No-Arbitrage: The Single Assumption Pricing Theory Rests On
No-arbitrage assumes that no position costing nothing today can deliver something for certain and never deliver a loss. Two arrangements delivering the same amounts in every outcome must cost the same today, or the assumption itself fails, so the assumption fixes a price whenever a payoff can be built out of things whose prices are already known.
Everything in this stretch of the subject comes out of one sentence, and the sentence is shorter than most people expect. The sentence says nothing about what anybody believes, nothing about what anybody prefers, and nothing about whether a quantity is expensive. The sentence rules out one arrangement and one arrangement only, and from that single prohibition an entire apparatus of pricing follows.
The argument begins with what it refuses to do. A payoff arrives: a list of amounts, one for each way the year can turn out. Whatever produced that list is never disclosed and never becomes known. The question is never what those amounts are worth to anybody; the question is what it costs today to arrange for exactly those amounts to arrive. The swap from worth to cost is the whole move, and the swap is the reason the answer does not depend on who is asking.
The everyday version involves nothing more exotic than a shelf of packets. A one kilogram packet sits beside a stack of half kilogram packets. If two of the halves together cost less than the single kilogram, then anybody who wanted a kilogram has two routes to it at two prices, and the cheaper route is not a matter of opinion. Nobody had to decide what a kilogram of anything is worth. The two routes deliver the identical thing, so the two prices have to line up, and if they do not, the gap is free to whoever notices. The shelf of packets is the entire argument. The rest of this guide is that argument written carefully enough to be checked.
What does no-arbitrage assume, stated precisely?
Most statements of the assumption are admired rather than checked, and the reason is that they are usually written loosely enough that no check is possible. So write it in a form where checking is a finite job. No-arbitrageThe assumption that no position costing nothing can deliver a certain gain with no possible loss. is a statement about what positions are permitted to exist, and it has exactly three parts, all of which must hold at once for a position to be the thing being ruled out.
The three parts are: the position costs nothing to set up today, it never loses in any outcome at the horizon, and in at least one outcome it gains. A position that costs nothing and does nothing is fine. A position that costs nothing and sometimes gains but sometimes loses is fine, and there are a great many of those. Only the combination is banned. The combination is banned because if such a position existed, anybody could take it a thousand times over and the prices that permitted it could not survive the attempt.
| \(V_0\) | what the position costs to set up today, in rupees, counting every leg of it |
| \(V_T\) | what the same position is worth at the horizon, one year later here |
| \(\mathbb{P}\) | the physical measure, the rule that weights the outcomes as they are believed to occur |
| \(T\) | the horizon, one year throughout this guide |
Two details in that line are doing more work than they look. The first is that the middle condition has to hold in every outcome, not on average and not usually. One outcome in which the position loses a single paisa is enough to take it out of the banned set entirely. The second is subtler and matters later: the two probability statements only ever ask which outcomes are possible and which are not. The assumption never asks how likely any outcome is, only whether it can happen at all, and that is why nothing that follows from it depends on anybody agreeing about probabilities.
Say that back to yourself. The independence of belief is the hinge of the whole sequence. Two people can disagree completely about how likely the process is to rise, and as long as they agree about which levels are reachable at all, they will agree about every price this argument produces. Belief drops out. Not because belief is unimportant, but because the argument never asked for it.
A position costs nothing today and gains in nine outcomes out of ten, losing a little in the tenth. Is it the thing no-arbitrage rules out?
Why is one assumption enough to fix a price?
The step from a prohibition to a price is where readers usually lose the thread, so it goes in three sentences, each then slowed down. First: suppose a payoff can be built out of things whose prices are already known. Second: then there are two arrangements that deliver identical amounts in every outcome, namely the payoff itself and the thing built. Third: if those two arrangements had different prices today, taking the cheap one and giving up the dear one would cost nothing, never lose and sometimes gain, and that is exactly what the assumption bans.
Notice what has and has not been established. Nothing at all has been said about the worth of the payoff, whether it is attractive, or whether anybody should want it. The argument produces a cost, not a valuation, and it produces that cost by inheriting the prices of whatever the payoff was built out of. Feed it different input prices and it returns a different cost, obediently, without complaint and without any view about which set of inputs is right.
The ban is worth seeing in rupees, as a ledger, before any of the machinery arrives. Taking the case below on trust for a moment: the payoff can be built for Rs 12.162285/-, and somebody has quoted a figure of Rs 13/- for the same payoff. Giving up the payoff at Rs 13/- and building it for Rs 12.162285/- leaves Rs 0.837715/- in hand before anything has happened. At the horizon the thing built pays exactly what is owed, in both outcomes, so the Rs 0.837715/- stays. The position cost nothing, cannot lose, and gains for certain.
The same ledger runs backwards just as cleanly. Had the quoted figure been Rs 11/- instead, the position would be built the other way round: take the payoff at Rs 11/-, give up the built version at Rs 12.162285/-, and keep Rs 1.162285/- on the same terms. So the ban bites from both sides, and it leaves exactly one figure standing: the build cost itself. Leaving one figure standing is what it means to say the assumption fixes a price rather than merely constraining it.
Before the machinery arrives: how many numbers does it take to write a replicating portfolio down?
How to Set Up a Discrete-Time No-Arbitrage Example: what are the four lines?
Everything above is general and therefore slippery, so it goes onto a case small enough to hold in mind. The case is the standard process, an invented quantity written S with a time subscript, starting at Rs 100/- exactly and observed over one year. The standard process is not a company, not an instrument and not an observation of anything. The standard process exists so the arithmetic can be checked.
The setup takes four lines and the check takes one. Line one names the two future values the process is allowed to take. Line two names how cash grows over the same stretch. Line three states the horizon. Line four names the payoff. The payoff arrives as a list of amounts with no explanation attached, and the argument neither needs nor uses one.
- Name the two values the process can reach
On a one step latticeA setup in which a quantity is allowed exactly two values at the horizon, an up value and a down value, with nothing in between. the process moves to Rs 122.140276/- or to Rs 81.873075/- and nowhere else. The two values come from the volatility of 20 per cent a year through the up factor 1.221403 and its reciprocal 0.818731.
Nothing in between is permitted, and that restriction is what makes the example finite enough to check by hand.
- Name how cash grows
A risk-free rate of 5 per cent a year compounded continuously grows cash placed today by a factor of 1.051271 over the year. The discount factorThe reciprocal of the growth of cash, used to turn an amount at the horizon into an amount today. running the other way is 0.951229.
One growth factor and its reciprocal. Cash growth is the only place a rate enters the whole build.
- State the horizon
One year, one step, and no intermediate trading. Everything is set today and read once at the horizon.
Without a horizon there is nothing to discount over and the two future values have no date attached.
- Name the payoff, as amounts and nothing else
The payoff pays Rs 22.140276/- if the process reaches the upper value and nothing if it reaches the lower one. The two amounts are the entire specification, and whatever produced them plays no part in the argument.
Asking what the payoff represents means the argument has already been left behind. The payoff is two numbers.
Now the check, and it is a single inequality. Cash must grow by less than the up move and by more than the down move. Written out: 0.818731 is less than 1.051271, and 1.051271 is less than 1.221403. Both halves hold, so the setup is admissible and the argument may proceed. If either half failed, the example would be broken before any payoff was named, and no amount of correct arithmetic afterwards would repair it.
| \(S_0\) | the starting value of the standard process, Rs 100/- exactly, invented |
| \(S_T\) | its value at the horizon, one of exactly two numbers |
| \(u,\ d\) | the up and down factors, 1.221403 and 0.818731, reciprocals of one another |
| \(B_t\) | the cash account, worth one rupee today and \(e^{rT}\) rupees at the horizon |
| \(r\) | the risk-free rate, 5 per cent a year continuously compounded, decimal 0.05 |
| \(T\) | the horizon, one year, decimal 1.0 |
What is a Replicating Portfolio, and How to Build a Replicating-Portfolio Teaching Example?
A replicating portfolioA holding of the process together with an amount of cash that delivers the payoff in every outcome. is two numbers. The phrase sounds like it names an elaborate object, and it does not, so the flat statement is worth making. The two numbers are how many units of the process to hold and how much cash to borrow or lend, and once they are in hand the whole thing fits on one line and can be checked by multiplying out.
Finding the two numbers is two equations in two unknowns, and there is no cleverness in it anywhere. The holding is worth its size multiplied by the value of the process, in each of the two outcomes. The cash borrowed has to be repaid with growth, the same amount in both outcomes because cash does not care what happened. The total is set equal to the payoff in the up outcome, then equal to the payoff in the down outcome, and the pair is solved.
| \(\Delta\) | the holding, how many units of the standard process are held |
| \(b\) | the cash amount, negative where cash is borrowed rather than lent |
| \(V_u,\ V_d\) | the two payoff amounts, Rs 22.140276/- and nothing, handed over as numbers |
| \(S_0u,\ S_0d\) | the two values of the process at the horizon, in rupees |
| \(e^{rT}\) | the growth of cash over the horizon, 1.051271 |
The cash term is the same number in both lines, so subtracting the second from the first makes it disappear. The holding is left all on its own, and the holding is the difference between the two payoffs divided by the difference between the two values of the process. Nothing else survives the subtraction. In particular the rate does not, and neither does anything about how likely either outcome is.
| \(\Delta\) | the holding, 0.549834 units of the standard process |
| \(b\) | the cash amount, minus Rs 42.821115/-, so Rs 42.821115/- is borrowed |
| \(e^{-rT}\) | the discount factor, 0.951229 over the one year horizon |
| \(S_0u - S_0d\) | the spread of the two values of the process, Rs 40.267201/- |
| \(V_u - V_d\) | the spread of the two payoff amounts, Rs 22.140276/- |
So the two numbers are 0.549834 units held and Rs 42.821115/- borrowed. Read the holding for a moment: it is a ratio of two spreads, and it says that when the process moves by a rupee the payoff moves by about fifty five paise. The reading is worth carrying. The same reading survives every refinement of the setup later in the subject.
Now the part that makes it a teaching example rather than an assertion. Multiply the two numbers out in each outcome and check them. In the up outcome the holding is worth 0.549834 multiplied by Rs 122.140276/-, or Rs 67.156876/-, and the borrowing has grown to Rs 45.016600/-, leaving Rs 22.140276/- exactly. In the down outcome the holding is worth Rs 45.016600/- and the borrowing has grown to Rs 45.016600/-, leaving nothing exactly. Both outcomes land on the payoff to six decimal places, and neither of them was arranged to; they fell out of two linear equations.
The cost of setting it up is now arithmetic. The holding costs 0.549834 multiplied by Rs 100/-, or Rs 54.983400/-, and Rs 42.821115/- of that arrives from the borrowing, so Rs 12.162285/- of the builder's own money goes in. The build cost is the first of the two routes, and the build cost was produced without a probability appearing anywhere.
In the down outcome the holding is worth Rs 45.016600/- and the borrowing repays Rs 45.016600/-. Why do those two match to the last paisa?
What is the Pricing Measure, and where did it come from?
Now the move that gives the sequence its name, and it is worth watching closely because almost every account of it introduces the answer first and derives it afterwards. Nothing is introduced here. The cost just computed has the two solved numbers substituted back into it, and the result is rearranged. No new idea enters at any point.
The cost is the holding at today's value plus the cash amount. Put the solved holding and the solved cash amount into that expression and collect the terms. A discounted weighted average of the two payoff amounts emerges without being asked for, and the weights depend on the up factor, the down factor and the growth of cash. On nothing else at all.
| \(V_0\) | what building the payoff costs today, in rupees |
| \(\Delta S_0\) | the cost of the holding at today's value of the process |
| \(b\) | the cash amount, negative here because cash is borrowed |
| \(e^{-rT}\) | the discount factor, 0.951229 |
Collect the coefficient of the up payoff and the coefficient of the down payoff separately. The algebra is dull and takes three lines on paper. The collecting produces the next block, and the only thing to watch is where each of its terms came from: every one of them was already in the setup.
| \(q\) | the weight on the up outcome, 0.577493 with the locked parameters |
| \(1-q\) | the weight on the down outcome, 0.422507 |
| \(u,\ d\) | the up and down factors, 1.221403 and 0.818731 |
| \(e^{rT}\) | the growth of cash, 1.051271 |
| \(V_u,\ V_d\) | the two payoff amounts, unchanged from the setup |
The two weights are the pricing measureThe weights under which discounted prices behave consistently, arrived at by solving rather than by assuming., written Q throughout this subject, and the important word in that sentence is arrived at. Nobody chose 0.577493. Nobody argued for it, fitted it, or preferred it. The weight is what the two linear equations produce when they are rearranged, and it would be the same number if every reader on earth thought the process was certain to fall.
Compute the weight: 1.051271 less 0.818731, over 0.402672, giving 0.577493. Then the discounted average of the two payoff amounts is 0.951229 multiplied by 0.577493 multiplied by Rs 22.140276/-, and that comes to Rs 12.162285/-. The average lands on the same figure the build produced, and it agrees to ten decimal places, not to two.
Reading the second route as a check on the first is tempting, and that reading is wrong in a way that matters. The two routes are not two methods that happen to agree. The second route is the first route rearranged, so the algebra settles the agreement rather than the arithmetic confirming it. The agreement is the pricing argument, and every later technique in this subject is a way of computing that same weighted average when the setup is too rich to solve by hand.
One property of those weights is worth stating now because the rest of the subject leans on it. Apply them to the process itself rather than to the payoff. The weighted average of Rs 122.140276/- and Rs 81.873075/- under 0.577493 and 0.422507 comes to Rs 105.127110/-, and discounting that at 0.951229 gives Rs 100.000000/- exactly, the value the process started from. Under these weights, and only under these weights, the discounted process is a martingaleA process whose expected future value, given what is known today, is its value today..
| \(\mathbb{Q}\) | the pricing measure, the weights solved for above |
| \(\mathbb{E}^{\mathbb{Q}}\) | the average taken under those weights rather than under believed ones |
| \(\mathcal{F}_0\) | the information available today |
| \(S_0,\ S_T\) | the standard process today and at the horizon |
| \(e^{-rT}\) | the discount factor, 0.951229 |
Here is the test that settles whether the weights are beliefs in disguise. The standard process has a drift of 8 per cent a year under the physical measure P, and that number appears nowhere in the price. Change it. At 2 per cent a year the price is Rs 12.162285/-. At 8 per cent it is Rs 12.162285/-. At 15 per cent it is Rs 12.162285/-. Meanwhile the discounted average of the same payoff under believed weights walks from Rs 10.537280/- to Rs 13.836779/- to Rs 17.944940/-, and not one of those three is the price of anything.
Was the pricing measure assumed, or did it fall out of the argument?
How to Check No-Arbitrage in a Simple Pricing Model, and where are the edges?
Checking is not the same job as building, and it is much cheaper. The question is whether the setup permits the banned arrangement, asked before anything is computed with it, and on a one step setup the whole check is the single inequality from earlier. Cash must grow by less than the up move and more than the down move.
Each failure direction has a different position behind it. Take the two in turn. Suppose cash grew by more than the up move, say by a factor of 1.3 against an up factor of 1.221403. Then holding cash beats holding the process in both outcomes. Giving up the process and holding cash is then a position that costs nothing and gains in every outcome. Now suppose cash grew by less than the down move, say by 0.7. Then the process beats cash in both outcomes and the same trick runs the other way.
Because the two factors are fixed by the volatility, the check translates directly into a band on the rate. The up factor is the exponential of the volatility multiplied by the square root of the horizon, and the down factor is its reciprocal. With a volatility of 20 per cent a year and a horizon of one year, that band on the continuously compounded rate runs from minus 20 per cent a year to plus 20 per cent a year, and 20 per cent is the volatility exactly. The coincidence is not a coincidence: it is what taking logarithms of the inequality does.
| \(\sigma\) | the volatility of the standard process, 20 per cent a year, decimal 0.20 |
| \(T\) | the horizon, one year, so the square root of \(T\) is 1 and the band edges sit at \(\pm\sigma\) |
| \(r\) | the risk-free rate, continuously compounded |
| \(u,\ d\) | the up and down factors, built from the volatility as an exponential and its reciprocal |
The same edges show up in the weight, and the weight is the more useful way to read them in practice. At a rate of minus 20 per cent a year the weight is exactly nought; at plus 20 per cent it is exactly one; outside the band it leaves the interval altogether and turns negative or exceeds one. A weight outside that interval is not a weight, and the discounted average it produces is not a price. So the practical check on any pricing setup is to look at the weights it produces and confirm every one of them lies between nought and one.
Cash grows by more than the up move. What has gone wrong?
The rate is about to move. Before the control below is touched: does the holding of 0.549834 units change?
Move the rate and watch the two routes stay welded together
The two future values stay at Rs 122.140276/- and Rs 81.873075/-, the payoff stays at Rs 22.140276/- and nothing, and the horizon stays at one year with a volatility of 20 per cent. The only thing that moves is the rate. At the default of 5 per cent both routes read Rs 12.162285/-, the figure worked above. The holding stays at 0.549834 units at every rate on the scale, and only the borrowing moves. The band edges sit at minus 20 and plus 20 per cent a year.
What is a Theoretical Value, and what is it not?
The figure the argument returns has a name and the name is doing a lot of quiet work. A theoretical valueWhat a payoff costs to build, given the prices of the things it is built from. is what a payoff costs to build, given the prices of what it is built from. Read that definition twice. The second clause is the one that gets dropped, and dropping it is how a perfectly correct number gets used to reach a wrong conclusion.
The everyday version is a kitchen scale. The scale reports a number and the number is a reading, not the weight of the object. The number is what the object weighs according to this scale, with this zero setting, on this surface. If the zero was set with something already on the pan, the reading is wrong while the mechanism is fine, and staring harder at the display will not reveal which of the two situations obtains. A theoretical value is a reading, and every input price is part of the setting that produced it.
So here are the four things a theoretical value is not, stated flatly. Worth was the input the argument refused, so a theoretical value is not what the payoff would be worth to anybody. No probability of anything went in, so a theoretical value is not a forecast. A market price is what somebody actually paid, and no such observation entered the argument anywhere, so a theoretical value is not a market price either. And a theoretical value is not evidence about any observed price: the reading and the observation are two separate things, and either can be the one that is off.
The failure: reading the theoretical value as what the payoff is worth
The arithmetic is correct and the conclusion is still wrong. A failure of that shape is durable. Somebody computes Rs 12.162285/-, sees a different figure elsewhere, and concludes that the other figure is out of line. But the Rs 12.162285/- inherited every input it was built from, including a volatility of 20 per cent a year that was put in by hand.
Change that single input to 22 per cent and nothing about the payoff moves. The payoff was handed over as two numbers, and two numbers do not have a volatility, so it stays Rs 22.140276/- in the up outcome and nothing in the down one. But the two future values move to Rs 124.607673/- and Rs 80.251880/-, the holding falls to 0.499152 units, the borrowing falls to Rs 38.104228/-, and the theoretical value becomes Rs 11.810956/-. A gap of Rs 0.351329/- has appeared, and it came entirely from the input.
The cost of the mistake is a conclusion drawn in the wrong direction. The arithmetic was correct throughout, so it offers no help at all in spotting it. A reader who treats the theoretical value as a valuation will look outward when the input, not the observation, was what moved. The discipline is to hold both possibilities open: the observation may be out of line, or an input may be, and nothing in the arithmetic decides between them.
A theoretical value disagrees with a figure observed elsewhere. What are the two possibilities?
What is a Complete Market, and what is an Incomplete Market?
Everything above rested on a phrase that slipped past unexamined: whenever a payoff can be built. The condition is not automatic, and whether it holds is the single thing separating the two kinds of setting this subject keeps returning to.
A complete marketA setting in which every payoff can be built out of what is already priced, so every payoff has exactly one price. is one where every payoff can be built out of what is already priced. An incomplete marketA setting in which some payoffs cannot be built, so the argument returns a range of prices rather than one. is one where some cannot. The one step setting in this guide is complete, and counting shows why: two outcomes, two instruments to build with, and two linear equations in two unknowns that always solve.
The everyday version is measurement again. One ruler measures any distance along a line, so a straight corridor is fully covered. A second direction leaves one ruler unable to reach every point, and a second ruler becomes necessary. Instruments are rulers, outcomes are directions, and completeness is having at least as many independent rulers as directions.
The construction breaks with nothing but the numbers already given. Allow the process a third value at the horizon, Rs 100/- exactly, sitting between the two it already had. The payoff still pays Rs 22.140276/- at the top and nothing at either of the other two. There are still only two instruments. Three equations, two unknowns, and in general no solution. The payoff cannot be built.
The assumption still gives something. The assumption rules out every price that would permit the banned arrangement, and that leaves an interval rather than a point. Working the bounds out on this case gives a lowest admissible price of Rs 4.877058/- and a highest of Rs 12.162285/-, an interval Rs 7.285227/- wide. The upper bound is the two outcome price from earlier, and the lower bound is the starting value less the discounted middle value, or Rs 100/- less Rs 95.122942/-.
| \(\mathcal{M}\) | the set of weightings that keep the discounted process a martingale, which holds one member when the payoff can be built and many when it cannot |
| \(\underline{V}_0,\ \overline{V}_0\) | the lowest and highest admissible prices, Rs 4.877058/- and Rs 12.162285/- on the three outcome case |
| \(V_T\) | the payoff amounts at the horizon, unchanged |
| \(e^{-rT}\) | the discount factor, 0.951229 |
A payoff cannot be built out of what is already priced. What does the argument give?
What is a Pricing Error, and what is a Calibration Error?
The last distinction here is the one the rest of this subject leans on hardest, and it is routinely collapsed into a single word. A pricing errorA wrong figure produced by a correct model that was fed a wrong input. is a wrong output from a correct model fed a wrong input. A calibration errorA wrong input, arrived at by fitting the model to observations badly or by fitting the wrong thing. is a wrong input arrived at by fitting badly. The two errors are measured in different units, found by different tests, and fixed in different places.
The kitchen scale separates the two cleanly, so return to it. The reading says 500 grams and the packet holds 490. The ten gram gap is the pricing error: it is measured in grams, on the output, and it is what somebody downstream actually suffers. Now ask how the wrong setting got there. If somebody zeroed the scale with the pan already loaded, the fault sits in the zeroing procedure, and that is the calibration error: it is measured in the setting rather than the reading, and re-weighing the packet will never find it.
The relationship between them is one directional and worth stating precisely. A calibration error produces a pricing error. A pricing error does not imply a calibration error. The wrong input might have been typed in wrongly, carried over from last year, or quoted on a convention nobody checked. So the pricing error is what is observed and the calibration error is one of several possible causes. The fix for one is never automatically the fix for the other.
The numbers already given make it concrete. Fed a volatility of 22 per cent, the correct argument returns Rs 11.810956/- against Rs 12.162285/-, so the pricing error is Rs 0.351329/-, in rupees, on the output. Whether 22 per cent was a calibration error depends entirely on how it was arrived at, and that question is not answered by anything in the pricing arithmetic.
One illustration of how a fit can go wrong without anybody making a mistake, using the invented observation set this subject carries. Fitting a single volatility to the same five invented observations gives an answer that depends on what the fit was asked to make small: 19.82 per cent if the loss is squared price error, and 20.70 per cent if the loss is squared implied volatility error. Identical observations, two answers, 0.88 points apart, and nothing about the observations chose between them. Calibration is examined properly later in this subject.
A model is correct and the input it was fed is wrong. Which kind of error is that?
How does somebody building a model actually use this?
None of this is used by deriving it again. The argument is used as a short list of checks that can be run against any pricing model, somebody else's or one's own, in the order that fails fastest first. A quantitative researcher reviewing a specification runs them; so does a risk reviewer reading a valuation report, and so does anybody validating code against the closed form it claims to implement.
The first check is the band, and it costs one line. Compute the implied weights and confirm every one sits between nought and one. A weight of 1.04 is not a rounding problem and is not fixed by refining the grid: the setup permits the banned arrangement, and every figure downstream of it is meaningless rather than slightly off. The band check catches a whole class of misconfigured setups before any pricing runs at all.
The second check is the agreement, and it costs one extra run. Compute the same figure both ways, once by building and once by averaging, and confirm they agree to many decimal places rather than to two. Agreement to two decimals is not agreement; it is two different numbers that round the same way, and on a real setup that difference usually means an inconsistent discount factor somewhere. The two routes are algebraically identical, so any visible gap between them is a coding fault rather than a modelling choice.
The third habit is the input register, and it is the one that separates a reviewer from a calculator. Every theoretical value is a reading, so what it was read on is the question. The register lists every input, marks each one as observed or fitted or assumed, and notes the date it was set. When the figure later disagrees with something, that register is what shows where to look, and building it after the disagreement is too late to be honest about.
One note on universality. Readers sometimes look for a rule block at this point. No regulator sets this assumption, no exchange publishes it, and no jurisdiction alters it. No-arbitrage is a statement about arrangements of amounts, so it holds identically everywhere and nowhere in particular. Clearing and contract conventions are a separate matter, and the exchange or the clearing corporation is the place to confirm them.
References
| Source | Document | Where |
|---|---|---|
| arXiv Quantitative Finance | Preprint repository for statements of the no-arbitrage condition, replication and the pricing measure | arxiv.org |
| Social Science Research Network | Working paper repository for the same material | ssrn.com |
| Cox, Ross and Rubinstein, 1979 | The lattice construction whose one step case is worked here, named because the name is part of the term, with no text reproduced | named in the text only |
| Black, Scholes and Merton, 1973 | The replication argument in continuous time, named here as the setting this one step case is the discrete cousin of | named in the text only |
| Hull, Shreve and Wilmott | Standard texts, consulted for notation and ordering only, with nothing copied | named in the text only |
The standard process, its parameters and the payoff worked above are invented.
Educational material. Not advice on any investment, tax, budget or market position.
