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(Solved): Let {X(t)}t>0 be a standard Brownian motion and let Y (t) = R t 0 Sign(Y (s))ds+ X(t) where Sign( ...



Let {X(t)}t>0 be a standard Brownian motion and let Y (t) = R t 0 Sign(Y (s))ds+ X(t) where Sign(x) = 1 for all x ? 0 and Sign(x) = ?1 for all x < 0. A. Then the process {Y (t)}t?0 is a Brownian motion B. Then the process {Y (t)}t?0 is Gaussian but is not a Brownian motion. C. Then the process {Y (t)}t?0 is Markov but is not a Brownian motion. D. None of the above is true.



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