# options/BS/MC

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## options/BS/MC

 Hi I have a general question that I've not been able to figure out. vol<-.1 strike<-1. library(RQuantLib) EU<-EuropeanOption(type="call",underlying=1.,strike=strike,dividendYield=0.,riskFreeRate=.02,maturity=1., volatility=vol) print(EU) Concise summary of valuation for EuropeanOption   value   delta   gamma    vega   theta     rho  divRho  0.0502  0.5987  3.8667  0.3867 -0.0303  0.5485 -0.5987 stocks<-exp(rnorm(10000, mean=.02-vol^2/2, sd=vol)) callvals<-pmax(0.,stocks-strike) summary(exp(-.02)*callvals)    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 0.00000 0.00000 0.01329 0.04987 0.08314 0.47050 So that's all as expected, the mean discounted call value under the risk-neutral measure and the BS price are close. However we are told that the option price doesn't depend on the mu of the stock, and there are stocks which have consistently higher mu's (BRK, GOOG) than other stocks (YHOO) for a similar level of vol.  So let's say I have a high-mu stock: mustocks<-exp(rnorm(10000, mean=.15, sd=vol)) mucallvals<-pmax(0.,mustocks-strike) print(summary(mustocks))    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  0.8155  1.0850  1.1630  1.1680  1.2440  1.7400 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 so i can buy this call for the BS price of \$.05, which equals its discounted payoff in RN, but now on this high-mu underlying i get a mean discounted value of .165, ie. 200% mean roi, with the lower quartile giving a paltry 60% return. what am i doing wrong? thanks! -neal         [[alternative HTML version deleted]] _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.
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## Re: options/BS/MC

 Well, a good discussion on why your second approach is completely meaningless can be found here: http://www.wilmott.com/messageview.cfm?catid=8&threadid=84115Thanks and regards, _____________________________________________________ Arun Kumar Saha, FRM QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST Visit me at: http://in.linkedin.com/in/ArunFRM_____________________________________________________
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## Re: options/BS/MC

 In reply to this post by neal smith 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 _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.