== {1, 2, 3} and transition probability matrix 4. Stationary Distribution Let (Xn)nzo be a Markov chain with state space S P = 0 (1/4 0 3/4\ 1/2 1/2 Compute the stationary distribution π. 2/3 1/3 0
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- 2. Stationary Distributions Recall that a stationary distribution of a Markov chain is a probability distribution P(X) such that starting with it and applying the transition probabilities does not change it. That is, it satisfies P(X = x) = Σ P(X₁+1 = x | X₁ = x') P(X = x'). (a) Give an example of a Markov chain with two states and deterministic transitions where applying the mini forward algorithm does not necessarily converge to a stationary distribution (b) Show, nevertheless, that your example has a stationary distributionHi there Could you please help solve this? ThanksThe random variables W1, W2,... are independent with common distribution k 1 2 3 4 Pr( W = k) 0.1 0.3 0.2 0.4 Let Xn max (W1,..., Wn) be the largest W observed to date. Determine the transition probability matrix for the Markov chain {Xn}.
- • General notation for Markov chains: P(A) is the probability of the event A when the Markov chain starts in state x, Pμ(A) the probability when the initial state is random with distribution µ. Ty = min{n ≥ 1 : Xn = y} is the first time after 0 that the chain visits state y. px,y = Px(Ty < ∞) . Ny is the number of visits to state y after time 0. 3. Each year an auto insurance company classifies its customers into three categories: Poor, Satisfactory, and Good. No one moves from poor to good or from good to poor in one year. The status of a driver can be modeled by a Markov chain {Xn : n ≥ 0} with state space S = {P, S, G} and transition matrix PS G P/1/2 1/2 0 S 1/5 3/5 1/5 G 0 1/5 4/5 (a) Compute the invariant distribution for this Markov chain. (b) Assume that the average prices (per year) of insurance policies for drivers with Poor, Satisfactory, and Good ratings are $600, $300, $200, respectively. In the long run, how much does a driver pay for the insurance per year? (c) For a…• General notation for Markov chains: P(A) is the probability of the event A when the Markov chain starts in state x, Pμ(A) the probability when the initial state is random with distribution μ. Ty = min{n ≥ 1 : X₂ = y} is the first time after 0 that the chain visits state y. Px,y = Px(Ty < ∞) . Ny is the number of visits to state y after time 0. 2. A bank classifies loans as paid in full(F), in good standing (G), in arrears (A), or as a bad debt (B). Loans move between the categories according to the transition matrix: P = F G A B F 1 0 0 0 G .1 .8 .1 0 A .1 .4 .4 .1 B 0 0 0 1 What fraction of loans in good standing are eventually paid in full? (That is, starting at state G, what is the probability that the chain eventually visits F?)What are the hypothesis that we need to verify to apply our analysis in M/M/1. Select all correct answers. The arrival follows a Poisson distribution The arrival follows an exponential distribution The service follows a logarithmic distribution The arrival follows a Markovian process We have one server that serves the customers We have one customer who can arrive every unit of measure [Choisir] [Choisir] [Choisir] [Choisir] [Choisir] [Choisir]