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final-practice-solution - Mathematics 136 Final Winter 2007...

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Mathematics 136, Final Winter 2007 Write your name and sign the Honor code in the blue books provided. This is a closed material exam, but you may use six pages (2 sides each) of notes. You have 180 minutes to solve all four questions, each worth points as marked (maximum of 100). Complete reasoning is required for full credit. You may cite lecture notes and homework sets, as needed, stating precisely the result you use, why and how it applies. 1. (5x5) Provide the complete, accurate and rigorous definitions for the following five concepts. (a) The finite dimensional distributions of a stochastic process X ( t ), a consistent collection of finite dimensional distributions, and a stationary stochastic process X ( t ). ANS: Definitions 3.1.5, 3.1.13, and 3.2.23. (b) A random walk, a simple random walk, and a stochastic process V n that is previsible for a filtration F n . ANS: Definitions 3.1.2 and 4.1.11. (c) A right-continuous filtration F t , t 0, a stopping time τ for a filtration F t , and its stopped σ -field F τ . ANS: Definitions 4.2.10, 4.3.11, and 4.3.21. (d) A continuous-time Markov process X ( t ), its state space S , its transition probabilities p t,s , and its initial distribution π . ANS: Definitions 6.1.10 and 6.1.11. The definition of the initial distribution of a Markov process can be found just after Definition 6.1.10 on page 115. (e) A Poisson law of parameter μ , a Poisson process { N t } of rate λ , and its jump times { T k } . ANS: Example 1.1.4 and Definition 6.2.1. The definition of the jump times of a Poisson process can be found on page 118 at the beginning of section 6.2, 2. (5+4+6+5+4+4) Parts a)–f) of this problem can be solved independently of each other! Consider the stochastic process Y t = W 2 t , with its canonical filtration G t = σ ( Y u , u t ), where W t a Brownian motion. (a) Explain why Y t , t 0 is a homogeneous Markov process. Provide its state space S , its initial distribution π ( · ) and its stationary regular transition probabilities p t ( A | x ). ANS: Intuitively, the Markov property of W t is not lost in the squaring operation because the information contained in the sign of W t holds no value for predicting future values of Y t . The state space S of Y t is [0 , ) and its initial distribution is the unit mass at 0 (since W 0 = 0). To get the regular transition probability, note that if Y s = x then W s = ± x , with each choice of the sign of W s equally likely, while Y s + h = ( W s + ξ ) 2 for ξ which is a Gaussian 1
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random variable, independent of W s having mean zero and variance h . Consequently, the regular transition probability for Y is given by the formula, p t ( A | x ) = 1 2 πt Z e - y 2 / 2 t 0 . 5( I ( x + y ) 2 A + I ( - x + y ) 2 A ) dy .
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final-practice-solution - Mathematics 136 Final Winter 2007...

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