it stays at 0 with probability 1-a and moves to 1 with probability a. From state 1 it moves to 2 with probability a, to 0 with probability b, and stays with probability 1-a-b. From state 2 it moves to 1 with probability b and stays at 2 with probability 1-b. Find the stationary distribution.概率中等derivation未尝试面试订阅2909A Four-State Ring with Resets to ZeroConsider states \ 0,1,2,3\ . From state i, the chain moves to i+1 \pmod 4 with probability 1/2 and resets to 0 with probability 1/2. Find the stationary distribution.概率困难derivation未尝试面试订阅2911Custom Three-State Reversible ChainConsider a chain on states \ 1,2,3\ with transition matrix \[ P=\begin pmatrix 0.6 & 0.4 & 0\\ 0.2 & 0.5 & 0.3\\ 0 & 0.6 & 0.4 \end pmatrix . \] Find the stationary distribution.概率中等derivation未尝试面试订阅3081Two Missing Days Before a Print ArrivesA local-level model satisfies x t=x t-1 +w t with w t\sim N(0,1), and observations have noise variance 2. After the last filtered state N(3,4), there are 2 consecutive missing observations. Then you observe a new value y=6. Compute the variance just before the new observation and the updated mean/variance after processing it.统计中等derivation未尝试面试订阅3082One Missing Observation Then UpdateA local-level model satisfies x t=x t-1 +w t with w t\sim N(0,3), and observations have noise variance 5. After the last filtered state N(-1,9), there are 1 consecutive missing observations. Then you observe a new value y=2. Compute the variance just before the new observation and the updated mean/variance after processing it.统计中等derivation未尝试面试订阅3096Tomorrow Variance After a Large ShockIn a GARCH(1,1) model with \omega= 1 10 , = 1 5 , and = 7 10 , suppose the current squared return is r t 2=4 and the current conditional variance is h t=2. Compute h t+1 .统计简单derivation未尝试面试订阅3099Volatility Update from a Moderate ReturnIn a GARCH(1,1) model with \omega=1, = 3 20 , and = 3 5 , suppose the current squared return is r t 2=4 and the current conditional variance is h t=5. Compute h t+1 .统计简单derivation未尝试面试订阅3111Does This GARCH Have a Finite Long-Run Variance?For a GARCH(1,1) model with \omega= 1 5 , = 1 4 , and = 3 4 , decide whether the model has a finite unconditional variance. If it does, compute it.统计中等derivation未尝试面试订阅3221A Confidence Interval Cannot Price a One-Off Launch DecisionA PM sees a frequentist 95% confidence interval for next-month strategy edge of [-0.1, 0.4] and asks, "So what is the probability the true edge is positive for this launch decision?" Why can't the interval answer that question by itself, and what Bayesian quantity would answer it?统计中等essay未尝试面试订阅3237Why a Posterior Mean Can Move While the MLE Does NotA desk observes only 4 new defaults for a rare event. Using a strong historical Beta prior, the Bayesian posterior mean default rate is much lower than the sample proportion, while the frequentist MLE equals the sample proportion exactly. Explain why these two answers can legitimately differ and what each one is conditioning on.统计简单essay未尝试面试订阅