1 assumption in BSM

Schweser Page 265 - I did not get this assumption. Assumption of Black-Scholes-Merton “The volatility of the underlying asset is constant and known” . . . . So if the volatility of IBM stock is CONSTANT and is KNOWN (5%), so say current price of IBM stock is 100 and as the volatility is 5%, we know that the price range of IBM stock is going to be 95 - 105, then why the heck would anyone buy a call on IBM with a strike of 150 or a Put on IBM with a strike of 35??

Is this volatility on the BSM model the same as the up-down price in the Binomial model? OR Ist this volatility the sigma used to calculate the d1 and d2 used in the BSM Model?

It’s the sigma used to calculate d1 and d2 and then N(d1) and N(d2)… They say sigma = volatility of continuously compounded returns on the stock. And in the Limitation to this assumption they say “In practice the volatility is not known and must be estimated. The bigger problems is that the volatility is often not constant over time and the BSM MODEL IS NOT USEFUL in these situations. More sophisticated models have been developed to handle this issue, but fortunately out of L2… bla”

First, 5% volatility is unreal. does not exist. even the most stable stocks of good, reliable companies have volatility of 20% and usually higher. volatilities in the range of 40% to 80% are most common for the majority of mid-sized companies. i’m excluding past few quarters of craziness where volatility has been well over 100%. second, why do you say “we know the price range will be 95-105”. you know the price range will be from 0 to infinity. the lognormal distribution “assigns” probabilities to any particular range. so for a starting stock price of 100, risk-free rate 1%, volatility 5% and remaining term 1 year, the probability that the stock price will be in the range 95-105 is still about 67%, and there is about 12% chance it will be below 95. the BS formula prices exactly that, the weighted average payoff given how likely it is the stock price will end up at a certain level

MS I hear you with the usage of lognormal distribution instead of standard normal distribution; since the lower bound of a stock price can be 0 tops and max can be infinity, which is very well explained by a lognormal distribution (as against the normal distribution which has the range of –infinity to + infinity) Also don’t know the prevalent volatility numbers or the stock prices as I am not from the industry and don’t deal with it on day-to-day basis, so forgive my ignorance on the same. I would like to know more on how you got the 67% and 12% figures (if possible in simple layman’s’ terms)

swaptiongamma, to your original question, volatility is constant means that throughout the priod (say one year), IBM’s volatility remains at 5% or whatever fixed number you have. The model does not work if the volatility is 5% the first 3 months, then jumps to 15% next 2 months, and back again to 5% later in the year.

if S is the stock price (lognormally distributed, stochastic variable) and K is the fixed strike price, then P(S>K) = (Probability that stock price will end up above K) = N(d2). as a side note, these are risk-neutral probabilities so don’t interpret their meaning too directly

> why the heck would anyone buy a call on IBM with a strike of 150 or a Put on IBM with a strike of 35?? You wouldn’t unless it’s really really cheap based on the calculated probabilities as pointed out by MS.

> P(S>K) = N(d2) I thought it was equal to N(d1), no? > these are risk-neutral probabilities so don’t interpret their meaning too directly Yes - Correct. They are not the true probabilities, but the ones similar to pi(u) and pi(d) from the Binomial Option pricing models (as per user expectations) So N(*) of something is like the z-score which gives the cumulative normal probability under the curve. Do we have a linear relationship between N(d1) and N(d2), like we have in probability [probability of snowing in June is 0.1, so probability of not snowing in June is 0.9] Also please look at this link: http://www.editgrid.com/bt/frm2007/LO_34.x_BlackScholes Given d1 they get N(d1), and given d2 they get N(d2). How do we get the values – using some tables? >swaptiongamma, to your original question, volatility is constant means that throughout >the priod (say one year), IBM’s volatility remains at 5% or whatever fixed number you >have. The model does not work if the volatility is 5% the first 3 months, then jumps to >15% next 2 months, and back again to 5% later in the year. >You wouldn’t unless it’s really really cheap based on the calculated probabilities as >pointed out by MS. Hi Dreary – I got what you are saying with volatility being CONSTANT over the holding period. But isn’t the whole point of ‘derivatives’ is to have some volatility in the underlying asset, so that people could speculate and/or hedge?

There is volatilty, but the model assumes there is no volatility to the volatility itself, if I may put it this way.

haha Dreary - this one makes more sense to me. But don’t you feel this model was a utter waste by assuming such irrational stuff to come down to a complex un-intuitive formula? Is it really used in the industry to calculate the price of the option?

swaptiongamma Wrote: ------------------------------------------------------- > haha Dreary - this one makes more sense to me. But > don’t you feel this model was a utter waste by > assuming such irrational stuff to come down to a > complex un-intuitive formula? > > Is it really used in the industry to calculate the > price of the option? Yes, it is used. It is not an utter waste at all. Look man…models are just that…MODELS. We cannot always describe a process perfectly with mathematics, so we try to estimate the process. Sometimes to make things practical, simplifying assumptions are made, irrational or not. The irrationality is not the point, but rather how that irrationality will effect the output, and the fact that the irrationality is known by its users who understand its limitations. For short time periods, constant volatility is not that far fetched, nor is constant interest rates. The pricing effects are rather minimal. For longer maturities, we have a bit of a problem. Hence…Heston’s stochastic volatility model. If you think BS for vanilla options is complex, do yourself a favor and stay as far away as possible of the models for stochastic vol and exotics.

Sorry bro - I think you got hurt. I was just venting out my frustration for not understanding basics and nothing more.

Nahhhh…I don’t get hurt by internet forums. You just seem to be ignoring the comments on the thread answering your question. Sometimes you need a thorough explanation, and sometimes harsh. No offense.

swaptiongamma Wrote: ------------------------------------------------------- > > P(S>K) = N(d2) > I thought it was equal to N(d1), no? > No, P(S>K)=N(d2). N(d1) is something else, it is related to partial expected values > > these are risk-neutral probabilities so don’t > interpret their meaning too directly > Yes - Correct. They are not the true > probabilities, but the ones similar to pi(u) and > pi(d) from the Binomial Option pricing models (as > per user expectations) > > So N(*) of something is like the z-score which > gives the cumulative normal probability under the > curve. > > Do we have a linear relationship between N(d1) and > N(d2), like we have in probability N(*) here in the black-scholes formula is the cumulative standard normal distribution. You get it with any excel spreadsheet or tables. The solution to the BS formula is written in terms of N(*) - the normal distribution, even though the probability distribution of the stock price is lognormal. the two distributions are very related

swaptiongamma Wrote: ------------------------------------------------------- > > P(S>K) = N(d2) > I thought it was equal to N(d1), no? “Interpretation: N(d1) and N(d2) are the probabilities of the option expiring in-the-money under the equivalent exponential martingale probability measure (numéraire = stock) and the equivalent martingale probability measure (numéraire = risk free asset), respectively. The equivalent martingale probability measure is also called the risk neutral probability measure. Note that both of these are “probabilities” in a measure theoretic sense, and neither of these is the true probability of expiring in-the-money under the real probability measure.” http://en.wikipedia.org/wiki/Black-Scholes

this explanation from wikipedia is clear as mud to somebody unfamiliar with the material. N(d1)=delta, N(d2)=P(S>K)=(risk-neutral probability to finish in the money) is clearer

Thanks guys! maratikus - I hope L3 is not as bad as L2? how’s prep going on?