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blackforesty (nothing) 于 (Tue Jan 30 13:20:04 2007) 提到:
d(X_t)=\mu*X_t*dt+\sigma*X_t^a*dB_t, B_t is Standard Brownian Motion.
What value of "a" will make X_t a Martingale?
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wusuowei (wuxuowei) 于 (Tue Jan 30 13:23:41 2007) 提到:
no value.
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wybeer (bbb_bbw_bw3_bmw) 于 (Tue Jan 30 14:43:17 2007) 提到:
what about
a = -mu/sigma*dB_t + y
where y is independent of B_t.
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