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422Expected Maximum of Two UniformsLet X 1, X 2 be independent Uniform (0,1) random variables. Compute E[\max(X 1, X 2)].概率简单数值题未尝试免费424Expected Gap Between the Two Largest UniformsLet X 1, \ldots, X 6 be iid Uniform (0,1). Find the expected gap between the largest and second-largest values: E[X (6) - X (5) ].概率中等数值题未尝试免费425Ratio of the Two Smallest Exponential Order StatisticsLet X 1, X 2 be independent Exp (1) random variables with order statistics X (1) \le X (2) . Define U = X (1) / X (2) .概率困难multi part未尝试面试订阅429Geometric Number of Geometric TrialsA gambler plays a sequence of rounds. In each round, he flips a biased coin with P( heads ) = p repeatedly until he gets heads; the number of flips in that round is Geom (p). The number of rounds he plays is itself Geom (q) (independent of the coin flips), where 0 < q < 1. Let S be the total number of coin flips across all rounds. Using the memoryless property of the geometric distribution, show that S \sim Geom (pq) and compute E[S].概率中等derivation未尝试免费432Asymmetric Penalties in an Exponential RaceTwo independent alarms go off at Exp (4) and Exp (6) times respectively. If alarm 1 fires first you pay \3; if alarm 2 fires first you pay \5. After the first alarm fires, the remaining alarm is reset (memoryless restart) and you pay an additional \1 when it fires. Find the expected total payment.概率中等数值题未尝试免费434Second Failure in a Memoryless Component ArrayA system has 4 independent components, each with lifetime Exp (2). When a component fails, it is removed and the remaining components continue operating. By memorylessness, surviving components' residual lifetimes are still Exp (2). Find the expected time until the second component fails.概率中等数值题未尝试免费439Sequential Elimination RaceThree players with independent lifetimes X 1 \sim Exp (1), X 2 \sim Exp (2), X 3 \sim Exp (4) compete. The first to "die" is eliminated, then the two survivors continue (by memorylessness, their residual lifetimes are fresh exponentials with the same rates). The second to die is eliminated, and the last survivor wins. (a) Find the probability that the elimination order is X 3, X 1, X 2 (i.e., player 3 dies first, then player 1, then player 2). (b) Find the expected total time until only one player remains.概率困难multi part未尝试面试订阅444Full Ordering Probability for Four Competing ExponentialsFour independent exponential random variables X 1 \sim Exp (1), X 2 \sim Exp (2), X 3 \sim Exp (3), X 4 \sim Exp (6) represent task completion times. Using iterated applications of the memoryless property, find P(X 4 < X 3 < X 2 < X 1) — the probability that the tasks complete in the specific order 4, 3, 2, 1.概率困难数值题未尝试面试订阅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455Geometric Mean of Random Gains via LLN and CLTLet X 1, X 2, \ldots be i.i.d.\ with P(X i = 1) = \tfrac 1 2 and P(X i = 0) = \tfrac 1 2 . Define the geometric-mean-like quantity Y n = (\prod i=1 n (1 + X i) ) 1/n . **(a)** Find \lim n Y n almost surely. **(b)** For n = 200, use the CLT to approximate P(Y 200 > 1.45). You may use: \ln 2 \approx 0.6931, \Phi(1.02) \approx 0.8461.概率困难derivation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467Symmetric Random Walk Displacement via CLTA particle performs a symmetric random walk: at each step i, it moves X i = +1 or X i = -1, each with probability 1/2, independently. After n = 400 steps, the position is S 400 = \sum i=1 400 X i. **(a)** What does the LLN say about S n / n as n ? **(b)** Using the CLT, approximate P(S 400 > 10). You may use \Phi(0.50) \approx 0.6915.概率中等数值题未尝试免费469Geometric Mean of Random Multipliers via LLN and CLTAn investment grows by a random factor each year: in year i, the portfolio is multiplied by X i, where X i are i.i.d.\ with P(X i = 2) = P(X i = 4) = 1/2. After n years, the annualized growth factor is the geometric mean G n = (\prod i=1 n X i ) 1/n . **(a)** Find \lim n G n almost surely. **(b)** For n = 100, use the CLT to approximate P(G 100 > 3). You may use \ln 2 \approx 0.6931, \ln 3 \approx 1.0986, and \Phi(1.70) \approx 0.9554.概率困难derivation未尝试免费474Sample Size for a Tail Probability GuaranteeLet X 1, X 2, \ldots be i.i.d.\ Exp (1) (mean 1, variance 1). A system designer requires that the sample mean X n exceeds 1.1 with probability less than 1\%. Using the CLT, find the smallest n satisfying P( X n > 1.1) < 0.01. You may use \Phi(2.33) \approx 0.9901.概率中等数值题未尝试免费476Hitting Time on a Four-State ChainConsider a Markov chain on states \ 0, 1, 2, 3\ with transition probabilities: p(0,1) = 1, \quad 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 , \quad p(3,3) = 1. Starting from state 0, compute the expected number of steps to reach state 3 for the first time.概率简单数值题未尝试免费478Lazy Random Walk Hitting TimeA particle moves on \ 0, 1, 2, 3, 4, 5\ . At each step from state i (0 < i < 5), it stays at i with probability \tfrac 1 3 , moves to i-1 with probability \tfrac 1 3 , and moves to i+1 with probability \tfrac 1 3 . State 0 is reflecting: from 0 the particle moves to 1 with probability \tfrac 2 3 and stays at 0 with probability \tfrac 1 3 . State 5 is absorbing. Starting from state 0, derive the expected number of steps to reach state 5.概率中等derivation未尝试免费