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.

  • 1
    $\begingroup$ amazon.com/Heard-Street-Quantitative-Questions-Interviews/dp/… look at question 2.49. You will get an explanation without any mathematics. (Girsanov theorem might be overkill here) $\endgroup$ Commented Jun 18, 2013 at 23:02
  • $\begingroup$ Why drift? If a $W_t$ drifts then it is no longer a martinagle, contradict with the fundamental assumptions. $\endgroup$
    – High GPA
    Commented Jun 23, 2017 at 3:46
  • $\begingroup$ @Lcrmorin is there any way for you to share an online pdf of this or post the response as I can't seem to access it. $\endgroup$
    – math111
    Commented Feb 11, 2019 at 12:57

16 Answers 16


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)

  • 6
    $\begingroup$ -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?". $\endgroup$ Commented Sep 22, 2015 at 12:36
  • 2
    $\begingroup$ @HenryHenrinson, I invite you to provide your own answer if you believe it adds value above and beyond the existing answers. $\endgroup$
    – Matt Wolf
    Commented Sep 23, 2015 at 3:02
  • 1
    $\begingroup$ @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. $\endgroup$
    – UmaN
    Commented Sep 24, 2015 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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Matt Wolf
    Commented Jun 19, 2013 at 1:39
  • 1
    $\begingroup$ 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 $\endgroup$
    – Matt Wolf
    Commented Jun 20, 2013 at 5:09
  • 1
    $\begingroup$ ...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. $\endgroup$
    – Matt Wolf
    Commented Jun 20, 2013 at 5:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – Matt Wolf
    Commented Jun 20, 2013 at 7:03
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2013 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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Matt Wolf
    Commented Jun 19, 2013 at 0:45
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Nov 5, 2013 at 14:12
  • $\begingroup$ @AndrewDabrowski: Check out the Girsanov Theorem. It lets you transform one brownian motion into another. $\endgroup$
    – fbt
    Commented Apr 22, 2014 at 22:56
  • $\begingroup$ This hits the quintessence but it is formulated in a misleading way: $\tilde W$ is a Brownian motion with respect to a different probability measure than $W$. $\endgroup$ Commented Apr 17, 2022 at 10:50

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.

  • 1
    $\begingroup$ 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? $\endgroup$ Commented Nov 5, 2013 at 14:09
  • $\begingroup$ 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?" $\endgroup$ Commented Sep 22, 2015 at 13:46
  • $\begingroup$ I don't understand this argument, if the market drifts upwards massively, it is very likely that the forward price will change, and since call-put parity tells us (C-P)=F (assuming zero rates), why couldn't 8-3 = 5, move to to 10-2 = 8. $\endgroup$
    – Quantuple
    Commented Mar 23, 2016 at 11:41
  • $\begingroup$ @HenryHenrinson : I believe I am answering just that. Rephrased - I assume, that any type of dependence on drift should move call and put prices in opposite directions (one option being more probable to be in the money than other). But that violates CP parity, because for constant price of underlying, all changes to call and put prices have to happen in same direction. $\endgroup$
    – airguru
    Commented Apr 14, 2016 at 14:32
  • $\begingroup$ @Quantuple : Yes, but we should assume everything else being equal. So imagine two markets with the same volatility, one non-drifting and just randomly entering price F, and second one drifting and just passing the same price F at this moment. Question is, why these two situations yield the same option prices even though it seems that either calls or puts have higher probability of ending up in the money for drifting market. I tried to rephrase it again in previous comment. $\endgroup$
    – airguru
    Commented Apr 14, 2016 at 14:36

"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.

  • 2
    $\begingroup$ +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. $\endgroup$
    – vonjd
    Commented Sep 23, 2015 at 18:21

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.


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.



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.


The key is the replicating portfolio. Let’s make one step back: a forward is the price paid today to own the underlying asset at a certain time ‘t’ in the future. For pricing a forward we use today’s price, subtract the present value of all future cash flows, add the interests accrued until ‘t’. Any other price would create an arbitrage (free lunch), in other words we could create a self financing ‘static’ portfolio that will generate a profit. In real life the replicating portfolio may incur in additional costs, e.g. the asset can be shorted but for a fee or the interest has a bid offer spread, however the forward has to be within a range.

Let’s suppose an equity index is trading at 100 with a 2% dividend yield and a 2% flat interest rate, we would expect the, say 10 years, forward to be 100. Empirically we know that the equity risk premium is between 3 and 6%, so our real expectation would rather be 130+, but we have to carry risk for 10 years. In a risk neutral environment the drift has to drop out.

When it comes to options the same concept applies, if and only if a ‘dynamic’ replicating portfolio is possible continuously at any time then the drift can drop out. No need to say that this is not possible in real life because:

  • a. the markets are closed at certain times, hence subject to opening gaps
  • b. there is a bid offer spread that has a cost and reduces the hedging frequency
  • c. the size on the bid or offer is limited.

No need to mention that many other hypothesis are violated: rates and volatility are not constant over time, nor independent with the underlying diffusion process.

Why the BS model is so popular then? Because it is elegant and simple, hence easy to tweak. Being the trader unable to capture the drift, he will increase the implied volatility to address that bias (although wrong that is the only legitimate way to offset it) contributing to another well established phenomenon called ‘volatility risk premium’, the fact that ATM implied volatility is generally higher than historical volatility. The longer the maturity of the option, the greater the bias, the higher its implied volatility.

I hope the reader will forgive the absence of any maths, my experience as an option trader is that maths sometimes prevent critical thinking.


For intuitive answer it's better to consider simple binomial model with drift and no interest rates.

  • Lets say your stock today is 2\$ (and equal to your call option strike)
  • Tomorrow with very high probability p (say 0.9999) it goes up to 3\$ and with very low probability (1-p) = 0.0001 it goes down to 1\$ i.e. very high drift up

  • seller of this option will want to hedge his risks. It happens that he can do it by constructing portfolio of the option minus some amount of stock $Call - \Delta S $

  • Expected value of portfolio tomorrow is

$ (1\$-\Delta3\$)p + (0\$-\Delta 1\$) (1-p) = p(1-2\Delta) = 0.9999(1-2\Delta) - \Delta$

by choosing $\Delta$ to be 0.5 seller can eliminate dependence on this high probability completely (and thus removing the high drift) and make his future portfolio worth $\Delta \$ $ whatever the probability to go up (i.e. drift).

This choice of delta happens to be $ \frac{\delta Call}{\delta S} $ as prescribed by Black Scholes.

i.e. if probabilities are higher it is true the option will be worth more, but it does not matter because the seller of the option eliminated his risk(to payout you more) by hedging and choosing correct $\Delta$.


I believe the simplest way to understand this intuitively is to look at the case of the forward contract instead of an option. If the drift of the stock is higher should not the forward be higher ? No of course because the forward can be statically replicated with a stock and a bond. Therefore the forward contract will grow at the risk free rate with its expiry date and not the stock drift.

Note that repo and dividend impact the forward though because those are actual cash flows which cannot be just replicated at inception so they impact the forward rate.


Everyone is giving the same type of answer..here 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.


Say the stock follows geometric brownian motion as you said

\begin{equation} dS = \mu Sdt+\sigma Sdz \tag 1 \end{equation}

Then Ito's Lemma says that the process followed by a function $c$ of $s$ and $t$ (such as the price of a call option) is

$$dc = \left( \frac{\partial c}{\partial S}\mu S + \frac{\partial c}{\partial t}+\frac12 \frac{\partial^2c}{\partial S^2}\sigma^2S^2 \right) dt + \frac{\partial c}{\partial S}\sigma Sdz \tag 2 $$ with the same source of uncertainty $dz$

We can then create a portfolio $\Pi$ consisting of

  1. $-1$ derivative
  2. $+\frac{\partial c}{\partial S}$ shares

The value of this portfolio is then $$\Pi = -c + \frac{\partial c}{\partial S}S \tag 3 $$

The change in the value of the portfolio in some small time interval $dt$ is then $$d\Pi = -dc+\frac{\partial c}{\partial S}dS \tag 4 $$ substituting $(1)$ and $(2)$ into $(4)$ gives $$d\Pi=\left(-\frac{\partial c}{\partial t} -\frac12 \frac{\partial^2c}{\partial S^2}\sigma^2S^2\right)dt \tag 5$$

Note that this portfolio is risk free because the $dz$ term disappeared. (More technically, this portfolio is delta neutral) Assuming a market without arbitrage, such risk free portfolio must earn a risk-free rate of return,

$$d\Pi = r\Pi dt \tag 6 $$ where $r$ is the risk free interest rate

substitute $(3)$ and $(4)$ into $(6)$ and we have $$\left(\frac{\partial c}{\partial t}+\frac12 \frac{\partial^2c}{\partial S^2}\sigma^2S^2\right)dt = r\left(c - \frac{\partial c}{\partial S}S\right)dt $$

so that

$$\frac{\partial c}{\partial t}+\frac12 \frac{\partial^2c}{\partial S^2}\sigma^2S^2 \frac{\partial c}{\partial S}S = rc$$

which is the Black-Scholes differential equation. Solving for c gives the Black-Scholes formula, but that's for another day. The key takeaway is that the $\mu$ term is hedged away when we create the delta neutral portfolio $\Pi$

  • $\begingroup$ I find your explanation of explicitly showing that the drift disappears from the hedge portfolio dynamics very intuitive. However, you reproduced the original Black and Scholes (1973) derivation which has an error (that happens to average out to zero). Your portfolio is always short one unit of the derivative and thus in all interesting cases not self-financing. See for example Section 3.1.10 in Musiela and Rutkowski's "Martingale Method in Financial Modelling" for details on this. $\endgroup$ Commented Nov 10, 2016 at 3:30

If drift is driven by earnings retention policies then the value of an option does depend on drift! I'm 99% on the following reasoning and would welcome input from others to tighten this up.

Consider what happens with Save Co., a hypothetical company that owns a pile of cash sitting in a savings account earning 1% APY. Suppose Save Co. is required to leave it's cash in this savings account according to it's by-laws so that management has discretion over whether to pay dividends and nothing else. Assume interest rates do not change over time.

To further simplify matters consider two scenarios. In scenario A management regularly pays out all earnings out as dividends. In scenario B management pays no dividends and allows interest to compound.

A share of stock in Save Co. will be worth more under scenario A than under scenario B. The reason is that investors can get a return above 1% by reinvesting their earnings in US treasuries (ignore tax considerations).

On the other hand a 10-year, slightly out of the money call option will be worth more under scenario B than under scenario A. The reason for this is that the cash pile will accumulate under scenario B and therefore earnings will increase over time so that the stock will be worth more after 10 years under scenario B but will be worth exactly the same after 10 years under scenario A.

This example is borrowed from Warren Buffett's 1985 letter to shareholders, the section titled "Three Very Good Businesses (and a Few Thoughts About Incentive Compensation)".


As several responses mention, the easiest to see it, is the replication. Please allow me not to use BS formula for options, it is easier to price a forward, and it will contain the important effect that you are trying to understand.

If I enter into a forward contract with you, I am obligated to give you 1 stock for $\\\$K$ at time T (both the exchange of the stock and payment of $\\\$K$ happens at time T). What is my cost today to be sure I can fulfill my obligation at time T?

I am going to buy the stock today for $\\\$X_0$ and receive $\\\$K$ at time T from you. To be able to do this, I borrow $\\\$X_0$ from the bank today that I will repay with interest at time T. My total cost at time T is (if today is t=0):

$$ X_0 e^{r T} - K $$

This total cost is the money I have to pay the bank at time T, minus the money I receive from you at time T. Therefore the cost at time t=0 is:

$$ e^{-r T}\left(X_0 e^{r T} - K\right) = X_0 - K e^{-r T}. $$

What I mean by the cost at time t=0, is that if I receive this money at t=0, I can put it into a bank account to grow with interest to $X_0 e^{r T} - K$ by time T to cover my costs at time T.

So, it doesn't matter what the drift of the stock is, the price I need to charge you at t=0 is $X_0 - K e^{-r T}$. Even though a higher drift stock is more likely to land above K, this formula is independent of the drift, it only depends on the interest rate r, and the t=0 price of the stock. If someone offered to sell it for a lower price, I can make money on it (arbitrage) by buying that contract and shorting the stock. If someone offered it on a higher price, they will make money on it (arbitrage) as they ask for more than their cost. So the fair (arbitrage free) price of this contract is exactly $X_0 - K e^{-r T}$ which is independent of the stock's drift. The price is always the cost to replicate the contract's payoff.

As soon as you understand this concept, the next interesting question is, why does the price of a call option depend on the volatility? And the short answer is, in order to replicate the payoff of the call option, your cost to do it depends on the volatility.

P.S.: These forward contracts are typically structured in practice in such a way that no cost occurs at time t=0. So they choose K to be $X_0 e^{r T}$ which is also called the forward price. Note, that the forward price is not the price of the forward contract, which is zero, but it is the choice of the strike that makes the contract worthless at t=0 (called the fair strike). I also assumed the borrowing and lending rate are both r, but this is only for the sake of simplicity, and it becomes irrelevant as soon as K is chosen to be the forward price.

In Black and Scholes derivation of their PDE (Journal of Political Economy, Vol. 81 No. 3, 1973, p.642) they tacitly assumed $w_1 = 0$. This resulted in the disappearance of drift from their PDE. Correcting for that ommission, restores drift to its rightful place in the (now more complicated) PDE that results. For more details see my answer to my question "Black-Scholes PDE derivation."

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.