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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.

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1… look at question 2.49. You will get an explanation without any mathematics. (Girsanov theorem might be overkill here) –  Were_cat Jun 18 '13 at 23:02

9 Answers 9

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)

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-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?". –  Henry Henrinson Sep 22 at 12:36
@HenryHenrinson, I invite you to provide your own answer if you believe it adds value above and beyond the existing answers. –  Matt Wolf Sep 23 at 3:02
@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. –  UmaN Sep 24 at 12:57

Being on the sell side and selling options you can intuitively think of it like this:

An option is like any other product that is being produced out of ingredients and because of the competitive situation of the producer this is done by the cheapest possible production process.

The ingredients are in a simple (Black Scholes) setting a stock and and a risk free bond. So this is where the name "derivative" comes from: It is derived from other, simpler products (the underlying).

Now to make it even simpler think of a call with a zero strike on the underlying. How would you price such a call? Well, you would price it like the stock (up to a correction for the risk free interest rate)! Here the same question could be asked: Shouldn't it be priced higher with the underlying having higher drift? The answer is no, because this is already included in the price of the underlying! You would kind of double count this effect.

With "real" options the reason is the same, the only thing that changes is that you have to adapt the proportion of the stock when the price of the underlying changes. But the original reasoning stays the same: The drift does not enter into the price of the derivative because it is already included in the price of the underlying.

A very readable intuitive paper on this issue is by a giant of the field, Emanuel Derman. It can be found here: The Boy's Guide to Pricing & Hedging

See also my answer to a similar question here: How does the "risk-neutral pricing framework" work?

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I love this short read you referenced, +1 alone for this one. Shows me time and again that the top researchers in this and related fields all strive for simplicity because reputation is earned by explaining stochastic calculus to an ape rather than how to peel a banana to an astronaut. –  Matt Wolf Jun 19 '13 at 1:39
I take a different view from yours: For me there is a fundamental difference between the price of a derivative and the value of the same. What risk neutral pricing does is assisting in finding the price of a derivative. The value may vastly differ depending on the inputs to the pricing model. That differential for me explains why some people buy, others sell. Of course each participant has a slightly different pricing model though most are surprisingly similar. What differs is the estimation of value. That for me is where buy and sell side guys differ in their view –  Matt Wolf Jun 20 '13 at 5:09
...and which ultimately causes each individual to buy when he thinks price < value and sell if price > value. The pricing algorithm itself is equally applied to buy and sell side demand and supply. Given I traded today on the sell side, and tomorrow I was trading on the buy side, I would apply the exact same pricing models and derivations, nothing changes there. What changes is my risk transfer needs and thus where I see value. As risk seller I see value in the spread I can provide, as risk buyer I see value in the alpha generated from underlying assets vs hedge risk protection. –  Matt Wolf Jun 20 '13 at 5:10
In a sense yes. Though Q != P, the pricing in both, Q and P, must agree because of the no-arbitrage requirement. But assigning a specific probability measure to either pricing or valuation could lead down a dangerous path. Purely intuitively speaking though I like your comparison and agree. –  Matt Wolf Jun 20 '13 at 7:03
This is argument is fine, but it presupposes the existence of continuous self-financing portfolios. Since those do not and will probably never exist in the real world, as an intuitive argument I don't think this works. –  Andrew Dabrowski Nov 5 '13 at 14:45

$$dS / S = \mu dt + \sigma dW \\ \\ dS / S -r dt= \mu dt - rdt + \sigma dW \\ \\ dS / S -r dt= [\frac{(\mu - r)}{\sigma}dt + dW]\sigma \\$$ Then, Girsanov tells us that, as long as the risk premium is bounded from below, we can write $[\frac{(\mu - r)}{\sigma}dt + dW]\sigma$ as $\sigma d\tilde{W}$ where $\tilde{W}$ is simply another brownian motion with mean 0 and variance equal to its time step. Hence, $$dS / S = rdt + \sigma d\tilde{W} \\$$

Another way to think about this is that, in the Black-Scholes world of option pricing, markets are assumed to be friction-less and hence, all assets can be perfectly hedged. If we can perfectly hedge our underlying asset, then in theory, it should have no volatility and should grow at the risk free rate, r.

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+1 Great answer. –  SRKX Jun 18 '13 at 15:49
Did you mean to say that in a world where we can perfectly hedge the risk premium has to be equal to the risk free rate? I am asking because the asset clearly has volatility even after changing measure to a risk neutral probability measure. In fact volatility is identical before and after. –  Matt Wolf Jun 19 '13 at 0:45
I'm puzzled by something: unless $\mu=r$ the mean of ${(μ−r)\over σ}dt+dW$ is not 0, so how can it be equivalent to anything with mean 0? –  Andrew Dabrowski Nov 5 '13 at 14:12
@AndrewDabrowski: Check out the Girsanov Theorem. It lets you transform one brownian motion into another. –  fbt Apr 22 '14 at 22:56

Recently I came across an interesting intuitive explanation:

Suppose driftless market. Market price is 105, strike price is 100. Call option costs 8, put option 3. (intrinsic value of call is 5, time value of both is 3)

Now the market starts drifting upwards massively. You say, that you would probably price call higher, e.g. at 10. Would you also price put higher? Probably not, since it's less probable that it will end up in the money. You would probably price put even lower, e.g. at 2.

But in that case, you clearly violate put/call parity, and it's possible for me to buy the call, and completely hedge it with the put, some cash and some stocks and earn riskless profit from you. Due to put/call parity, the price of put must rise together with price of call (they must have the same time value), which is inconsistent with dependence on drift.

I disregarded interest rates for clarity.

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This is basically Derman-Taleb's argument that put-call parity implies the BS equations. I like it! But Merton apparently did not, in fact he sicced some attack dogs on DT. Anyone more knowledgeable care to step in and arbitrate? –  Andrew Dabrowski Nov 5 '13 at 14:09
The market moving is not the same as there being sensitivity to the underlying drift. You're talking about sensitivity to the underlier price. OP asks for: "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?" –  Henry Henrinson Sep 22 at 13:46

If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for high volatility stocks. You are able to bound your losses while allowing for potentially large gains. If a stock has a super high rate of return (several times the volatility, for example) then there is almost no chance of the stock price going down. So an insurance policy is not very valuable. We are better off just buying the amazing stock.

Our intuition tells us that if $\mu$ is high, the price of the option should be high because it is more likely to finish in the money. At the same time, our intuition tells us that the option (insurance policy) is less valuable because there is less chance of the stock finishing out of the money.

The math tells us that $\mu$ actually just doesn't affect the price of options.

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Starting with two Differential Equations (one has stochastic term)

$ \frac{dS_t}{S_t}=μdt+σdW_t$ $ \frac{dD_t}{D_t}=-rdt$

I am assuming $\mu$, $\sigma$, and $r$ to be constant as is done in most introductory financial math courses and otherwise it would be more difficult to solve. Notice: $D_t = e^{-rt}$ is the solution to the ODE above which is our continuous discounting factor. $$-$$$$-$$ Remember this is theoretical and quite different from reality

Intuition Why should $D_tS_t$ be a martingale? To think about this simply and with a business sort of mindset. One way to think about it is, is to say that the discounted stock price should not move because all the information is already priced into the stock. This is the efficient markets hypothesis.

Since we have given some possible idea (there is a more difficult arbitrage argument involving continuous hedging as well) about why $D_tS_t$ should be a martingale.

Back to the Math

We want $D_tS_t$ to be a martingale. Let's try to make it so.

$d(D_tS_t) = D_tdS_t + S_tdD_t$

$= D_tS_t(μdt+σdW_t) + S_t(-rD_t)dt = D_tS_t[(μ-r)dt + σdW_t]$

Well let's choose a $dW_t = d\tilde{W}_t - \theta dt$ such that the drift term is eliminated in $D_tS_t$ so that it becomes a martingale (all Ito processes w/ no drift are martingales).

$\theta = \frac{μ-r}{σ}$ leads to

$d(D_tS_t) = D_tS_t[(μ-r)dt + σ(d\tilde{W}_t-\frac{μ-r}{σ}dt)] = σD_tS_tdW_t$

Now it's a martingale !


$\frac{dS_t}{S_t} = μdt + σdW_t = rdt + σd\tilde{W}_t$

$\therefore \frac{dS_t}{S_t}=rdt+σdW_t$ under risk neutral measure

Much of the time we think in default by the risk neutral measure. So we may ignore Girsanov's change of measure from the physical measure to the risk-neutral measure. Not that anyone actually knows the physical measure anyway.

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"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?"

All these other answers are focusing on the wrong aspect of the question - it is true that the maths makes the drift drop out from the BS PDE, but that doesn't explain why intuitively that feels wrong.

The main reason for this is that you are confusing the investment thought process with the pricing thought process.

To understand it better, look at it this way: imagine you have two stocks $S_1$ and $S_2$, with different underlying Brownian motions, same sigma, but different drifts $\mu_1$ and $\mu_2$ (say $\mu_1 > \mu_2$). This does not introduce arbitrage, as the underlying Brownian motions are different. Since everything but the drifts is the same, the BS PDE ends up being the same, so

  1. the call options on $S_1$ and $S_2$ have the same price, which is the same as saying the drift has no impact on the price.

Your intuition, though, tells you that:

  1. the call on $S_1$ should be worth more because there's a higher chance of it expiring in the money (equivalent to saying the drift matters).

Both 1 and 2 are correct, and there is no contradiction between the two. 1. is given by the usual BS logic, where the price is set such that there is no arbitrage and the derivative is perfectly hedged. 2. is given by the process followed in the investment process where what matters is the trade off between risk and reward (or the market price of risk). Because of hedging, the option price does not care about what the trade off between risk and reward is in the market, it is oblivious to any market prices of risk. Since the market price of risk is directly tied to the drift, this should make it clearer why drift does not matter in this world.

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+1: This is why some authors differentiate between option pricing and option valuation. The first being on the sell side, the latter on the buy side. –  vonjd Sep 23 at 18:21

If someone wants simple intuition, here is what happened to the drift. It did go into the formula, believe it or not, but it came into the B-S math sort of in two ways so it cancelled out in the end.

It disappears because Black-Scholes assumes people may have different preferences for risk, but at least everyone is consistent on their own preference.

There is a drift in Black-Scholes. There needs to be some way to say how much return (or drift) you personally must get to take a certain amount of risk. More accurately, we must know how much return above the risk-free rate (or risk premium) you must get to accept one $\sigma$ of return risk.

So, imagine we had a formula that required us to somehow calibrate it for your personal tastes. How do we calibrate you, then? If you want to buy an option on stock A, we need to know how much excess return you need to suffer the risk of a stock just like stock A. If we could find another stock, B, that had identical return and risk prospects, then somehow we could "plug in" that stock's numbers to the formula for you.

But wait - why look for an identical stock? Stock A is out there and we could calibrate you with it. Stock A is the very stock to use then!

Sounds a bit fishy? Not really. Presumably, you are neither buying nor selling stock A today because you don't believe it is either overvalued or undervalued. Thus, whatever you believe in your head about the risk and return to price an option on A can use stock A as the calibration. Imagine doing just the drift term: if you think stock A is going up 15%, well, then, 15% is the right drift to use for stock A options for you. As long as you are consistent it all falls out and we don't need the drift.

This makes more sense (and this is the one of the hedges that keeps getting talked about) if you imagine that you think you have secret information that a company's stock will jump 10% later today. If you really though that you would think the stock is really, really undervalued right now, true? But, you could also buy a call and sell a put that expires tomorrow with identical strikes. That would also go up 10% later today if you are right!

If you try the example I just gave in your head, you can see that, if you are right, the call options are woefully undervalued and puts overvalued) by 10%. So, the accumulation of everyone else's preferences and beliefs - which go into both the stock and the options prices - are very different from yours. You, in fact, think that stock A has a lot more return at very low risk to occur compared to everyone else...

It doesn't spit out by giving you your own price for the option (we all share one price), but it gives a price based on a collective estimate of drift - though, oddly, we don't know what that estimate is. However, option prices would seem too low if you estimate a higher drift or lower volatility, so you would load up on options.

It definitely can fall out of the math a couple of ways, via risk-neutral pricing and Girasnov's theorem, or via a Taylor series expansion that can't ignore one of the second order terms due to quadratic variation, etc. It can also be thought of via a hedge, but that is also hard to understand. None of these is intuitive.

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Everyone is giving the same type of is a different perspective. Put call parity can actually be violated in special cases where a positive drift is established ,which is equal , mathematically speaking, to reflected Brownian motion about some floor value. I've seen a few options that have a risk-free profit that exceeds the interest rates. For example, the Jan 2017 Berkshire Hathaway BRK/B ATM 130 dollar options slightly violate put call parity with a discrepancy of $1. Someone could theoretically go long the underlying, sell the call, and buy the put to get a guaranteed return. But this would tie up a lot of capital , so a fund believes they can get better return elsewhere, the inefficiency can theoretically remain open provided the opportunity cost remains high enough. A possible reason why the discrepancy exists into the reflected Brownian motion, in that Warren Buffet has made pledges to not let the stock fall below its 'book value' which acts as a reflected barrier. Black Scholes Formula makes the assumption stocks can go zero (or very close to it), which in the case of BRK is very unlikely. If the barrier is very close to zero and the expiration is near, the violation would too small to notice and within the rounding error. This barrier concept could also explain why Warren Buffet sold 100-year puts on index futures, believing that BS were overpricing them on mathematical premise the S&P 500 could go to zero. If the S&P 500 has a reflecting barrier of even 5% its present price, the 100-year puts are way overpriced according to BS. You would need nuclear war or something awful like that to get the S&P 500 to fall to 5% its present value, in which case the counter-party would probably default anyway.

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