Indeed., Can't beat homotopy!Says all you need to know!The book is " Homotopy Analysis Method
in Nonlinear Differential
Equations" p. 432
Indeed., Can't beat homotopy!Says all you need to know!The book is " Homotopy Analysis Method
in Nonlinear Differential
Equations" p. 432
I read it a long time ago, if i remember correctly its just notation G(S , [$]\tau[$] , E ) = V( S , [$]\tau[$] , E ) = value of the option, nothing special, as far as i can remember. It is however an important paper!I could be wrong but I did not find definition of what is American option price in Merton 1973 paper. From definition of the contract it follows that we know only the value of G ( S , 0 , E ) . What does it mean G ( S , [$]\tau[$] , E ) is unknown. I will appreciate if somebody correct me.
unfortunately your thread (one of many) has been derailed / nuked by list1. Probably best to re-read all the posts except those that contain 'list1' and then decide whether you did get an answer or not.Im sort of confused did someone answer my question?
yes they didIm sort of confused did someone answer my question?
see Paul's post, he answered itthe terminal condition above can be simplified to
$$\mathop {\lim }\limits_{\tau \to 0} V(S,\tau) = 0$$
This is not so intuitive . How can the value of the option be equal to zero in this case?
(note: in the range) $${\Sigma _1} = \left\{ {(S,\tau )|B(\tau ) \le S < + \infty ,0 \le \tau \le T} \right\}$$
in the black-scholes model, the original statement is correct for r>q and incorrect for r<qSo it's saying that the option is zero at expiration above the exercise boundary.
That's only true if the exercise boundary is at or above the strike at expiration.