第 5 / 23 页
非代码面试题
显示 20 / 453 道匹配题目
答题状态：未尝试未正确已正确
ID题目领域难度题型进度权限
443Series System Replacement Costs via Competing ExponentialsA machine has two critical components in series: component A with lifetime Exp (3) and component B with lifetime Exp (5), independent of each other. When either fails, the entire machine stops, the failed component is replaced (cost \20 for A, \50 for B), and both components restart fresh (the survivor restarts by memorylessness, the replacement is new). Find the expected replacement cost per unit time in the long run.概率中等数值题未尝试免费448Threshold Exceedance for the Minimum of Two ExponentialsLet X \sim Exp (2) and Y \sim Exp (3) be independent. Define M = \min(X, Y) and fix a threshold c = 1. (i) Find P(M > 1). (ii) Given M > 1, find E[M - 1 \mid M > 1] and P(X < Y \mid M > 1) (the probability that X is the minimum, given both survived past 1).概率中等数值题未尝试免费452Sample Mean of Exponential Service TimesA server processes 100 independent requests. Each request takes Exp (1) time (mean 1 second). Let T = 1 100 \sum i=1 100 T i be the average processing time. **(a)** State what the Law of Large Numbers guarantees about T as n . **(b)** Using the CLT, approximate P( T > 1.2). You may use \Phi(2) \approx 0.9772.概率简单数值题未尝试免费453Call Center Overflow via Poisson CLTA call center receives calls according to a Poisson process with rate = 4 calls per minute. The center operates for an 8-hour shift (480 minutes). The center can handle at most 2000 calls per shift before service quality degrades. Using a suitable normal approximation, estimate the probability that the total number of calls in a single shift exceeds 2000. You may use the following: \Phi(1.83) \approx 0.9664.概率中等数值题未尝试免费457Sum of Uniform Random Variables Exceeding a ThresholdLet U 1, U 2, \ldots, U 60 be independent Uniform (0,1) random variables, and define S = \sum i=1 60 U i. Using the CLT, approximate P(S > 35). You may use \Phi(2.24) \approx 0.9875.概率中等数值题未尝试免费458Empirical Frequency Accuracy via the CLTA biased die shows a six with probability p = 1/3. You roll it n = 900 times independently and record p = ( number of sixes )/n. **(a)** State what the Law of Large Numbers guarantees about p as n . **(b)** Using the CLT, approximate P(| p - 1/3| < 0.02). You may use \Phi(1.27) \approx 0.8980.概率中等数值题未尝试免费459Continuity Correction in the Normal ApproximationLet S \sim Bin (200, 0.45). Use the CLT with a continuity correction to approximate P(S = 85). Recall that for a discrete integer-valued random variable, P(S = k) \approx \Phi\! ( k + 0.5 - ) - \Phi\! ( k - 0.5 - ). You may use: \Phi(-0.64) \approx 0.2611, \Phi(-0.78) \approx 0.2177.概率中等数值题未尝试免费462Estimating Pi by Monte Carlo and the Law of Large NumbersTo estimate , you draw n = 10 , 000 points (X i, Y i) independently and uniformly on the unit square [0,1] 2. Define Z i = 1 (X i 2 + Y i 2 \le 1), and let = 4 Z where Z = 1 n \sum i=1 n Z i. **(a)** Explain why E[ ] = and why almost surely. **(b)** Using the CLT, find an approximate 95\% confidence interval for given that the observed Z = 0.7854. You may use \Phi(1.96) \approx 0.975 and \approx 3.1416.概率简单数值题未尝试免费463Inventory Stockout Probability via Poisson CLTA warehouse stocks 240 units of a product each week. Weekly demand follows a Poisson (225) distribution. Using a normal approximation, find the probability that demand exceeds supply in a given week. You may use \Phi(1.00) \approx 0.8413.概率中等数值题未尝试免费465Berry-Esseen Bound for a Sum of Uniform Random VariablesLet U 1, \ldots, U n be i.i.d.\ Uniform (0,1) and S n = \sum i=1 n U i. The Berry-Esseen theorem states \sup x |P\! ( S n - n/2 n \le x ) - \Phi(x) | \le C\, 3 n , where 2 = Var (U i), = E[|U i - 1/2| 3], and C \le 0.4748. **(a)** Compute = E[|U i - 1/2| 3] exactly. **(b)** Evaluate the Berry-Esseen bound for n = 50. **(c)** How large must n be for the bound to guarantee the CLT error is below 0.01?概率困难derivation未尝试免费466Election Poll Margin of Error via the CLTA pollster surveys n = 1 , 600 voters to estimate the proportion p supporting a candidate. Suppose the true proportion is p = 0.5. Using the CLT, find the probability that the sample proportion p falls within 0.02 of the true value, i.e., approximate P(| p - 0.5| < 0.02). You may use \Phi(1.60) \approx 0.9452.概率简单数值题未尝试免费468Aggregate Insurance Claims via the CLTAn insurer has 300 independent policyholders. Each policyholder files a Poisson (3) number of claims per year. Let T = \sum i=1 300 N i be the total number of claims. Using the CLT, approximate P(T > 960). You may use \Phi(2.00) \approx 0.9772.概率中等数值题未尝试免费472Sample Size for Desired Estimation AccuracyA random variable X has mean = 5 and standard deviation = 2. You observe n independent copies X 1, \ldots, X n and compute X n. Using the CLT, find the smallest n such that P(| X n - 5| > 0.3) < 0.05. You may use \Phi(1.96) \approx 0.975.概率简单数值题未尝试免费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490Martingale Construction for Hitting Time on a Non-Symmetric ChainA Markov chain on \ 0, 1, 2, 3\ has transition probabilities: p(1, 0) = \tfrac 1 3 , \quad p(1, 2) = \tfrac 2 3 , \quad p(2, 1) = \tfrac 1 2 , \quad p(2, 3) = \tfrac 1 2 . State 0 is absorbing and state 3 is reflecting: p(3, 2) = 1. Let T = \inf\ n \ge 0 : X n = 0\ . **(a)** Find a function f : \ 0,1,2,3\ R and a constant c > 0 such that M n = f(X n \wedge T ) - (n \wedge T) c is a martingale. Use this to compute E[T \mid X 0 = 2]. **(b)** Find a function g : \ 0,1,2,3\ R such that N n = g(X n \wedge T ) - (n \wedge T) d is a martingale for an appropriate constant d, and use the optional stopping theorem to compute E[T 2 \mid X 0 = 2]. Hence find Var (T \mid X 0 = 2).概率困难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