# ² m ² 1 y t j ² m ² 2 y tm 2 but with breaks

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² m ² 1 ' , y t , j ² m ² 2 ' ,..., y tM ' , 2   But: with breaks always in the same place, y t ' k   would be too correlated with y t ' k ² 1   Solution(!): (1) For k ± 1, set 2 k " 1   and y ' k " 1   to initial guesses (2) Set j ± 0 (3) Draw m ' L Poisson( 5 ) If m ' ² 1 ³ M , set m ± M otherwise, set m ± m ' ² 1 (4) Draw y tj ' k   , y t , j ² 1 ' k   ,..., y t , j ² m ' k   | y , 2 , y ti ' ± y t , i ' k " 1   for i ³ £ j , j ² 1,..., j ² m ¤ and t ± 1,2,..., T (5) Set j ± j ² m ² 1. If j ³ M , go to (6) else, go to (3)

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7 (6) Draw 2 k   | y , y ' k   (7) Set k ± k ² 1 and go to (2) How to do step (6)? Conditional on y ' , this is a more standard problem Example: Cox-Ingersol-Ross model dy t   ± ¡ ) " * y t  ¢ dt ² @ y t   dW t   Let x t   ± log y t
8 Ito’s Lemma: if dy t   ± 6 y t   , t   dt ² @ y t   , t   dW t   x t   ± f y t    with f .   twice continuously differentiable, then dx t   ± a y t   , t   dt ² b y t   , t   dW t   a y t   , t   ± 6 y t   , t   ² f y   ² y y ± y t   ² 1/2  ¡ @ y t   , t  ¢ 2 ² 2 f y   ² y 2 y ± y t   b y t   , t   ± @ y t   , t   ² f y   ² y y ± y t   Here f y   ± log y dy t   ± ¡ ) " * y t  ¢ dt ² @ y t   dW t   ² f / ² y ± 1/ y ² 2 f / ² y 2

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