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Cuchulainn
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Re: Terminal Condition for American Put Option

January 26th, 2017, 1:31 pm

The book is " Homotopy Analysis Method
in Nonlinear Differential
Equations" p. 432
Says all you need to know!
Indeed., Can't beat homotopy!
 
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list1
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Re: Terminal Condition for American Put Option

January 26th, 2017, 5:07 pm

I looked at  "Alternative Characterizations of American Put Options" Peter Carr, Robert Jarrow, Ravi Myneni and could not understand B t  definition.  Eq (2) , (3) do not define B t  . I will appreciate if somebody help me to understand.  
On the other hand they defined price of AO as one that present maximum payoff where max is taking with respect to stopping time. In such case maximum payoff corresponds to maximum price of the put. As I mentioned earlier it looks more reasonable to consider rate of return and to chose stopping moment that performs maximum rate of return for AO
 
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list1
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Re: Terminal Condition for American Put Option

January 26th, 2017, 10:39 pm

In Carr, Jarrow, Myneni paper it is written "Merton 1973 showed that while the Black-Scholes European option pricing methodology applied to American call options on non-dividend paying stocks, it did not apply to American put options" . BS method is quite clear uses as definition either put or call BSE solution as it eliminates risky term in hedged portfolio. To be formally correct we can say that there exists function C ( t , S ; T , K ) which satisfies some properties. This why we call this function call option price. Merton uses notation [$] G ( S , \tau , E )[$] for American put. This notation used informally as far as American put price is undefined. It is difficult to accept or reject any of the properties their as far as we do not have a formal definition of the American put price.
Subjectively taking BS approach we need perform reduction of the American put to European put. Of course I could be wrong but subjectively this reduction looks different than it commonly used.
 
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Paul
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Re: Terminal Condition for American Put Option

January 26th, 2017, 11:56 pm

List, see a specialist. I'm worried about you.
 
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list1
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Re: Terminal Condition for American Put Option

January 27th, 2017, 1:46 am

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.
 
MAYbe
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Re: Terminal Condition for American Put Option

January 27th, 2017, 2:05 am

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.
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!
 
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Paul
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Re: Terminal Condition for American Put Option

January 27th, 2017, 7:31 am

list, why don't you start your own thread with your own questions? That way we might be able to help you learn something in a structured manner. I'm afraid that your current knowledge is too little, too shallow, and too disjointed for you to be able to help anyone here.
 
MAYbe
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Re: Terminal Condition for American Put Option

January 29th, 2017, 12:46 pm

Im sort of confused did someone answer my question?
 
frolloos
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Re: Terminal Condition for American Put Option

January 29th, 2017, 1:12 pm

Im sort of confused did someone answer my question?
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.
 
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ppauper
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Re: Terminal Condition for American Put Option

January 29th, 2017, 3:08 pm

Im sort of confused did someone answer my question?
yes they did
the 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) $${\Sigma _1} = \left\{ {(S,\tau )|B(\tau ) \le S < + \infty ,0 \le \tau \le T} \right\}$$
see Paul's post, he answered it
So 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.
in the black-scholes model, the original statement is correct for r>q and incorrect for r<q