Simple Black Scholes and volatility questions
Two short questions, guys --
How would you answer to a question, "tell me about the black scholes model" during the interview? The way that I answered was listing several different factors(strike price, maturity, current price...), some assumptions, and giving an extra emphasis on the volatilty issue. Which led to the follow-up question, "then if the option is already in the money, would volatility(or the price of the call option) be much of an issue?" to which I said "no, since the delta is always high enough."
Would there be better or more correct ways to answer these type of questions?
of course volatility still matters because you can still trade the vol/gamma. it depends on how in the money i guess, if very deep in the money then you wouldn't be able to make much off gamma or the convexity
The other issue to consider would be the theta of the option. After all, the vols for an ITM option 1 week from expiry is radically different from one 1 year from expiry - suffice to say, the vols will be a lot higher for the former, which offers more opportunities for gamma trading (ideally, the option should fluctuate incessantly between ITM/OTM close to the ATM level).
"The other issue to consider would be the theta of the option. After all, the vols for an ITM option 1 week from expiry is radically different from one 1 year from expiry - suffice to say, the vols will be a lot higher for the former, which offers more opportunities for gamma trading (ideally, the option should fluctuate incessantly between ITM/OTM close to the ATM level)."
What do you mean the "vols" will be higher for an option with 1w vs 1yr to expiry? That's not true. The gamma should be higher, but not the vega.
I'm pretty certain vega should be higher as time until expiry increases.
What am I saying? ;)
Yes, I meant to say "the vols for an ITM option 1 week from expiry is radically different from one 1 year from expiry - suffice to say, the GAMMA will be a lot higher for the former, which offers more opportunities for gamma trading (ideally, the option should fluctuate incessantly between ITM/OTM close to the ATM level)".
This would naturally be the case since for a shorter-dated option, the probability of it to reverse its direction is lower than that of a longer-dated option. The VEGA, however, is the reverse (shorter-dated option has lower vols since increasing the vols would not be helped by the limited trading timespan, thereby limiting its potential to be traded in a wider range).
Hope that clarifies. Thanks for pointing that out, Jimbo.
"I'm pretty certain vega should be higher as time until expiry increases."
Exactly a 1yr option has higher vega than a 1 week option
For those who understand Black Scholes.. (Originally Posted: 11/17/2007)
..or the general idea behind option valuation. I am still trying to wrap my mind around the fact that Black Scholes assumes that stock price grows at the risk free rate (or whatever rate it is possible for the counterparty to hedge the option). I understand that if I receive X for selling an option and dynamically hedge it, then Black Scholes would say X should be exactly equal to the cost of hedging it over the life of the option, given volatility. Essentially, the model determines the fair price of options for those who hedge the options.
How do people who buy options without hedging fit into the picture? Clearly, if I buy an out of the money call, I am hoping and expecting the stock to grow at more than simply the risk free rate. If I assume the stock grows at 10% instead of 5%, isn't there a disconnect between how much I value an option versus how much a dynamic hedger would value it? (i.e. I would be willing to pay more).
Thanks for those of you who made it through my rambling.
perhaps this will help you along...
as a buyer, you probably want to focus on the implied volatility associated with the current option price.
there are other things to look for as well (gamma, etc.)
Not sure if this is what you're asking, but of course there is a disconnect bt what you value the option at and what the hedger values it at. You think the option is worth more than you're buying it for but you're taking on risk by doing so. If you're right, you make a profit on the option; if you're not, you lose money.
You'd get better responses if you posted this in the S&T forum.
Yes that's exactly right. The thing is though, because it's possible to perfectly hedge it (in the BS world), that's the no-arbitrage price and hence that is the price under all measures.
Well if you are going to pay more for it, then you are providing the seller (who can hedge it) with risk free profits, i.e. arbitrage (similarly, nobody would sell it to you for less than the BS price).
There are two people selling cans of soda. You are really thirsty. The one guy charges 85cents. The other guy charges 90 cents. Which one are you going to buy from? What if you are REALLY thirsty?
Anytime a payout can be created in mutliple ways, rational people will choose the cheapest way.
It's like saying you think the dollar will strengthen against the yen at the 2yr point...you're not going to accept any worse level than the 2yr fwd, b/c you can recreate it.
through arbitrage pricing in a risk neutral world.
What that means is that, using the soda analogy Jimbo gave above, let's assume there are 2 objects in question:
which from the customer's perspective both are the same; they provide soda. Assume that 1 is more expensive than 2. What I will do then, is to sell a contract of 1 to the customer, and proceed to engage 2. the difference between the two prices is an arbitrage, or riskless profit.
The risk free rate is in the equation due to the fact that at any two points in time, X amt of dollars will be wort Xe^rt t years from now, where r is risk free and t is time in years.
What a dynamic hedger will do is to really rebalance what you call this replicating portfolio which will end up with the same value as the option, thus implying that option can be valued this way.
One of the basic assumptions of Black Scholes is that stock returns are normally distributed. Your assumption of 10% does not fit that of Black Scholes, which uses brownian motion to model stock prices. So the answer is yes, if you assume stock prices are growing at 10% then yes you would value it differently because your assumptions are different from that of the model. Hope this helps.
Black scholes has a number of unrealistic assumptions, in reality by the way. but it's a useful reference point. the assumption is that the underlying process is lognormal.
That's not correct. Why isn't the assumption of the stock prices growing at 10% compliant with the BS world?
dSt/St = 10%dt + sigma.dWt
or even
d(lnSt) = 10%dt + sigma.dWt
Regardless of what the drift coefficient is under the real world probability measure, the price is still the same.
"Regardless of what the drift coefficient is under the real world probability measure, the price is still the same."
This is the great power of black-scholes....the drift drops out.....
Hi I'm not the expert on this, correct me if I'm wrong:
That drift equation you gave, 10% was the mean in the drift, rather than absolute growth of 10%(i.e. instead of drift of exp(miudt+sigmaZ*sqrt(dt)) it should now be 1+10% which is multiplied by the previous stock price, and this violates the normal distribution assumption of stock returns/lognormal distribution in stock prices. I assume his dt is 1 year which is also different from that in Black Scholes. Of course we can reduce his 10% to continuous time.)
So I am saying the deviation of his assumptions will cause the change in his valuation, if what his expectation of the real world is true. The value given by the BS equation only holds if the basic BS assumptions holds, which I assume his assumptions are violating. This is why I say he would value it differently given his expectations. But the world may hold on to BS.
"and this violates the normal distribution assumption of stock returns/lognormal distribution in stock prices"
No it doesn't.
would you care to elaborate?
the main difference is that i see his assumption as instead of having a drift term which is normally distributed he is substituting it with 1+r.
That is to say, instead of stock prices following the wiener process as assumed in BS [ Sexp(miudt+sigmaZsqrt(dt))] he is now assuming stock prices grow at 10% (not as the mean of the normal distribution, but 10% annually) which gives us S*(1+r) where r=10%. A stock growing at 10% annually (again i stress, not the mean of the normal distribution of stock returns) can hardly have it's growth described as normal. His assumption actually implies a constant growth.
I think most of you are thinking of 10% as the mean of the drift, in that case it is debatable.
If you think my interpretation of is assumption is too stiff, looking back, I just might agree with you ;)
I think it's pretty clear that he was referring to a case where the drift coefficient if 10%.
How can we assume that the return on a stock is a deterministic 10%? If that were the case go short the risk free asset (assuming that its growth rate is less than 10%, otherwise reverse the trade) and long the stock: arbitrage.
I just assumed that was his own expectation (which may or may not happen; not deterministic, thus no arbitrage opportunities; though yes i agree that anything not deterministic should not be modeled that way, that my bad). I was thinking when darkxfriend posed the question he did not know what the drift coefficient was.
At the end of the day it would be in which way he wants to adjust for his assumptions. Thanks all for this discussion. Very insightful. Great posts as always.
Wow. Thanks for all the comments, although I'll admit that a lot of it is over my head.
One more question I did have and that seems to have been grazed is:
On my desk (not a trading desk, but we deal with equity derivatives), everybody talks about BS assuming that stock prices grow at the risk free rate. I recently learned about normal distribution of continuous return and lognormal distribution of price at maturity. Correct me if I'm wrong, but is it true that the median of the lognormal distribution is 1 (assuming $1 stock), while the mean is a function of the assumed volatility? From what my desk says, I have long assumed that the mean is supposed to be simply current price multiplied by 1+risk free rate to the power of t where t is time until maturity, but I'm not so sure now, although nobody has ever said volatility affects the mean. At the same time, I don't feel like the guys on the desk were referring to the median.
Let me put this another way. Assume a stock has a volatility of 0%. Isn't the mean of its price distribution always $1? But if its volatility is 80% or so, its mean is much higher. How is that reconciled with the basic assumption that stocks grow at the risk free rate? And does this have anything to do with the notion that volatility erodes the mean? (Admittedly I don't even know what that means).
Would greatly appreciate feedback on this. I should just sit down and read Hull's textbook at some point.
The black scholes model is a derivatives pricing model that takes the inputs: current stock price, strike price, risk free rate, volatility, time to expiration, and dividend yield to estimate a fair price for the derivative. The insight of the Black Scholes model (and what won them the Nobel Prize) was the discovery that you could model the path of a derivative's price based on an assumption of risk-neutral behavior and then discount at the risk free rate. This made derivatives pricing extremely more useful, since there was no need to estimate the proper rate of return and discount rate.
If the option is in the money, volatility is still very much an "issue" because the more volatile the option, it is more likely to go even more in the money or perhaps go the other way. Also, just knowing that an option is in the money tells me nothing about how long the option has till expiration. The main drivers of an options price is volatility and time to expiration. These two factors will always have a major effect on an option's price.
I misunderstood that you meant 1w vega
Thanks all for the input. Your explanations made things much more clear.
But I still don't understand why the MD nodded when I answered that question.... haha
Just to boot, also: higher vol = higher option value = higher cost of carry (in normal interest rate enviornments)
I know you mentioned something about assumptions but that was a bit vague.
You could have brought it further by talking about how Black Scholes differ from practice, eg things like volatility smiles, time structure of volatility, etc
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