What I understood, you are saying that present value of your payoff is:

exp(-.02)*mucallvals)

but, in this case, atleast you can not discount your payoff by the risk free
rate, i.e. you can not use the expression "exp(-.02)" for discounting
purpose. Your return distribution i.e. "mustocks<-exp(rnorm(10000, mean=.15,
sd=vol)) " is in real world, right?
_____________________________________________________

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
_____________________________________________________


On Sun, May 15, 2011 at 5:05 AM, neal smith [via R] <
ml-node+3523400-2031256334-124430@n4.nabble.com> wrote:

> Hi
>
> 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
> arbitrage-free price.
>
> 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))
> mucallvals<-pmax(0.,mustocks-strike)
> print(summary(exp(-.02)*mucallvals))
>    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
>
> (.25x-(.25(100-x)))*binomial(p=.99,n=100,x)=(.5x-25)*binomial(.99,100,x).
>
> 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?
>
> thanks
> -neal
>
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