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This is the Past Exam of Stochastic Processes which includes Probability Generating, Non-Negative Integers, Non Negative Integers, Probability Generating Function, Markov Chain, Probability etc. Key important points are: Discrete Random Variable, Non Negative Integer Values, Probability Generating Function, Non Negative Integer, Finite Valued Random, Independent, Means, Markov Chain, Conditional, Initial Condition
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PART II (Third or Fourth Year) MATHEMATICS & STATISTICS Math 354 : Stochastic Processes 2 hours
You should answer ALL Section A questions and TWO Section B questions. In section A there are questions worth a total of 50 marks, but the maximum mark that you can gain there is capped at 40. SECTION A
A1. Let Z be a discrete random variable which can take non-negative integer values. Note Z may take the value ∞. Define the probability generating function of Z. If G(z) is the probability generating function of X, what are G(1) = limz↑ 1 G(z) and G(0)?
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SECTION A continued
A2. Let N be a non-negative integer finite-valued random variable and let {W 1 , W 2 , · · · } be a col- lection of independent identically distributed, and non-negative integer finite-valued random variables, which are independent of N. Let Y =
i=
Wi,
which means that Y = 0 when N = 0 and Y = ∑Ni=1 Wi when N ≥ 1. Let GY (s) = E[sY^ ], GW (s) = E[sWi^ ] and GN (s) = E[sN^ ]. By conditioning on N , show that GY (s) = EN^ [(GW (s))N^ ]^ = GN (GW (s)). Consider the following Markov chain, (called a branching process). Conditional on the value of Xn− 1 , Xn is given by Xn =
X ∑n− 1 i=
Zn,i where {Zn,i} are a collection of independent identically distributed, and non-negative integer finite-valued random variables. The initial condition is X 0 = 1. Let Gn(s) = E[sXn^ ], and G(s) = E[sZn,i^ ]. Show that Gn(s) = Gn− 1 (G(s)) = G(Gn− 1 (s)). (^) [15]
A3. Consider two urns A and B containing a total of N balls. An experiment is performed in which a ball is selected at random from the N balls in the two urns, at time t (t = 1, 2 , · · · ). Note that, each ball has probability 1/N to be selected. Then an urn is selected at random (A is chosen with probability p and B is chosen with probability q) and the ball previously drawn is placed in this urn. The state of system at each trial is represented by the number of balls in A. Determine the transition matrix for this Markov chain. [10]
A4. A particle performs a random walk Xt on the non-negative integers with a retaining barrier at 0. The transition probabilities are P [Xt+1 = 1|Xt = 0] = p, P [Xt+1 = 0|Xt = 0] = q and P [Xt+1 = i + 1|Xt = i] = p, P [Xt+1 = i − 1 |Xt = i] = q, where 0 < p < 1 and p + q = 1. Explain why the random walk is irreducible and aperiodic. [7] please turn over
B1. Consider two independent Poisson processes X(t) and Y (t) with rate λ and μ respectively. Let two successive events of X(t) occur at T, T ′^ where T ′^ > T , so that X(t) = X(T ) for T ≤ t < T ′^ and X(T ′) = X(T ) + 1. Let τ = T ′^ − T. (i) Explain why P [τ > t] = P [X(t) = 0|X(0) = 0]. [6] (ii) Using (i) to prove that the probability distribution of τ is exponential(λ) i.e. the pdf of τ is λ exp(−λt), t > 0. [6] Define N = Y (T ′) − Y (T ) = the number of events of Y (t) process in the time interval [T, T ′). (iii) Using the result of (ii), find the probability distribution of N i.e. P [N = m], m = 0, 1 , · · ·. [14] (iv) The transition probability rate matrix of the Poisson process X(t) is given by
hlim→ 0 P(h h)^ −^ I=^ Q^ = (qi,j^ ), where P(h) is the transition probability matrix of the Poisson process X(t). Calculate qi,i and qi,i+1. [4]
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SECTION B continued
B2. Consider the following game. You start with one token. You repeatedly toss a fair coin: if it is a head you gain a token, otherwise you lose a token. The game finishes if you ever have no tokens (you lose the game) or if you ever have three tokens (you win the game). Let Xt be the number of tokens you have after the tth coin toss. If before the tth coin toss you have either won or lost the game, then Xt = 3 or Xt = 0 respectively. Assume the game is fair. (i) Show E(Xt|Xt− 1 = i) = i. [4] (ii) By using E(Xt) = E(E(Xt|Xt− 1 )), show that E(Xt) = E(Xt− 1 ). Hence show that E(Xt) = 1 for all t. [4] (iii) What are the two potential eventual outcomes of the game, and their values for Xt? [4] (iv) Let p be the probability that you eventually win the game. Find the value of p. [10] (v) Now assume that you start with i tokens, and to win the game you ever have n tokens. Show that the probability of winning the game is i/n. [8]
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