4106-08-h7-solu

4106-08-h7-solu - HMWK 7 Solutions 1 Xn = Y1 Y2 Yn where X0 = 1 and the Yi are iid with P(Y = 0.5 = P(Y = 1 = P(Y = 1.5 = 1/3 Argue that Xn

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HMWK 7 Solutions 1. X n = Y 1 × Y 2 ×···× Y n where X 0 = 1 and the Y i are iid with P ( Y = 0 . 5) = P ( Y = 1) = P ( Y = 1 . 5) = 1 / 3. Argue that X n converges and find the limit. SOLUTION: Since E ( Y ) = 1, X n is a non-negative martingale, so it must con- verge wp1 to a rv X such that E ( X ) < by the martingale convergence theorem. Because R n = ln( X n ) = Σ n i =1 ln( Y n ) is a random walk with iid increments ln( Y n ), we conclude by SLLN that lim n R n n = E [ln( Y )] = 1 3 ln 3 4 < 0. So the random walk has negative drift; lim n R n = -∞ , wp1. Thus X n = e R n 0, wp1. 2. Consider the Gambler’s ruin Markov chain { X n } on { 0 , 1 , 2 , 3 , 4 } ( N = 4) with transition matrix P = 1 0 0 0 0 1 / 2 0 1 / 2 0 0 0 1 / 2 0 1 / 2 0 0 0 1 / 2 0 1 / 2 0 0 0 0 1 Suppose that X 0 = 1. Does X n converge wp1? Find the limiting rv X if so. Repeat for each initial condition X 0 = i, i = 0 , 1 , 2 , 3 , 4. SOLUTION: First we can easily check that { X n } is a non-negative martingale for each initial state X 0 = i, i ∈ { 0 , 1 , 2 , 3 , 4 } . This is because X n is the simple symmetric random walk (already a known MG) away from the boundaries (e.g., when X 0 = i ∈ { 1 , 2 , 3 } ), and at a boundary (e.g. X 0 = 0 or X 0 = 4) it remains constant. Thus X n converges wp1 to a rv X with E ( X ) < . From the gambler’s
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This note was uploaded on 10/20/2010 for the course IEOR 4106 taught by Professor Whitt during the Fall '08 term at Columbia.

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4106-08-h7-solu - HMWK 7 Solutions 1 Xn = Y1 Y2 Yn where X0 = 1 and the Yi are iid with P(Y = 0.5 = P(Y = 1 = P(Y = 1.5 = 1/3 Argue that Xn

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