nealsmith314 at gmail.com
Sun May 15 01:32:43 CEST 2011
I'm not saying the mean call value .165 with mu=.14 equals the call
price. i am not hedging the call in this case, and i am exposed to 0
payoff and hence loss of the price of the call if the call expires out
of the money. as the thread you mention explains, the call price
equals the price of the risk-free replicating/hedging portfolio, and
any other price exposes one to arbitrage. so i'm not going to pay
.165 for this call, i agree it's worth only .05, the BS risk-neutral
my question is, now if i go ahead and buy the call for BS/arb-free
price of .05 and hold it until expiration, the distribution of returns
seems to be very high and favorable:
mustocks<-exp(rnorm(10000, mean=.15, sd=vol))
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.00000 0.08153 0.15670 0.16590 0.23860 0.73860
i can ask this question on the tree also, suppose s0=1, and at t=1,
either s1=1.5 or s1=.5, and i have a call with K=1, and r=0, then as
the lyceum/hull/etc explain, i can hold the riskless hedging portfolio
of long .5 stock and short (borrow) $.25, so that i risklessly
replicate the payoffs, meaning the option is worth .5(1)-.25=.25. i
have no question about that, and that is well explained in the wilmott
thread mentioned along with hull etc.
instead, i want to specifically understand the real world probability.
suppose i move to the real world, and i know p(up)=.99 and
p(down)=.01. so i play 100 times, of course i am exposing myself to a
risk of loss, and my payoff if i have x ups (and hence 100-x downs)
in the 100 trials is given by
for instance why shouldn't i buy the call for .25, and then, on
average, 99 out of 100 times, pick up the.25*99=$24.75, and the
(typically) 1 time lose the .25, so i end up with $24.50 on average
(which equals the expectation of the above payoff)?
is this just the result of the fact that i am exposing myself to a
range of losses upto the maximum -$25 loss (wp .01^100)? is this
related to "market price of risk" etc?
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