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Stochastic Calculus & Derivative Pricing Theory
1Probability Foundations
The Probability SpaceRandom VectorsSigma-AlgebraExpectationSample Space and EventsDensity and Distribution FunctionsRisk-Neutral ProbabilityState Price Density vs…
2Stochastic Processes and Jumps
Properties of a Stochastic ProcessMartingaleBrownian Motion and Its PropertiesBrownian Motion vs Geometric…Stopping TimeThe Markov PropertyState VariablesTransition ProbabilityQuadratic VariationQuadratic Variation vs Ordinary…Submartingale and SupermartingaleMartingale RepresentationMarkov Process vs MartingaleOptional StoppingFiltrationJump ProcessesThe Poisson ProcessLevy ProcessesJump Diffusion
3Ito Calculus
The Ito IntegralThe Ito Integral vs the Riemann IntegralInfinitesimals in Stochastic CalculusQuadratic CovariationIto's LemmaHow to Apply Ito's…The Infinitesimal GeneratorIto Calculus vs Ordinary Calculus
4Stochastic Differential Equations
Stochastic Differential EquationsStochastic Differential Equation vs…Drift and DiffusionStrong and Weak Solutions ComparedDiscretisationGeometric Brownian Motion
5Pricing Theory and No-Arbitrage
No-ArbitrageGirsanov, Radon-Nikodym and Change…Physical and Risk-Neutral Measures…The Fundamental Theorems of…The Law of One PriceThe Pricing KernelDiscount Factors and Zero-Coupon PricesReplication vs HedgingComplete Market vs Incomplete MarketClearing Margin Architecture
6Option Pricing Theory
European and American OptionsMonte Carlo European OptionThe Black-Scholes PDEBlack Scholes and the GreeksThe Payoff FunctionThe Binomial ModelBinomial Option PricingDelta Hedging in TheoryBoundary, Initial and Terminal ConditionsThe Exercise BoundaryHow to Check Put-Call…
7Volatility Models
Constant, Local and Stochastic…Vasicek Model vs CIR ModelThe Heston ModelThe SABR ModelThe Volatility ProcessImplied VolatilityVolatility Smile vs Skew vs Surface
8Interest Rate Models
Interest-Rate DerivativesMean ReversionThe Zero-Coupon BondThe Ornstein-Uhlenbeck ProcessThe Discount CurveZero RatesShort-Rate Model vs Market Model
9Numerical Pricing
Closed Form and Numerical…Monte Carlo PricingEuler and Milstein Schemes ComparedTree MethodsFinite Difference MethodsNumerical Error and StabilityVariance Reduction
10Calibration and Model Risk
Model OverrideMarket Price and Model PriceCalibrationHow to Document a Pricing ModelThe Educational Illustration LabelMarket ConventionsModel Uncertainty and LimitationsBacktesting a Pricing ModelIdentifiabilityCalibrated ParametersThe Calibration Loss Function

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.

The representation
$$ M_t \;=\; M_0 \;+\; \int_0^{t} H_s \, d\tilde{W}_s \qquad \text{for every } t \in [0,T] $$
\(M_t\)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
What it says in wordsThe value of the martingale at any time equals the single number it started at, plus everything that a schedule of bets on the driving Brownian motion has accumulated since the start.

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.

The size condition on the bets
$$ \mathbb{E}^{\mathbb{Q}}\!\left[\int_0^{T} H_s^{\,2}\, ds \right] \;<\; \infty $$
\(\mathbb{E}^{\mathbb{Q}}[\cdot]\)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
What it says in wordsThe average of the total of the squared bet sizes across the year has to be a finite number, which is the technical price of the accumulation being a sensible object at all.

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.

ONE NUMBER, PLUS ONE RUNNING TOTAL. THERE IS NOTHING ELSE IN THE OBJECT. start Rs 10.450584/- the value today bets minus Rs 7.553659/- the accumulation end Rs 2.896925/- the value at the horizon on the locked path Educational illustration. The discounted value of the at-the-money contract, invented parameters throughout.
The discounted value of the at-the-money contract is its starting figure of Rs 10.450584/- plus one running total, and on the locked path that running total comes to minus Rs 7.553659/-, ending the year at Rs 2.896925/-.
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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.

The accumulation as a limit of ordinary sums
$$ \int_0^{t} H_s\, d\tilde{W}_s \;=\; \lim_{n \to \infty} \sum_{i=0}^{n-1} H_{t_i}\Bigl( \tilde{W}_{t_{i+1}} - \tilde{W}_{t_i} \Bigr) $$
\(t_i\)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
What it says in wordsThe accumulation is what an ordinary running total settles down to when the bet size is held fixed at the start of each step and the steps are made finer 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 STEP OF THE ACCUMULATION, MONTH TWO ON THE LOCKED PATH WHAT THE DRIVER DID +0.461880 the move in the month x THE BET SIZE, FIXED FIRST 11.217065 set before the move happened = WHAT THE STEP ADDED +Rs 5.180941/- into the running total REVERSE THE ORDER AND THE THEOREM DIES SIZE FIRST, THEN THE MOVE the size cannot see what is about to happen, so the running total stays a fair game MOVE FIRST, THEN THE SIZE any total at all could be manufactured, and no statement about a fair game survives Educational illustration. Every figure is computed from the invented locked path.
Each step of the accumulation is the bet size fixed at the start of the step multiplied by how far the driver moved during it, and reversing that order would let any running total be manufactured at will.

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.

Try it out

The result produces a bet size. Is that a single number or a schedule?

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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.

The condition doing the work
$$ \mathcal{F}_t \;=\; \sigma\bigl(\tilde{W}_s \,:\, 0 \le s \le t \bigr) $$
\(\mathcal{F}_t\)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 \(t\)
What it says in wordsThe information available at any time is precisely what the history of the driving Brownian motion supplies, so any quantity known at that time is already a question about the path of that one driver.

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.

EVERYTHING RANDOM COMES FROM ONE PLACE. THAT IS THE WHOLE CONDITION. INSIDE THIS BOUNDARY, BETS ON THE DRIVER CAN REACH EVERYTHING THE DRIVING BROWNIAN MOTION one source, and only one THE PATH OF THE STANDARD PROCESS every level it visits is a question about the driver THE INFORMATION AT EVERY TIME exactly what watching the driver reveals A SECOND GAUGE WOULD SIT OUT HERE bets cannot cross
The condition carrying the theorem is that one driving Brownian motion generates both the path and the information, so anything a second unreachable gauge caused would sit outside what bets can rebuild.
Try it out

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.

The worked instance, its integrand and its bet size in units of the process
$$ H_t \;=\; e^{-rt}\,\sigma\,S_t\,\Delta_t, \qquad \Delta_t \;=\; N\!\left( \frac{\ln\!\left( S_t / K \right) + \left( r + \tfrac{1}{2}\sigma^{2} \right)\left( T - t \right)}{\sigma \sqrt{T-t}} \right) $$
\(H_t\)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
What it says in wordsThe bet size at any moment is the discount factor times the volatility times the level of the process times its sensitivity to that level, and the sensitivity is the standard normal distribution function evaluated at a quantity built from the level, the strike, the rate, the volatility and the time still left.

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.

VALUING ANSWERS ONE QUESTION. REPRODUCING ASKS A DIFFERENT ONE AT EVERY LEVEL. WHAT VALUING GIVES Rs 10.450584/- one number, about today and about nothing else WHAT REPRODUCING NEEDS, AT ONE INSTANT, ON THREE PATHS 0.075875 at Rs 70/- 0.636831 at Rs 100/- 0.978941 at Rs 140/- units of the process Educational illustration. One year to the horizon in all three readings, volatility 20 per cent, rate 5 per cent.
Valuing the contract produces the single figure Rs 10.450584/-, while reproducing it needs a bet size at every level the process might reach, and at one instant those sizes run from 0.075875 to 0.978941.
Try it out

What does the result add to already being able to value a contract?

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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.

MonthLevel of the standard processTime still leftBet size, in units
StartRs 100.0000/-1.00000.636831
3Rs 107.0910/-0.83330.756391
7Rs 109.4235/-0.50000.811726
10Rs 91.6497/-0.25000.242912
12Rs 99.2884/-0.08330.490965
Range across the yearRs 91.6497/- to Rs 109.4235/-1.0000 to 0.08330.242912 to 0.811726
THE BET SIZE THROUGH ONE YEAR ON ONE PATH. IT IS NOWHERE NEAR CONSTANT. 1.00 0.50 0.00 0.636831 0.811726 0.242912 start month 7 month 10 horizon
Along one year of the locked path the bet size climbs to 0.811726 in the seventh month and collapses to 0.242912 in the tenth, so what the result promises is a whole schedule and never one figure.

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 RUNNING TOTAL ACROSS TWELVE STEPS OF THE LOCKED PATH 0 +Rs 3.342569/- minus Rs 10.129708/- ends at minus Rs 6.754284/- start horizon WHAT THE COARSE SUM RECOVERED, AGAINST WHAT ACTUALLY HAPPENED twelve step sum minus Rs 6.754284/- the true change minus Rs 7.553659/- Rs 0.799375/- Educational illustration. The shaded strip is what twelve steps could not reach.
A twelve step sum of the bets recovers minus Rs 6.754284/- of a true change of minus Rs 7.553659/-, and the Rs 0.799375/- strip left over is the coarseness of the partition rather than any failure of the result.
Try it out

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?

Try it out

The level is about to rise well above Rs 100/-. What happens to the bet size, before the control below is moved?

Play with it

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.

THE BET SIZE AGAINST THE LEVEL. THE WHOLE CURVE IS THE SCHEDULE. 1.00 0.50 0.00 0.636831 Rs 70/- Rs 100/- Rs 140/- solid line, one year left. long dashes, six months left. short dashes, one month left. THE BET SIZE AT THE ONE YEAR HORIZON, READ ON A SCALE FROM ZERO TO ONE UNIT 0.075875 at Rs 70/- 0.636831 at Rs 100/- 0.978941 at Rs 140/- Educational illustration. Volatility 20 per cent a year, rate 5 per cent a year, strike Rs 100/-, one year to the horizon.
Level of the process
Rs 100/-
Bet size, in units
0.636831
Curve traced so far
0 per cent
At a level of Rs 100/- with one year still to the horizon the bet size is 0.636831 units of the standard process, and 0 per cent of the schedule has been traced so far.
Educational illustration. Every reading is computed from the formula rather than sampled, so the default reproduces the worked example on every reload. The three static readings are 0.075875 at Rs 70/-, 0.636831 at Rs 100/- and 0.978941 at Rs 140/-. The representation result says a schedule of this kind exists at every moment on every path. The result produces none of these three numbers, and offers no method for finding them.
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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.

TWO KINDS OF RESULT. THIS ONE IS THE LEFT COLUMN. THE REPRESENTATION A CONSTRUCTION Does a bet schedule exist? YES YES Is it essentially unique? YES YES Does it supply a formula or a method? NO, NOTHING YES Does it give the size at Rs 91.65/-? NO, NOTHING YES Two rows agree and two rows do not, and the two that do not are the whole cost of the work.
An existence result and a construction agree that the schedule is there and essentially unique, and part company completely on whether anything supplies the size at a given level.
Try it out

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.

WHERE 0.636831 ACTUALLY CAME FROM, AND WHERE IT DID NOT ROUTE THAT DELIVERED IT write the value as a formula in the level, then differentiate that formula ROUTE THAT DELIVERED NOTHING the representation result, which says only that some such number is there THE CELL EVERYBODY READS 0.636831 the closed form exists here only because this case happens to have one remove it and the theorem is unchanged Educational illustration. The figure is a computed consequence of invented parameters.
The number 0.636831 arrived by differentiating a closed form that this particular case happens to possess, and the representation result contributes no route to it at all.
Try it out

A piece of work is planned on the basis that the representation result settles the hedging problem. What has been mispriced?

The theorem hands over existence and never the schedule itself. See what that leaves.

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.

The same quantity with a second driver present
$$ M_t \;=\; M_0 \;+\; \int_0^{t} H_s \, d\tilde{W}^{(1)}_s \;+\; \int_0^{t} G_s \, d\tilde{W}^{(2)}_s $$
\(\tilde{W}^{(1)}\)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
What it says in wordsWith a second independent driver present the quantity splits into two accumulations, and bets on the first driver alone can reach only the first of them, leaving the second unreproduced on every path.

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.

ONE QUESTION DECIDES WHETHER THE RESULT APPLIES AT ALL IS EVERY SOURCE OF RANDOMNESS IN THE PAYOFF GENERATED BY THE DRIVING BROWNIAN MOTION? YES NO INSIDE THE RESULT the martingale is its starting value plus one accumulation of bets on the driver a schedule exists at every moment and on every path, and is essentially unique OUTSIDE IT ENTIRELY the quantity splits into two accumulations and bets reach only the first of them nothing is wrong with the mathematics, and the reproducibility has simply gone The boundary is sharp. There is no partial credit and no approximately applies on the right hand branch. Educational illustration. What the right hand branch is called, and what can be done inside it, are covered separately.
Whether every source of randomness in the payoff comes from the driving Brownian motion decides the whole question, and a second independent driver puts the payoff outside the result with no partial credit.
Try it out

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

  1. 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.
  2. 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.
  3. 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.

The Ito integral that the representation is written with is set out under Ito calculus. How the bet schedule is computed for a payoff without a closed form is a method question covered separately and much later. Carrying out a hedge in practice is covered separately. The name for a situation with a second driver, and what can be done inside it, is covered separately. Contract payoffs are covered separately.
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References

SourceDocumentWhere
arXiv Quantitative FinancePreprint repository for martingale methods and representation results in pricingarxiv.org
Social Science Research NetworkWorking paper repository for the same materialssrn.com
ItoThe representation and the integral that carry his name, named here for structure onlynamed in the text, no text reproduced
Black, Scholes and Merton, 1973The closed form whose derivative supplies the sensitivity used in the worked instancenamed in the text, no text reproduced
Hull, Shreve and WilmottStandard texts, consulted for notation and ordering only, with nothing reproducednamed 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.

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