# solution1.pdf - Assignment 1 Solutions IEOR173 Spring 2019...

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Assignment 1 Solutions IEOR173, Spring 2019 February 7, 2019 Problem 1 (a) By conditioning, μ = p 0 E [ X | X = 1] + (1 - p 0 ) E [ X | X 6 = 1] = p 0 + (1 - p 0 ) E [ X | X 6 = 1] . Therefore, E [ X | X 6 = 1] = μ - p 0 1 - p 0 . (b) Again, by conditioning, E [ X 2 ] = σ 2 + μ 2 = p 0 E [ X 2 | X = 1] + (1 - p 0 ) E [ X 2 | X 6 = 1] = p 0 + (1 - p 0 ) E [ X 2 | X 6 = 1] . Hence, E [ X 2 | X 6 = 1] = σ 2 + μ 2 - p 0 1 - p 0 , and var( X | X 6 = 1) = E [ X 2 | X 6 = 1] - ( E [ X | X 6 = 1]) 2 = σ 2 + μ 2 - p 0 1 - p 0 - μ - p 0 1 - p 0 2 . Problem 2 (a) By taking expectation on the both sides of the recursive equation, we have E [ X n +1 ] = c + ρE [ X n ] . 1
Because | ρ | < 1, { E [ X n ] } converges. Let α be the limit. Then, by taking the limit on both side, α satisfies α = c + ρα ⇐⇒ α = c 1 - ρ . (b) Similar to the previous problem, by taking the variance of both sides, we have var( X n +1 ) = ρ 2 var( X n ) + σ 2 . { var( X n ) } converges because | ρ | < 1. Let β be the limit. Then, β satisfies β = ρ 2 β + σ 2 ⇐⇒ β = σ 2 1 - ρ 2 . Problem 3 In this problem, the answer is not unique. An answer gets full credit if it is convincing enough. The followings are examples of possible answers. (a) No. It seems the future development depends on the trend, which means the transition property varies on the past few periods. To make it Markov, we need to keep track of the several past periods. (b) Yes. When an agent comes to the auction, she only observes the current bid and decides the next bid. Thus, the bid process does not depend on the history and only on the current state. Problem 4 (a) No. The intuition for this question is that knowing just the summation X n + Z n does not offer enough information, as there may be many possible combinations. Combining (i) the