Why Drifts are not in the Black Scholes Formula
This question has puzzled me for a while.
We all know geometric brownian motions have drifts $\mu$:
$dS / S = \mu dt + \sigma dW$
and different stocks have different drifts of $\mu$. Why would the drifts go away in Black Scholes? Intuitively, everything else being equal, if a stock has higher drift, shouldn't it have higher probability of finishing in-the-money (and higher probability of having higher payoff), and the call option should be worth more?
Is there an intuitive and easy-to-understand answer? thanks.
http://www.amazon.com/Heard-Street-Quantitative-Questions-Interviews/dp/0970055234 look at question 2.49. You will get an explanation without any mathematics. (Girsanov theorem might be overkill here)
Why drift? If a $W_t$ drifts then it is no longer a martinagle, contradict with the fundamental assumptions.
Here couple pointers that may make it clearer:
Drift can be replaced by the risk-free rate through a mathematical construct called risk-neutral probability pricing.
Why can we get away with that without introducing errors? The reason lies in the ability to setup a hedge portfolio, thus the market will not compensate us for the drift above and beyond the risk free rate under risk-neutral probability pricing.
As long as such hedge exists and couple other conditions are met (please look up Girsanov's Theorem) we can introduce a risk-neutral measure so that when applying it to the differential equation and through application of Ito Calculus the drift term vanishes, which greatly simplifies the underlying math.
It is not easy to come up with a way to reliably measure drift as it is not well known and hard to estimate in reality. Similarly, real probabilities are different for every underlying asset and are difficult to estimate because investors require different risk premiums for each and every asset. Hence, being able to take the detour through risk neutral pricing greatly simplifies the derivation not only mathematically but also intuitively. Please note that the volatility of the risk neutral paths and paths under real-world probabilities are identical, what differs is the drift term vs risk-free rate.
Keep in mind that not all derivative functions can be derived through risk-neutral pricing, in fact I venture to estimate that less than 20% of all traded derivatives by investment banks can be priced through risk-neutral pricing (in terms of number of different instruments not trading volume).
In order to understand the technical reason of the elimination of drift through risk-neutral pricing you need to walk through the mathematics and one of the cleanest and most intuitive approaches I have seen was in Steven Shreve's book, Stochastic Calculus for Finance II, page 218-220 (2004 edition)
-1, this completely ignores the question "Intuitively, everything else being equal, if a stock has higher drift, shouldn't it have higher probability of finishing in-the-money (and higher probability of having higher payoff), and the call option should be worth more?".
@HenryHenrinson, I invite you to provide your own answer if you believe it adds value above and beyond the existing answers.
@HenryHenrinson no it doesn't ignore that question. The key is the assumption of the possibility to set up a hedging portfolio. If that is possible, then there is no risk (in that hedged portfolio) no matter how risky the stock is. There are countless of papers discussing the problems with dynamic hedging.