options/BS/MC

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options/BS/MC

neal smith
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

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Re: options/BS/MC

Arun.stat
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=84115

Thanks 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

neal smith
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

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Re: options/BS/MC

Arun.stat
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] <[hidden email]> 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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Re: options/BS/MC

Rex-2
In reply to this post by neal smith
neal smith <[hidden email]> [2011-05-14 09:23]:

>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?

Nothing, unless you expect the values to be equal. The BSM value is
based on a hypothetical riskless hedge, but buying the option is
anything but riskless, and the expected future price of a high-mu
stock (and hence the call option) will be far different from the
expected future price of a stock with a mu equal to the riskless rate.

I found the Wilmott thread mostly unhelpful, and so I've put up a page
using R to illustrate the issues. I hope it's useful.

http://www.nosyntax.net/cfwiki/index.php/Trend_vs_option_value

Anyone may edit the page or the discussion page, and I encourage
comments on the discussion page. There may well be typos, but I hope
there are no conceptual blunders.

-rex
--
Democracy is the art and science of running the circus from
the monkey cage. --H. L. Mencken

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