# put call or binomial ?

There was a question in schweser (its from the study notes so cant cut and paste) which went something like this Stock price = 60 Time period = 1 year standard deviation = 10% risk free rate = 5% call value = 4.09 what is the put value ? i dont remember the options but i know the answer.

strike price?

Yea, if you had the strike price you would use pt-call parity: c + x/(1+r)^t = s + p

sorry yes, there was a strike price of 60… question is why wont you use binomial tree ?

Probably because they were asking for risk-free-arbitrage-put-value? Which will be better suggested by solving the Put Call Parity, than estimating using the binomial lattice (where we estimate volatility and stuff…) c + x/(1+r)^t = s + p 4.09 + 60/1.05 = 60 + P P = 57.142857 + 4.09 - 60 = 1.232857??

yes the answer was around 1.16 if i remember correctly … this is btw the first challenge problem from schweser option readings, in case any one has the study notes with them. Hmm… so just because they gave us the call value we should assume they are asking for a no-arbitrage situation and so we ignore the SD and use the put-call ?.. The question just said what is the most likely put value … no mention of arbitrage anywhere.

1.2329 agreed no binomial tree because it’s over 1 year, and there’s no 'chance of up is x rate, chance of y rate is 1/x)

allépourpêcher Wrote: ------------------------------------------------------- > 1.2329 agreed > > no binomial tree because it’s over 1 year, and > there’s no 'chance of up is x rate, chance of y > rate is 1/x) i thought thats what standard deviation tells us… up 10% down 10% . No ? and time priod is 1 year. couldnt get what the issue is here ?

use put-call parity whenever you can