If the Ti's are independent and satisfy P(Ti =d)=p=1 P(Ti =u), we have E(Tn+1|Fn)=E(Tn+1)=pd + (1

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If the Ti's are independent and satisfy P(Ti =d)=p=1 − P(Ti =u), we have E(Tn+1|Fn)=E(Tn+1)=pd + (1 − p)u=1 + r, and thus ( ˜ Sn) is a P-martingale, according to Question 1.

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Introduction To Stochastic Calculus Applied To Finance

ISBN: 9781584886266

2nd Edition

Authors: Damien Lamberton, Bernard Lapeyre

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Question Posted: Oct 26, 2024 06:01 AM

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If the Ti's are independent and satisfy P(Ti =d)=p=1 P(Ti =u), we have E(Tn+1|Fn)=E(Tn+1)=pd + (1 p)u=1 + r, and thus ( Sn) is a P-martingale, according to Question 1.
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If the Ti's are independent and satisfy P(Ti =d)=p=1 P(Ti =u), we have E(Tn+1|Fn)=E(Tn+1)=pd + (1

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Introduction To Stochastic Calculus Applied To Finance

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