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solution_hw3

# solution_hw3 - 0.1 Exercises 1 A function f is said to be...

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0.1 Exercises 1. A function, f , is said to be Lipschitz-continuous on [0 , T ] if there exists a constant C > 0 such that for any t, t 0 [0 , T ], | f ( t ) - f ( t 0 ) | < C | t - t 0 | . Show that a Lipschitz-continuous function has bounded variation and zero 2-variation over [0 , T ]. Proof. Let the partition π n : 0 t 0 < t 1 < ... < t n = T . Then, sup π n n X i =1 | f ( t i ) - f ( t i - 1 ) | ≤ sup π n C n X i =1 | t i - t i - 1 | = cT < and n X i =1 | f ( t i ) - f ( t i - 1 ) | 2 C 2 n X i =1 | t i - t i - 1 | 2 C 2 ˆ sup 1 i n | t i - t i - 1 | n X i =1 | t i - t i - 1 | ! C 2 T sup 1 i n | t i - t i - 1 | → 0 2. (a) Show directly that tB t is not a martingale. (b) Using the Doob’s decomposition theorem, it can be show: if X t is a martingale and has a bounded variation, then X t must be a constant in t . Use the above result to show that R t 0 B s ds can not be a martingale. ( Remark : By Doob’s decomposition theorem, any sub-, super-martingale X t has the unique decomposition X t = M t + A t , where M t is a martingale and A t is of bounded variation with A 0 = 0. So if X t is already a martingale, then A t = 0,

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