Martingale Representation: Why Hedging Is Possible at All
The martingale representation result says a martingale driven by one Brownian motion can be written as its starting value plus an accumulation of bets on that same Brownian motion. A payoff can therefore be reproduced by trading rather than only valued. The result says the bets exist at every moment. The result never says what they are.
Two halves of that are doing entirely different jobs, and separating them is most of the work of this guide. The first half is a statement about shape: a certain kind of random quantity turns out to be a fixed number plus a running total, and nothing else. The second half is a statement about what the result withholds. Almost every misreading of this theorem comes from hearing the first half and forgetting the second, and what that misreading costs is not a wrong figure written down somewhere. The cost is work planned as though the difficult part had already been finished by somebody else.
What does the result actually say, before any notation?
Start with something that has no randomness in it at all. A household electricity meter has a reading today and a reading in a year. The difference between them is not a mystery: it is everything that flowed through the meter in between. The reading in a year is the reading today plus the accumulated flow, and that sentence is true whatever the flow did, whether it ran steadily or spiked every evening.
Now make the flow random. A quantity that wanders unpredictably has no reason, in general, to be describable that neatly. Such a quantity could be pushed by many separate causes at once, could drift, and could jump. The martingale representation result identifies a class of random quantities for which the meter sentence survives intact, and it is a surprisingly large class.
The result says this: if a quantity is a martingale, and if the only randomness anywhere in the situation is one Brownian motion, then that quantity equals its starting value plus an accumulation of bets placed on that same Brownian motion. The bets have a size at every instant. The size is allowed to depend on everything known up to that instant, so it can react to the whole history. The size cannot look forward, and that restriction is what stops the statement from being trivially true of anything.
| \(M_t\) | the martingale, its value at time \(t\) |
| \(M_0\) | its starting value, one ordinary number known today |
| \(H_s\) | the bet size at time \(s\), known by time \(s\) and never earlier |
| \(\tilde{W}_s\) | the driving Brownian motion under the risk-neutral measure Q |
| \(\int_0^t\) | the accumulation of all those bets from the start up to time \(t\) |
| \(T\) | the horizon, one year on every worked instance here |
Two conditions ride along with that statement and neither is decoration. The bet size must be adaptedKnown by the time it happens, never earlier. A quantity that needs tomorrow's reading to be worked out today is not adapted., meaning it is worked out from information already available. And the bets must not be so wild that the accumulation stops making sense, a condition on the expected total of their squares. Drop either and the sentence stops being a theorem.
| \(\mathbb{E}^{\mathbb{Q}}[\cdot]\) | the average taken under the risk-neutral measure Q |
| \(H_s^{\,2}\) | the square of the bet size at time \(s\) |
| \(\int_0^{T} \cdot\, ds\) | the total of those squares across the whole horizon |
| \(<\infty\) | finite, which is the whole content of the condition |
Put the invented worked instance underneath it so the shape is not abstract. The standard processThe single invented traded quantity this reading order runs on, starting at Rs 100/-, with a volatility of 20 per cent a year. starts at Rs 100/-, carries a volatility of 20 per cent a year, and sits in a world with a risk-free rate of 5 per cent a year. Take the at-the-money contract on it, strike Rs 100/-, one year. Under the pricing rule, the discounted value of that contract is a martingale, and its starting value is Rs 10.450584/-. On the one path this reading order draws, the locked path, that discounted value finishes the year at Rs 2.896925/-. The representation says the whole of the Rs 7.553659/- fall in between is the running total of a schedule of bets.
What does it mean to accumulate bets on a source of randomness?
The word bet needs pinning down before it misleads anybody. Here it means a size, a number of units held against the driving randomness, and nothing about intent, conviction or wagering. The word carries the same sense in which a bathroom scale has a sensitivity: turn the dial by one notch and the reading moves by so much. Nobody is being told to hold anything.
The accumulation itself is built the way any total is built. Chop the year into steps. In each step, take the bet size as it stood at the beginning of that step, multiply it by however far the driving Brownian motion moved during the step, and add the products up. Then make the steps finer and finer and see what the totals settle to.
| \(t_i\) | the start of step \(i\) in a partition of the interval into \(n\) pieces |
| \(H_{t_i}\) | the bet size fixed at the start of the step, never at the end of it |
| \(\tilde{W}_{t_{i+1}} - \tilde{W}_{t_i}\) | how far the driving Brownian motion moved during the step |
| \(n\) | the number of steps in the partition, made larger without limit |
The detail that carries the whole construction is that the bet size is fixed at the start of the step, before the driver has moved. Fixing the size before the driver moves is what makes it a bet rather than a report. If the size could be chosen after seeing the move, any total at all could be manufactured, and no theorem about a fair game could survive it. The machinery that makes this limit rigorous is the Ito integral, set out under Ito calculus.
Look at what one step actually does. On the locked path, in the second month, the driving Brownian motion moved by 0.461880. The integrandThe bet size at each moment, which is the quantity the accumulation is built out of. at the start of that month stood at 11.217065. Multiply them and the step contributes plus Rs 5.180941/- to the running total. The product is one term in a sum of twelve, and the theorem says that as the steps shrink, this sort of sum is the entire story of the martingale.
One sentence separates this result from the one line summary people carry away from it. The bet size is not a number. The bet size is a whole schedule. A step counter carried through a day makes the difference plain. The day ends with one figure, say eleven thousand steps, and that figure says nothing whatever about the pace at four in the afternoon. The pace at four in the afternoon is a different kind of object from the day's total, and the representation result is a statement about the pace, not the total.
The result produces a bet size. Is that a single number or a schedule?
Which condition is doing all the work here?
Statements of this theorem usually carry a clause that reads like housekeeping, sitting between commas, easy to skim past. The clause says the information is the one generated by the Brownian motion. Far from housekeeping, that clause is the entire result.
The generated filtrationThe information built out of the driving Brownian motion and nothing else, so that knowing the driver up to a time is the same as knowing everything up to that time. means this: at any moment, everything that is known is exactly what watching the driving Brownian motion up to that moment reveals. No more, and no less. There is no other dial in the room, no second gauge, no separate reading anybody could take.
| \(\mathcal{F}_t\) | the information available at time \(t\) |
| \(\sigma(\cdot)\) | the collection of everything answerable from what is inside the brackets |
| \(\tilde{W}_s : 0 \le s \le t\) | the whole history of the driving Brownian motion up to time \(t\) |
Once the information is exactly what one driver generates, everything random anywhere in the situation is a question about that driver, and bets on the driver can therefore reach it. The intuition runs in one sentence: bets on a source of randomness can reproduce anything whose randomness came from that source. The theorem is the precise version of it, and the precise version is genuinely hard to prove, but the clause carrying the content is the one about the information.
Turning it around shows the force of the condition. Suppose the information were larger. Suppose there were a second gauge in the room that also moved unpredictably and that the payoff cared about. Then a quantity could be known at time t without being a question about the first driver at all, and no schedule of bets on the first driver could rebuild it. The clause between the commas is what shuts that door, and a second source of randomness opens it again.
Which condition is carrying the result?
Why does this make a payoff reproducible rather than only valuable?
Valuing something and reproducing it are different achievements, and the gap between them is the gap between a number and a schedule. Valuing the at-the-money contract on the standard process gives Rs 10.450584/-. The figure answers a question about today, and says nothing at all about what anybody would have to do tomorrow, or in the ninth month, or on a path that visits Rs 91.65/-.
ReplicationReproducing a payoff exactly by holding a changing quantity of something through time, rather than merely putting a value on the payoff. is the second achievement. Replication asks whether the payoff at the horizon can be arrived at from Rs 10.450584/- by adjusting a position through the year, with no top up and no shortfall on any path the process takes. Reproduction is a far stronger claim than a valuation, and it is the claim the representation result underwrites.
The link is short once the pieces are in place. Under the pricing rule the discounted value of the contract is a martingale. The information in this setting is what the driving Brownian motion generates. So the representation applies, and the discounted value equals Rs 10.450584/- plus an accumulation of bets on the driver. Because the process and the driver move together, an accumulation of bets on the driver is exactly what holding a changing quantity of the standard process produces. The payoff is therefore reachable from the starting value by trading, and that is what the theorem buys.
| \(H_t\) | the bet size against the driver at time \(t\), the integrand of the representation |
| \(\Delta_t\) | the same bet expressed in units of the standard process, 0.636831 at the start |
| \(S_t\) | the standard process at time \(t\), starting at Rs 100/- |
| \(K\) | the strike of the at-the-money contract, Rs 100/- |
| \(r\) | the risk-free rate, 0.05 a year, continuously compounded |
| \(\sigma\) | the volatility, 0.20 a year, so the variance rate is 0.04 |
| \(T-t\) | the time still left to the horizon, one year at the start |
| \(N(\cdot)\) | the standard normal distribution function |
At the start every one of those is known: the level is Rs 100/-, the strike is Rs 100/-, the rate is 0.05, the volatility is 0.20 and a full year remains. The bracket comes to 0.350000, and the standard normal distribution function at 0.350000 is 0.636831. So the sensitivity to the levelHow much the value of the contract moves when the level of the process moves, which at the start of this worked instance is 0.636831. is 0.636831 units of the process, and the integrand against the driver is 0.20 times Rs 100/- times 0.636831, giving 12.736613, the discount factor at the start of the year being one. Both describe the same bet.
What does the result add to already being able to value a contract?
What does the bet schedule look like on the locked path?
Abstractions about schedules are easier to trust once one has been written out. The locked path is the twelve step path this reading order draws whenever a path is needed, built rather than sampled so that it reproduces on every reading. Under the pricing rule its levels run from Rs 100/- at the start down to Rs 91.65/- in the ninth month and back to Rs 103.05/- at the horizon.
At every one of those months two things have changed, the level and the time still left, so the bet size is recomputed. The table below takes five of the twelve months. The bet size wanders between 0.242912 and 0.811726 across a single year on a single path. Nothing demonstrates more plainly that it is not a number.
| Month | Level of the standard process | Time still left | Bet size, in units |
|---|---|---|---|
| Start | Rs 100.0000/- | 1.0000 | 0.636831 |
| 3 | Rs 107.0910/- | 0.8333 | 0.756391 |
| 7 | Rs 109.4235/- | 0.5000 | 0.811726 |
| 10 | Rs 91.6497/- | 0.2500 | 0.242912 |
| 12 | Rs 99.2884/- | 0.0833 | 0.490965 |
| Range across the year | Rs 91.6497/- to Rs 109.4235/- | 1.0000 to 0.0833 | 0.242912 to 0.811726 |
One more thing is worth showing. Adding up the twelve steps of the accumulation on the locked path, taking the bet size at the start of each month and multiplying by that month's move in the driver, gives twelve products that come to minus Rs 6.754284/-. The true change in the discounted value over the year is minus Rs 7.553659/-. The sum is short by Rs 0.799375/-.
The shortfall is not a failure of the theorem; it is the coarseness of a twelve step partitionThe chopping of an interval into steps. A coarse partition has few long steps, a fine one has many short ones.. The accumulation in the theorem is a limit taken as the steps shrink, and twelve steps across a year is a very long way from that limit. The gap is the same effect as pacing out a coastline with a long ruler: the answer is not wrong arithmetic, it is the wrong ruler. Refining the steps closes the gap. Rigour for that refinement comes from the Ito integral, set out under Ito calculus.
The twelve step sum reaches minus Rs 6.754284/- against a true change of minus Rs 7.553659/-. What does the Rs 0.799375/- gap show?
The level is about to rise well above Rs 100/-. What happens to the bet size, before the control below is moved?
The schedule underneath every level
One control: the level of the standard process, from Rs 70/- to Rs 140/-, with one year still to the horizon throughout. The heavy line is the bet size at that horizon. The two faint lines behind it are the same schedule with six months and one month left, drawn to show that the schedule moves in time as well as in level. Every stretch of the curve visited is shaded in turn, so the shaded part is what has been traced and the unshaded part is what the result was quietly promising all along. The default sits at Rs 100/- and returns 0.636831, the figure in the worked instance.
Does the result hand over the schedule, or only its existence?
Only its existence. Existence alone is the whole practical content of the theorem, and the part most often lost.
An existence resultA statement that something exists, proved without any method for finding it or writing it down. says that somewhere there is an object with a property. A constructive resultA statement that comes with a recipe, so that following the steps produces the object itself. produces the object. In ordinary life the difference is easy to feel. Somebody can prove that among a thousand weights on a shelf there is a combination that balances a particular parcel exactly, without ever naming which weights to pick up. The proof is genuine, it is useful, and it leaves the whole afternoon of searching still to be done.
The martingale representation result is the first kind. Its proof establishes that the bet schedule is there, at every moment and on every path, and that it is essentially unique. The proof contains no formula for the bet size, no algorithm, and no way of reading one off, and no amount of restating the theorem will produce one.
Does the martingale representation result give the bet size?
The failure: reading an existence result as a construction
The representation is among the most quoted results in the subject, and it is quoted as though it delivered the hedge. The theorem delivers the fact that a hedge exists. The two sentences are not the same, and the distance between them is where the entire difficulty of the work lives.
The confusion is manufactured here in plain sight. The figure 0.636831 sits in the worked instance a few blocks above, right beside a statement of the theorem, and a reader moving quickly connects them. The theorem did not produce 0.636831 and could not have. The number came from writing down a closed form for the value of the contract in the level and differentiating it, and the closed form exists here only because this particular case happens to have one. Take away the closed form and the theorem is completely unchanged: it still says the schedule is there, and it still says nothing about the schedule itself.
Somebody who misses that will plan a piece of work on the assumption that the hard part is settled, and will be surprised twice. Once when no formula for the bet size turns out to be available for the payoff in front of them. And once more when the numerical method that produces one turns out to be the entire project rather than a finishing step. The cost is not an arithmetic error. The cost is a plan built on a theorem that was saying the opposite of what it was heard to say: not that the hard part is done, but that the hard part is worth attempting.
A piece of work is planned on the basis that the representation result settles the hedging problem. What has been mispriced?
Where does the result stop, and what is true outside it?
Everything above rests on one driver. Take that away and the result does not weaken gracefully; it stops applying.
Consider a room whose temperature is to be reproduced by working the heater dial. If the heater is the only thing affecting the room, then the room's history was made by the dial in the first place, so a schedule of dial settings can match any temperature history the room produces. Now a window opens onto an unpredictable draught. The room's temperature now moves for a reason the dial never caused, and no schedule of dial settings, however clever, can reproduce it. The shortage is not one of skill but one of an instrument.
A second source of randomnessAn extra unpredictable driver, independent of the first, that the bets have no way of reaching. does exactly that to the representation. If the payoff depends on a second Brownian motion independent of the first, then the martingale still decomposes, but into two accumulations rather than one, and only the first is reachable by bets on the first driver.
| \(\tilde{W}^{(1)}\) | the first driver, the one the standard process moves with |
| \(\tilde{W}^{(2)}\) | a second Brownian motion, independent of the first |
| \(H_s\) | the bet size against the first driver |
| \(G_s\) | the bet size against the second, which no position in the process supplies |
The result therefore reaches exactly as far as its driver does, and one step past that boundary it says nothing at all. Notice what has and has not happened here. The mathematics has not broken. The quantity is still a martingale, the second accumulation is a perfectly well behaved object, and everything is still true. Reproducibility has gone. The instrument available reaches only part of the randomness, and the name for that situation, along with what can be done inside it, is covered separately.
A payoff depends on a second source of randomness, independent of the driver. Does the result still apply?
How does somebody scoping a piece of work actually use this?
The result earns its keep as a question asked early, before anybody writes code. Somebody reviewing a pricing model, or planning a piece of quantitative research, or reading a specification written by somebody else, can put it to work in three steps that take an afternoon rather than a quarter.
The three questions, in order
- Is the quantity a martingale under the measure being used? If it is not, the result has nothing to say and no amount of arguing about instruments will change that. The property is never a property of the quantity alone, so naming the measure comes first, always.
- Is the information exactly what the traded driver generates? Step two decides everything and is skipped most often. List every source of randomness the payoff depends on, then check each against the driver. One item on that list that the driver does not generate moves the whole payoff to the other branch.
- Where is the schedule going to come from? If the first two answers are yes, the result says a schedule exists and stops. Everything after that is a method question, and it is the whole of the remaining work rather than a finishing touch on it.
The value of running those three in order is that the third one gets its true weight. A specification that says the payoff can be hedged because of the representation result has answered the first two questions and left the third completely blank. The specification still sounds as though it had answered all three. The most useful thing this theorem does for a piece of planning is show which of the problems is the real one.
There is a household version of the same discipline, and it is the same shape. Somebody is told that their monthly outgoings can, in principle, be met from what comes in. The promise is an existence claim about the total, and it is worth having. The promise says nothing at all about which week is tight, and the schedule of which week is tight is what actually has to be managed. Knowing that the year balances is genuinely useful and is nowhere near enough to run the year on.
One boundary on all of this, stated plainly. The sensitivity and the bet size named above are mathematical objects inside an invented worked instance, and neither is a position anybody holds. The failure described above is a failure of planning, not a failure of buying or selling.
No jurisdiction sets any of this. The result is a theorem, so it holds identically wherever it is read, and there is no regulator, standard or circular anywhere that defines it, amends it or supersedes it. Where any conduct duty attaches to work built on top of it, that duty comes from the activity rather than from the mathematics, and it must be confirmed at its own source.
References
| Source | Document | Where |
|---|---|---|
| arXiv Quantitative Finance | Preprint repository for martingale methods and representation results in pricing | arxiv.org |
| Social Science Research Network | Working paper repository for the same material | ssrn.com |
| Ito | The representation and the integral that carry his name, named here for structure only | named in the text, no text reproduced |
| Black, Scholes and Merton, 1973 | The closed form whose derivative supplies the sensitivity used in the worked instance | named in the text, no text reproduced |
| Hull, Shreve and Wilmott | Standard texts, consulted for notation and ordering only, with nothing reproduced | named in the text, no text reproduced |
The standard process, its four parameters, the locked path and the at-the-money contract written on it are invented.
Educational material. Not advice on any investment, tax, budget or market position.
