ts from lily 1. Let X(t) be the number of the lily is a continuous-time Markov chain, give its para f is required. (t) = j|X(0) = 1). Explain why P12(t) = equation to compute p₁1(t). P13(t Chapman-Kolmogorov equation, and prove that 5 P₁₁(t) = 1- P₁i(t). 7²11(t).
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- Question 3. Recall that the discrete stochastic process {Xn, n = 0, 1, 2,...} is a Markov chain if for each n, P(Xn+1 = j|Xn = i, Xn-1 = in-1,..., Xo = io) = P(Xn+1 = j|Xn = i) = Pij. Let {Xn, n = 0, 1, 2,...} be a Markov chain, does the following hold as well? P(Xn+2=jXn = i, Xn-1 = in-1, ..., Xo = io) = P(Xn+2 = j|Xn = i) Give a proof if you think it is true, otherwise give a counterexample. Hint: You may use the law of total probability for conditional probability without proof. Please step by step answer.1- The number of items produced in a factory during a week is known to be a randomvariable with mean 50● Using Markov's inequality, what can you say about the probability that this week'sproduction exceeds 75?● If the variance of one week's production is equal to 25, then using Chebyshev'sinequality, what can be said about the probability that this week's production isbetween 40 and 60?The figure above illustrates a continuous-time Markov process. Suppose the system is currently instate 2. After a small amount of time Δ, what is the probability that the system is in each state?
- If Kt = B2t - t, where B is standard Brownian Motion, show that Kt is a martingale, and a markov processA cellphone provider classifies its customers as low users (less than 400 minutes per month) or high users (400 or more minutes per month). Studies have shown that 80% of people who were low users one month will be low users the next month, and that 70% of the people who were high users one month will high users next month. a. Set up a 2x2 stochastic matrix with columns and rows labeled L and H that displays these transitions b. Suppose that during the month of January, 50% of the customers are low users. What percent of customers will be low users in February? In March?A frog is in a pond with 5 water lilies numbered from 1 to 5. With exponential rate 1, the frogleaves its current water lily, chooses a new one uniformly among the four others and jumps to it.We assume the frog starts from lily 1. Let X(t) be the number of the lily where the frog is at timet.a. Admitting that X(t) is a continuous-time Markov chain, give its parameters (i.e. the vi and pijof the course). No proof is required.b. Let p1j (t) = P (X(t) = j|X(0) = 1). Explain why p12(t) = p13(t) = p14(t) = p15(t) (nocomputations required).c. Write the forwards Chapman–Kolmogorov equation, and prove thatp′11(t) = 14 − 54 p11(t).d. Solve this equation to compute p11(t)
- suppose a continuous time markov process with three states s = {1, 2, 3} and suppose the transition rates q1,2, q2,3, q1,3, and q2,1 are non-zero, with all the other transition rates being zero. set up and solve the kolmogorov forward equations for this processAlan and Betty play a series of games with Alan winning each game independently with probability p = 0.6. The overall winner is the first player to win two games in a row. Define a Markov chain to model the above problem.Suppose that X0, X1, X2, ... form a Markov chain on the state space {1, 2}. Assume that P(X0 = 1) = P(X0 = 2) = 1/2 and that the matrix of transition probabilities for the chain has the following entries: Q11 = 1/2, Q12 = 1/2, Q21 = 1/3, Q22 = 2/3. Find limn→∞ P(Xn = 1).
- why is the covariance of a deterministic and a stochastic process 0? This relats to Arithmetic Bronian MotionIn the B&K model of Example 18.5-1, suppose that the interarrival time at the checkout area is exponential with mean 5 minutes and that the checkout time per customer is also exponential with mean 10 minutes. Suppose further that will add a fourth counter. Counters 1,2, and 3 will open based on increments of two customers and counter 4 will open when there are 7 or more in the store. (a) The steady-state probabilities, for all . (b) The probability that a fourth counter will be needed. (c) The average number of idle counters.Consider two machines that are maintained by a single repairman. Each machine functions for an exponentially distributed amount of time with rate λ=2 before it fails. The repair times for each unit are exponential with rate μ=1.The Markov chain model for this situation has state space indicating the number of machines that are in the repair shop: S={0,1,2}. Notice that you move from 0 to 1 if one of the two machines breaks down. You move from 1 to 0 when the machine in the repair room is repaired.