When I was young, the pricing of stock options and other derivatives seemed like an obscure black art. Then one day Don Brown showed me a simple example that made everything crystal clear. Today I’ll share an even simpler version of Don’s example.

Imagine a stock that sells for $10 today. A year from now it will be worth either $20 or $5. (Yes, I know that real-world stocks have a wider range of possible future prices. That’s why I called this a **simple** example.) What would you pay for an option that allows you to buy the stock next year at today’s $10 price?

You might think you’d need a whole lot more information to answer that question. You might expect, for example, that the answer depends on the **probability** that the stock price will go up to $20 rather than down to $5. You might expect the answer to depend on how much traders are willing to pay for a given dollop of risk-avoidance.

But the amazing fact is that none of that matters. The only extra bit of information you need is the interest rate.

Let’s assume, for example, that the interest rate happens to be 25%. (Yes, I know that’s unrealistic.)

Now let’s price the option. The key is to focus on my imaginary cousin Jeter, who never buys stock options. Jeter happened to wake up with $12 in his pocket today. Then he went out, borrowed $8, and used his $20 to buy 2 shares of stock.

A year from now, one of two things will happen. Either Jeter will get lucky, sell his 2 shares for $40, use $10 to repay his debt ($8 plus $2 interest), and pocket $30. Or he’ll get unlucky, sell his 2 shares for $10, use that $10 to repay his debt, and pocket $0.

**Edit**: Thanks to those who caught the typos in the above paragraph. I think they’re fixed now.

I, on the other hand, bought 3 stock options today. A year from now, one of two things will happen. Either I will get lucky and use my 3 options to buy 3 shares of $20 stock at a price of $10 each, pocketing a $30 profit. Or I will get unlucky and the stock price will plumment, in which case I will throw my option away and pocket $0.

In other words, Jeter and I are **guaranteed** exactly the same outcome next year. Either the stock price goes up, and we each pocket $30, or it goes down and we each pocket $0. In that strong sense, Jeter’s strategy and mine are perfectly interchangeable.

Jeter’s strategy costs him $12 out of pocket. Therefore my strategy must also cost $12 out of pocket — otherwise, nobody would ever pursue the pricier strategy. Since buying 3 stock options costs $12, the price of a single option must be $4. Problem solved.

Of course the real world presents far trickier scenarios, with stocks that can go up or down by any amount at any instant, and options that can be exercised at the time of your choice. But the trick for pricing these options is always the same: First invent an imaginary cousin Jeter who buys and sells only assets whose prices are already known. Devise a strategy for Jeter that mimics the value of the option under every conceivable circumstance. Devising that strategy can be a difficult technical exercise, but it’s not impossible. Now figure out what that strategy costs, and you’ve got the price of your option.

There are practical pitfalls galore. For starters, how do you account for the cost of trading, or for the fact that some of us have access to better interest rates than others? That’s why there’s still work to be done in this area. But ultimately, there’s just one big idea, and if you grasp this simple example, you’ve mastered it.

I don’t understand why Jeter ends up with 4 stocks when he is unlucky. Am I missing something?

That’s not the only error. Apparently Steve started with more than one set of numbers, and when he copied and pasted them into this post, he mixed his examples.

@Ryan : I think it’s a typo. It should read “sell his 2 shares for $10, use that to pay off his debt, and pocket $0″

When I first read this, I thought there had to be a mistake (besides the typo everyone jumped on :) ) in the part that says “You might expect, for example, that [the option price] depends on the probability that the stock price will go up to $20… But the amazing fact is that none of that matters.”

Surely, I reasoned, if there were a 99% probability that the stock would go up to $20 and only a 1% probability that it would fall to $5, then an option would mean you would (almost) certainly make $10 a year from now, which means, factor in the interest rate and you would pay (almost) $8 for the option today.

The answer to this objection, I think, is that the probability and the risk-averseness of other traders *do* matter, but only insofar as they are *reflected in the current price of the stock*. So in fact it would be impossible to have a 99% probability that the stock goes up to $20 a year from now, because otherwise the stock’s current price would be a lot more than $10 (unless traders were ridculously risk-averse). The current $10 price doesn’t tell you the probabilities of the $5 and $20, and it doesn’t tell you the risk-averseness of traders, but it does restrict you to certain *combinations* of probabilities-and-risk-averseness-of-traders. Is this right?

For example, suppose traders have no risk aversion (so that if the expected payoff of a bet is $x, it is worth exactly $x). The stock is worth $10. If you banked the $10, you’d have $12.50 a year from now. In market equilibrium where banking the $10 and buying the stock are equally attractive, the expected value of the stock would also have to be $12.50 a year from now, which means there’s *exactly* a 50% chance of it being worth $5 and *exactly* a 50% chance of it being worth $20. And that means that your option (to buy the stock a year from now for $10) has exactly a 50% chance of being worth $10 from now, so with no risk aversion, its expected value in a year is $5. Divide out by the interest rate and you would expect to pay exactly $4 for it today.

That shows that when traders have no risk aversion, that forces the probabilities of getting $5 and $20 to be 50/50 (in order to get a present-day stock price of $10), and hence that the current price of the option should be $4. Steve’s Cousin Jeter example is more general and shows that for *all* possible combinations of risk aversion and probabilities (that give you a present-day price of $10), the price of the option should always be $4.

Technically, could the correct answer to the question be something other than $4, since it asks “What would YOU pay for an option that allows you to buy the stock next year at today’s $10 price?”

$4 is what you’d pay if you have exactly the same risk aversion as all the other traders in the world. If you are less risk-averse, then you would be willing to pay more than $4, since the option is slightly more attractive to you if you don’t mind the gamble. And if you’re more risk-averse, you’d pay less than $4.

(Of course this isn’t a very deep insight, since the question “What would you pay for an apple?” could have correct answers from $0 to $100 depending on the wacky food tastes of the person answering the question :) )

Presumably if you ask “What should the market price be for the option?”, the answer to *that* question would be exactly $4.

Wow, you’re even able to make risk-neutral pricing sound simple. I’m impressed!! Good work.

“options that can be exercised at the time of your choice” – the ability to exercise the option prior to expiry does not increase the value of the option.

You should never exercise early because you are giving up the time value of the option….you’re better off selling the option to someone who will pay you the intrinsic value (equal to the profit on early exercise) plus the time value of the option (which you lose on early exercise).

I spent a lot of money to learn the binomial model to price options (at UR to boot) when I could’ve just read this!

If we only knew what to pay for the stock we’d be all set!

But the answer DOES depend on the probability that the stock goes to $20 rather than $5. It’s just that that probability is already baked into today’s stock price of $10, in combination with the interest rate of 25%.

So you need an additional (unstated) assumption: that the market is efficient, so that (1) today’s stock price of $10 does reflect the true probability, and (2) there is no arbitrage opportunity available.

You can even turn the question around. “How do you figure out what a share of Microsoft stock is worth?” “Look at the option price and the interest rate, and you can figure it out.”

I calculate that if the interest was 0%, the price should be $3.33. Is this right?

As has been pointed out, this is the “market price” assuming you have no insider information.

Bennett_Haselton and Phil_Birnbaum are right. It would be more accurate to say that “the probability distribution on its future price is irrelevant except to the extent it influences the current price”. Or, in the language of causal networks (see Judea Pearl), “Current prices screen off future prices”, or “Once you know the current current price, you learn nothing else upon being told the probability distribution on the future price”.

Assume I — being god or Krugman — know the stock will be worth 5. This is not a guess or an opinion; it is certain. (Say the stock is on a bet to find a new integer between 12 and 13 and a five dollar bill). Is the option worth $4 to me? I think not. Why then is the stock worth $10 to others? There seems to be an assumption lurking here that is related to either symmetric information or the value of the option being calculated forinvestors like Jeter not like omniscients such as Paul Krugman.

A number of comments here make use of the assumption that current prices are perfect reflections of our most accurate expectations for the future. The strong form of the efficient market hypothesis. It’s a wonderful assumption, because it yields numbers for everything, and it should even be pretty good a lot of the time.

But it’s often wrong too. Without getting into the debate over how often, I think it’s worth calling out that assumption by name when we make it.

Steve, are you saying that pricing an option should not ever require taking into account the probabilities, or is that just the case for this simple example? I think I understand the example, but I do not see how you could apply your strategy when there are more than two possible outcomes.

The arbitrage argument outlined here serves only to fix the relationship between the option price and the current share price. Attitudes toward risk etc., determine the levels of both securities. If, for example, there is a 99% chance the stock price will rise to $20 a year from now, both the option and current stock price are very undervalued for any reasonable assumption about risk aversion.

Tom- risk neutral pricing has nothing to do with assuming that stock prices are efficient. This stuff has been understood for a few decades. You all should try to understand this stuff instead of immediately concluding that SL is confused.

Sorry for being dense, but why don’t the parameters of the thought experiment influence the option price? For example, if I assume the two outcomes are $30 and $5, instead of $20 and $5, I seem to get different results. Inserting new values into the problem:

A year from now, one of two things will happen. Either Jeter will get lucky, sell his 2 shares for $60, use $10 to repay his debt ($8 plus $2 interest), and pocket $50. Or he’ll get unlucky, sell his 2 shares for $10, use that $10 to repay his debt, and pocket $0.I, on the other hand, bought 2.5 stock options today. A year from now, one of two things will happen. Either I will get lucky and use my 2.5 options to buy 2.5 shares of $30 stock at a price of $10 each, pocketing a $50 profit. Or I will get unlucky and the stock price will plumment, in which case I will throw my option away and pocket $0.

. . .

Jeter’s strategy costs him $12 out of pocket. Therefore my strategy must also cost $12 out of pocket — otherwise, nobody would ever pursue the pricier strategy. Since buying 2.5 stock options costs $12, the price of a single option must be $4.80. Problem solved.But my parameters shouldn’t result in a different option price from Steve’s. I must be missing something in the assumptions.

@Tom: Yes or something like it. The EMH would rule out my counter example. As would the weaker assumptions I suggested. But your point — that Steve is relying on some such unstated assumption — is spot on.

@Anon: neither Tom nor I assume Steve is confused. We argue that he has not stated all the assumptions. I gave a counter example. It can be ruled out of bounds by such things as the EMH or weaker assumptions. But unless it is ruled out the conclusion Steve presents is contradicted by it. (This counter-example refutes a theorem stuff has been understood for decades. You should try to understand these comments before assuming anyone is confused.)

Al V., I think the answer is that the cost of the option DOES depend on the assumptions about what the stock price might be in a year. Nobody is claiming otherwise.

The point of the post, I think, is that you don’t have to figure out the respective probabilities of those outcomes if you know the outcomes themselves, the current stock price, and the interest rate.

I am probably missing something about this example. I hope someone can set me straight . . .

Suppose I have $10 plus some other amount $C, which is the cost of a stock option. My baseline option for this money is to loan it at the 25% rate, which gives me 12.50 + 1.25C dollars in one year.

Instead of the baseline, I could purchase the stock for 10 dollars and loan C dollars at the 25% rate. Two situations can happen in a year:

good: I have 20 + 1.25C dollars, or +7.50 dollars relative to the baseline.

bad: I have 5 + 1.25 dollars, or -7.50 dollars relative to the baseline.

As an alternative, I could purchase an option for C dollars and loan 10 dollars at the 25% rate. Two situations can happen in a year:

good: I have 20-10+12.50 dollars, or 10-1.25C dollars relative to the baseline.

bad: I have 12.50 dollars, or -1.25C dollars relative to the baseline.

To make both ‘good’ situations the same, I need 10-1.25C = 7.50 dollars, or C = $2.

To make both ‘bad’ situations the same, I need -1.25C = -7.50 dollars, or C = $6.

The average of these costs is $4.

Did I make an obvious mistake in logic or math?

If not, is it a coincidence that the average is $4?

If I didn’t make a mistake, then why shouldn’t we average the two costs with respect to their associated probabilities?

Thanks to anyone who can help me learn a little more about this.

(This will likely confirm my practice of relying on other people to manage my investments.)

BTW, you can figure out the cost of the option using probabilities too, if you want.

The stock costs $10. A year from now, you have $20 with probability X, and $5 with probability (1-X). Taking present values, $20 and $5 become $16 and $4 today.

So $10 = $16X + $4(1-X). X equals 0.5.

What’s the stock option worth? Well, a year from now, you have $10 profit with probability 0.5, and $0 profit with probability 0.5. That works out to $5. Taking present value gives $4, which was the answer to the original question.

In response to Thomas Bayes (1:05pm), I think the reason the calculations average to $4 is that in this problem, the probability just happened to work out to 0.5 for each case. I suspect that if you weight the two values by their probability, you get the option value. I’m too lazy to check or prove that, though. :)

Thomas Bayes: I do not see any justification for averaging your 2 and 6 to get 4.

The point is not to find one strategy that makes both good situations the same and another that makes both bad situations the same. The point is to find a *single* strategy that makes both good situations the same AND makes both bad situations the same.

As an exercise, it might be instructive to figure out what happens if the “high” stock price is, say, $30 instead of $20. You need solve for two variables, namely the number of shares Jeter buys and the amount he borrows. Equality of outcomes in the two scenarios gives you two equations in those two unknowns. This gives you Jeter’s strategy; now proceed from there.

In this toy model if I (and I suppose everyone else to) somehow knew the price was going up to $20 then I’d imagine the interest rate would be 100% for the year not 25%. Does the 25% interest rate tell us the probability of it going up?

@ Tom and KenB

Steve isn’t assuming any version of the EMH. The only assumption is the existence of able and vigiliant arbitrageurs. The only assumption then is that prices of interdependent securities are consistent with each other. Say in his example the option was worth only $3. I could turn nothing into something in the following way:

Short 1 share to raise $10, buy 1.5 options for $4.50, and spend the remainder of $5.50 on a bond at 25%. At the end of the period I collect my $6.875 from the bond investment. The remainder of my portfolio has two possibilities:

A. Stock price falls to 5 – I have a short stock position worth negative $5 and the options expire valueless, leaving me with 1.875 after using part of the bond proceeds to close the short position.

B. Stock price rises to 20 – I have a short stock position worth negative $20, I exercise the options providing a positive cash flow of $15. With the $6.875 bond proceeds, I am left with $1.875 after closing the short position.

@Phil Birnbaum, Thanks for the explanation. What you’re saying is that the market has priced the probabilities into the current price. And if the dollars used are (as in my alternative example), $30 and $5, each with a 50% probability, then the current price would be $17.50, not $10, and the option price would be $6.75 (if I did the math right), not $4.80 from my prior example.

So the correct alternate example is:

A year from now, one of two things will happen. Either Jeter will get lucky, sell his 2 shares for $60, use $10 to repay his debt ($8 plus $2 interest), and pocket $50. Or he’ll get unlucky, sell his 2 shares for $10, use that $10 to repay his debt, and pocket $0.I, on the other hand, bought 4 stock options today. A year from now, one of two things will happen. Either I will get lucky and use my 4 options to buy 4 shares of $30 stock at a price of $10 each, pocketing a $50 profit. Or I will get unlucky and the stock price will plumment, in which case I will throw my option away and pocket $0.

. . .

Jeter’s strategy costs him $27 out of pocket. Therefore my strategy must also cost $27 out of pocket — otherwise, nobody would ever pursue the pricier strategy. Since buying 4 stock options costs $27, the price of a single option must be $6.75. Problem solved.And that foots out nicely. Thanks!

@Aaron G: I am unclear about how you are shorting the 1 share to raise $10. What is your precise contract with the other party?

Anon & Aaron —

Please check my comment again. I didn’t say that Steve is assuming EMH. I said that some of the comments make that assumption, and that we should be explicit about EMH when we do use it.

When Steve says an option is the same as so many shares of stock long or short and so many dollars plus or minus, we all understand that. No problem.

But when people say that the option probabilities are baked into the price of the stock, that statement does rely on EMH. The price of the stock is a result of supply and demand, and people’s feelings about future prices are only one of the forces driving supply and demand.

The thing I like about Steve’s example is that all this talk about probabilities being baked into the price of the stock, and the existence of arbitrageurs, and the EMH are all just kerfuffle.

He’s shown that if he’s considering buying 3 options for $12, he can obtain *exactly* the same result using Jeter’s strategy. He made no assumptions whatsoever about probabilities, arbitrageurs, EMH’s. We could be at the peak of a bubble or the trough of a bust, and his argument would still apply. All the other stuff may be true, but is not necessary to prove that the option should be priced at $4.

What if there were three possible share prices in a year’s time? What are the three variables you solve for to find Jeter’s strategy?

Steve,

I just looked this over again, and I missed the key point that the number of shares Jeter buys is different from the number of options you buy. I was making them the same.

When I make them different, then I have two equations and two unknowns, and I get an equation like this:

Cost per Option = X*(a-1)*(I-b)/(I*(a-b))

where X is the original cost per stock share, a is the proportional stock gain, b is the proportional stock decline, and I is the interest gain. (To derive this, I assumed that a is greater than one and b is less than one.) For your example, X=10, a=2, b=1/2, and I=5/4. When I use these values in the equation, I get a cost per option equal to 4, just like you said I would.

The other variable in my equation is the ratio of the number of stocks Jeter buys to the number of options you buy, and this ratio is determined by this equation:

r = (a-1)/(a-b)

which turns out to be 2/3, just like you said it would.

If, as you suggested, I make a=3, then the cost per option is $4.80 and the ratio of Jeter’s stocks to your options is 4/5. Does this make sense?

Harold asked about the case when the interest rate is zero (or I=1). Just as Harold suggested, I get 10/3 for the cost.

Let me know if you see an error here; otherwise, thanks for the lesson.

Ken B —

The solution being described here isn’t precisely option pricing. It’s the cost of synthesizing your own options, should you need them. (But the cost of synthesizing something tends to be intimately tied to how it’s going to be priced in the real world a lot of the time.)

Say we know the share price will be $5 in one year, with certainty. But somehow nobody else knows that, so the stock is currently trading at 10.

Even in that case, I can create three synthetic call options in the same way Steve described for Jeter, by putting up $10 collateral, buying two shares of stock and paying $2 in interest. When the year is up, Jeter’s position will be worthless, just like the $10 calls. It will cost $4 per option to synthesize them that way. Obviously a terrible idea if you plan to hold the options.

BUT what if some dreamer out there is bidding $6 for call options on the stock? Do the option arbitrage folks have to just

sitthere and watch this sucker give his money away to somebody else? No. In theory (and, I hear, to some extent in reality–I haven’t done this), they can sell options to him at 6, and hedge away the risk by creating the same options within their own portfolios, but at a cost of only 4.Now if you really are right that the stock is going to 5, then in principle you don’t need the hedge. You can keep the whole 6. Though your boss might be hard to convince.

Thomas Bayes: I agree with your $4.80, and with your 10/3.

Mike H: If there were three possible share prices, then with the method of this simple example, we’d have to introduce a third asset.

You might think, then, that because in the real world there are infinitely many possible share prices, we’d have to introduce an infinite number of additional assets, making this whole method thoroughly impractical. That turns out not to be the case, because instead of restricting Jeter to choosing some assets and holding them for a year, we can let him use strategies that involve continuously buying and selling stocks and bonds. Since he can adjust his portfolio at every instant, we get enough variation to solve the problem.

For example: Suppose Jeter buys options (making him hope the stock price will rise) and simultaneously sells short contracts (i.e. promises to deliver stocks at a future time at today’s price) (making him hope the stock price will fall). He sells just enough short contracts to make him, on balance, indifferent to whether the stock price falls or rises. At every instant, as the stock price changes, Jeter adjusts his short holdings to keep himself indifferent to what the stock price does in the future. This, for Jeter, is a risk-free strategy, and so must pay the same as lending at the risk-free interest rate (i.e. the rate on a Treasury bond). Set the return on this strategy equal to the the return on a Treasury bond and you can back out the price of the option.

How come no one has said “Black-Scholes” yet? Wikipedia oin “Black–Scholes”: “The fundamental insight of Black-Scholes is that the option is implicitly priced if the stock is traded” See also http://en.wikipedia.org/wiki/Rational_pricing#Options

“At every instant, as the stock price changes, Jeter adjusts his short holdings to keep himself indifferent to what the stock price does in the future”

I presume his holdings are the -dV/dS mentioned in http://en.wikipedia.org/wiki/Black-Scholes#Mathematical_model ?

@Ken B

When I say that I will short the stock, I am referring to a naked short – that is, I take the cash in payment for stock but do not actually deliver it. I would have to cover the short if the buyer decides to sell it, financed by shorting the stock to another party if I want to maintain the short position.

Naked shorting isn’t actually something that your typical market participant can engage in. If you set up a margin account with a broker and want to short a stock, you have to borrow (therefore incurring some cost) the share from another of the broker’s clients who is willing to lend it.

While nobody can actually short stocks with no transaction costs, investment banks and some other market insiders can (usually) arrange short sales at far lower transaction costs than the typical participant. Steve’s example only requires that such insiders generate enough trading volume to put appropriate pressure on the price of the option to push it back to the value consistent with the interest rate and stock price. Legal or other impediments to naked shorting can cause the no-arbitrage condition to relax considerably as transaction costs are much higher when shares actually have to be delivered.

@Aaron G: Thanks. However now I think under the terms of the problem your naked short is not quite allowed. The price of the share is $10. But you are only selling a promise, so the price should be less than 10. However knowing what you plan I will work through this assuming you raise 9.99 rounded to 10 :>

@Aaron G: Yes, this Dutch book argument is very nice. You show at a $3 per option I can make money. That does work. I assume the rest of the argument is anything > 4 loses bucks, anything < 4 makes bucks, ergo the value must be 4.

I’m not sure I stressed Ken B’s point enough in my last comment. In his post, Steve is not talking about option pricing exactly, he’s talking about the cost of synthesizing options fully hedged.

If we knew that the probability of the stock going to 5 was 100%, then we’d know that the $10 calls are worthless. We

couldsynthesize them for $4 hedged, but we could also synthesize them unhedged for much less than that, just interest on the collateral we have to put up in order to write them. And we’d know that their actual value is zero.Under those circumstances, of course it wouldn’t make sense to say that the correct price would be 4. So if we actually know the probabilities of future stock prices, that does affect option pricing in ways not discussed in the post here. This approach is for people who are agnostic about future prices. (Smart people, in other words!)

I am selling envelopes.

In each envelope is either a $5 bill or a $20 bill.

All envelopes are identical.

Current market price is $10.

Current interest is 25%.

I know the value of the envelopes.

Envelopes can be opened tomorrow.

What is the value to ME of an option to buy back the envelopes?

It is clear I think the value is not $4.

What on the other hand is the price the market will charge me to buy the options?

These are different questions.

Say I start buying envelope options.

My purchase will affect the price of the option.

First assume every knows I know the contents. We get a price.

Next assume no-one knows I know the contents. We get a price.

If these prices do not match we have an assumption rather like the Efficient Market Hypothesis lurking here.