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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

Closed Form and Numerical Approximation Compared

A closed form is a formula that hands back the answer directly, and one exists for a small number of problems and for none of the interesting extensions. A numerical method computes an approximation instead: it chops something continuous into finitely many pieces and does arithmetic on the pieces. Which thing it chops is what separates one method from the next.

Four methods stand in for a formula when no formula exists, and they are set out here side by side rather than one at a time. Each of the four is opened in full under its own subject: the Monte Carlo method, discretisation schemes, tree methods and the finite difference method. What the four share is worth having first. Each replaces one continuous thing with a finite thing, and each is set here against a single number that is known exactly. Any method met anywhere can then be read the same way, by asking which continuous thing it replaced and what the replacement cost.

One process, an invented one written S with a time subscript, runs through everything that follows. Call it the standard process. The standard process starts at Rs 100/-, carries a volatility of 20 per cent a year, is discounted at 5 per cent a year continuously compounded, and runs for a horizon of one year. On it sits one contract, the at-the-money one, struck at Rs 100/-. What that contract pays is set out under the instruments themselves; here the payoff is only the function whose value four methods are competing to produce. Every figure below is arithmetic on those parameters rather than a reading taken off any market.

What is a closed form, and when does one exist?

Closed-Form Solution: what the phrase promises and what it does not

A closed formA formula returning the answer directly, available for a small number of problems. Inputs go in at one end and the answer comes out at the other, with no repetition. is a finite expression built from the inputs and from functions already regarded as known. The inputs are substituted, the expression is evaluated, and the work is finished. There is no loop, no refinement, no step count to choose, and no question of how long to run it for. A closed form has no accuracy setting, because there is nothing inside it to set. The missing setting is the property readers most often underrate.

Here is the everyday version. The fence needed for a rectangular plot is twice the length plus twice the width, and that is the end of it. Twice length plus twice width is a closed form. The fence needed to follow the edge of a lake is another matter. There is no expression to substitute into. Somebody walks the edge with a measuring rod instead, counts the rods, and accepts that a shorter rod would have given a different and larger number. The difference between the plot and the lake is not that one is harder; it is that one has an expression and the other has only a procedure.

For the standard process and the at-the-money contract an expression does exist. The expression is the solution associated with Black, Scholes and Merton, 1973, and on the locked parameters it evaluates once and returns Rs 10.450584/-. That one figure does all the work in this guide and in everything that follows it, and the reason is worth stating plainly. Rs 10.450584/- is not useful because it is the price of anything real. Rs 10.450584/- is useful because it is known.

The closed form on the standard process
$$ C \;=\; S_0\,N(d_1) \;-\; K\,e^{-rT}\,N(d_2), \qquad d_1 \;=\; \frac{\ln(S_0/K) + \bigl(r + \tfrac{1}{2}\sigma^2\bigr)T}{\sigma\sqrt{T}}, \qquad d_2 \;=\; d_1 - \sigma\sqrt{T} $$
\(C\)the value the formula returns, Rs 10.450584/- on the invented parameters used here
\(S_0\)the starting value of the standard process, Rs 100/-
\(K\)the strike of the at-the-money contract, Rs 100/-
\(r\)the rate used for discounting, 0.05 a year, continuously compounded
\(\sigma\)the volatility of the standard process, 0.20 a year
\(T\)the horizon, 1.0 years
\(N(\cdot)\)the standard normal distribution function
\(d_1, d_2\)two intermediate quantities, 0.350000 and 0.150000 exactly here
What it says in wordsTake the starting value and weight it by one number between nought and one, take the strike discounted back to today and weight it by a second number between nought and one, then subtract the second from the first. Both weights are read off a single fixed table of the normal distribution, so once the six inputs are fixed there is nothing left to decide and nothing left to refine.

Substituting gives d1 of 0.350000 and d2 of 0.150000, both exact on these parameters, and the two weights read 0.636831 and 0.559618. A hundred times 0.636831 is 63.683065. The discount factor over the year is 0.951229, so the discounted strike is 95.122942, and 95.122942 times 0.559618 is 53.232481. Subtract, and 63.683065 less 53.232481 is 10.450584. Notice what did not happen anywhere in that paragraph: no choice, no step count, no tolerance, no second run.

So when does an expression exist? Almost never, and the honest way to say it is that the expression survives only while the problem stays simple in a very particular set of ways. Let the volatility itself wander and the expression goes. Let the contract look at the whole path rather than only its ending value and the expression goes. Let the holder choose when to stop rather than waiting for the horizon and the expression goes. Put three or four processes in at once and the expression goes. Every extension that makes a problem worth posing also removes the formula. Approximation is therefore the normal case, and the formula is the exception.

What the problem makes impossible decides the method. Nothing else does. DOES AN EXPRESSION EXIST FOR THIS PROBLEM? ask this first, because a yes ends the question YES NO USE IT. ALL FOUR ARE RULED OUT. Rs 10.450584/-, once, with no step count to argue about. ASK WHAT THE PROBLEM FORBIDS Each feature below knocks out one candidate and leaves the rest. PAYOFF WATCHES THE WHOLE PATH rules out a plain recombining lattice HOLDER CHOOSES WHEN TO STOP rules out a plain sampling method FOUR OR MORE PROCESSES AT ONCE rules out a grid, whose size explodes with them ANSWER NEEDED TO SIX DECIMALS rules out sampling: the spread will not shrink fast Nothing here is about which method is cleverer. Every branch is about what the problem will not allow.
Which method to reach for is decided by what the problem makes impossible: a formula rules out all four when one exists, a payoff watching the whole path rules out a plain recombining lattice, several processes at once rule out a grid, and a demand for six decimals rules out sampling.
Try it out

For how many pricing problems does a closed form exist?

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What does a numerical method compute instead?

Closed-Form Solution vs Numerical Approximation

A numerical methodA method computing an approximation by chopping something continuous into finitely many pieces and doing arithmetic on the pieces. does not compute the answer. A numerical method computes a different quantity instead, one reachable in finitely many arithmetic operations, and that different quantity sits close to the answer. How close depends on a setting, and the setting is the number of pieces the continuous thing was chopped into. The count goes by a different name in each method: steps, nodes, a space step, a count of draws. The name changes and the idea does not.

Back to the lake. Walking its edge with a rod one hundred metres long gives one number. The shorter rod follows more of the wiggles, so a rod ten metres long gives a larger number. A rod one metre long gives a larger number still. The length being sought is what those numbers approach as the rod shrinks. The length itself is never in hand at any moment. What is in hand is the length at the rod used. Every numerical method in this guide is that walk, and the rod is whatever the method chopped.

DiscretisationThe chopping itself: replacing something continuous, such as time or the level of a process, with finitely many representative values. is the name for the chopping. Accepting it means accepting a second quantity, the difference between the approximation and the truth, and that difference has a shape. The difference is not a random blemish. For every method here it behaves like a constant multiplied by the chopping size raised to some power. The power says what refinement buys, which makes it the single most useful thing to know about a method.

What an approximation is, and how its error behaves
$$ V_h \;\longrightarrow\; V \ \text{ as } \ h \to 0, \qquad e(h) \;=\; V_h - V \;\approx\; C\,h^{\,p} \qquad\Longrightarrow\qquad \frac{e(h)}{e(h/2)} \;\approx\; 2^{\,p} $$
\(V\)the answer, which the method never returns
\(V_h\)what the method does return, at a chopping size \(h\)
\(h\)the chopping size: a time step, a space step, or one over the count of draws
\(e(h)\)the signed error, which carries a direction as well as a size
\(C\)a constant belonging to the method and the problem together
\(p\)the order: how fast the error falls when the chopping is refined
What it says in wordsThe approximation settles on the answer as the chopping is made finer, and the distance between the two behaves like a fixed constant multiplied by the chopping size raised to a fixed power. Because of that shape, halving the chopping size divides the error by two raised to that power, so the order can be read off by halving twice and taking the ratio of the two errors rather than by trusting anyone about it.

Three of the four methods here are deterministicReturning the same answer every run. Three of the four methods here are deterministic; the sampling method is not.: run them twice at the same settings and the digits are identical. The fourth is not, and the difference is not a detail. Determinism changes what a method can hand back. A deterministic method returns a number wrong by an amount that is not known. A sampling methodA method replacing an average over everything with an average over finitely many draws. Its output is a centre and a spread rather than a single figure. returns a number wrong by an amount that can be estimated, but the number itself moves on every run.

How the accuracy of a sampling method scales
$$ \operatorname{sd}\bigl(\hat{V}_n\bigr) \;=\; \frac{s}{\sqrt{n}}, \qquad s \;=\; 14.719404, \qquad \frac{14.719404}{\sqrt{10{,}000}} \;=\; 0.147194 $$
\(\hat{V}_n\)the average over \(n\) draws, which is what the method returns
\(n\)the count of draws
\(s\)the standard deviation of one discounted payoff, 14.719404 on the invented parameters here, evaluated from the distribution rather than measured from a run
\(\operatorname{sd}(\cdot)\)the spread of the answer around the truth
What it says in wordsThe spread of a sampling answer around the truth is one payoff spread divided by the square root of the count of draws, so accuracy is bought at the square of its price: four times the draws for half the spread, and a hundred times the draws for a tenth of it. Every figure in this block is evaluated from the distribution itself rather than drawn at random, so no reading changes on a reload.

Sampled quantities are computed rather than drawn throughout, and the reason is worth one sentence. One of the four methods here works by drawing at random. A worked example that drew at random would come out differently on every reading, which leaves a method nothing to be checked against. So wherever a sampled quantity is needed, the formula for how that quantity scales is evaluated instead. The spread of 0.147194 at ten thousand draws is not the spread that some particular run happened to show; it is the spread the arithmetic says any run of that size will have.

Three methods return a dot. This one returns a band. the truth, Rs 10.450584/- 10,000 DRAWS one spread each side: 0.147194 40,000 DRAWS four times the work: 0.073597 1,60,000 DRAWS sixteen times the work: 0.036799 10,00,000 DRAWS a hundred times: 0.014719 Each band is half the one above it, and each one cost four times as much to obtain.
A sampling method at ten thousand draws answers with a spread of 0.147194 around the truth, so what it returns is a band and not a figure, and the band halves only when the count of draws is multiplied by four.
Try it out

What does a sampling method return at ten thousand draws?

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What are the four methods, and what does each one chop up?

Four methods are used against problems that have no formula. Introduced by their machinery, as the four usually are, they sound like four unrelated inventions. They are not four unrelated inventions. The four differ in exactly one respect: which continuous thing each of them replaced with a finite thing. Learn that and the machinery follows; learn the machinery first and the four stay four separate tricks for ever.

A latticeA method chopping time into steps and the level into a small number of moves at each step, so the whole future collapses to a finite set of nodes. chops time. The year becomes a fixed number of dates and nothing is said about what happens between them. A path scheme chops time as well, and then chops the randomness along with it. Once time comes in intervals, the random driving term has to be delivered in packets rather than continuously. A gridA method chopping both time and the level of the process, then solving the governing equation at every crossing point of the resulting mesh. chops both time and the level, laying a mesh over the whole region and solving at every crossing point. And the sampling method chops nothing whatsoever. The sampling method leaves the problem entirely intact and approximates something else instead: the averaging.

Monte Carlo deserves its own sentence, being the one readers file in the wrong drawer. Monte Carlo is not a coarse version of the problem. Monte Carlo is the exact problem, answered by taking an average over a handful of outcomes instead of over all of them. Think of counting a crowd. A grid draws chalk squares over the ground and counts every square. Monte Carlo counts six squares chosen at random and multiplies up. Neither has simplified the crowd; only one of them has simplified the counting.

The four differ in what they chop, not in how clever they are. A LATTICE chops TIME into steps, and nothing else start horizon one axis chopped, the other left whole A PATH SCHEME chops TIME, and the randomness with it one packet of randomness per interval chopping time forces the second chop A GRID chops TIME and the LEVEL together high low both axes chopped: a mesh, not a line A SAMPLING METHOD chops NOTHING at all the whole distribution is left intact the averaging is what gets approximated Four methods. Four different things chopped. That difference is the whole of the difference.
A lattice chops time, a grid chops time and the level, a path scheme chops time and the randomness with it, and a sampling method chops nothing at all and approximates the averaging instead.

Binomial Model vs Black-Scholes Model

The binomial lattice and the closed form are set against each other so often that the relationship gets lost, so it is worth stating once and plainly. The two are not rivals. The binomial lattice of Cox, Ross and Rubinstein, 1979, is a chopped-up description whose answer moves as the chopping gets finer. The solution of Black, Scholes and Merton, 1973, is the number that chopping is moving toward. One is an approximation with a dial on it; the other is the thing at the end of the dial. Where the formula exists, the lattice is a way of getting near it. Where no formula exists, the lattice is all there is, and no rival stands against it.

The lattice is only two rules. Between one date and the next the process is allowed exactly two moves, one up and one down, and the value at any node is the weighted average of the two values it leads to, discounted by one step. Two rules are the whole thing. Tree methods take the lattice apart properly under their own subject; the rule is quoted here once so that the phrase chopping timeReplacing a continuous stretch of time with a finite set of dates, so that nothing is described between one date and the next. has something concrete attached to it.

One backward step of a lattice
$$ V^{(m)}_j \;=\; e^{-r\Delta t}\Bigl[\, p\,V^{(m+1)}_j \;+\; (1-p)\,V^{(m+1)}_{j+1} \,\Bigr], \qquad u = e^{\sigma\sqrt{\Delta t}}, \quad d = \frac{1}{u}, \quad p = \frac{e^{r\Delta t} - d}{u - d} $$
\(\Delta t\)the length of one step, \(T/n\); at ten steps over one year it is 0.1
\(n\)the number of steps the year has been chopped into
\(V^{(m)}_j\)the value at node \(j\) on date \(m\), working backwards from the horizon
\(u, d\)the up and down factors; at ten steps they are 1.065288 and 0.938713
\(p\)the weight attached to the up move; at ten steps it is 0.523795
\(r, \sigma, T\)rate 0.05, volatility 0.20 and horizon 1.0, as everywhere in this guide
What it says in wordsThe value at any node is the weighted average of the two values it can lead to, pulled back by one step of discounting, and the weight is fixed by requiring that the process itself grows at the discounting rate on average. Nothing in the rule refers to anything between the two dates, which is exactly what it means to say that a lattice has chopped time and left everything between the cuts undescribed.

Monte Carlo vs Lattice Method

Now the sharper contrast, and it is the one this whole reading order is organised around. A lattice enumerates. A lattice writes down every state its own chopped-up description allows and visits each one exactly once. Running it twice therefore changes nothing. A sampling method does not enumerate. Drawing is what it does instead, and every draw is one outcome pulled from a distribution that has been left whole. The lattice approximates the problem and then solves it exactly; the sampling method keeps the problem exact and then solves it approximately.

Everything else about the two follows from that one sentence. The lattice is deterministic, so it returns a firm number and no error estimate. The sampling method is not deterministic, so it returns a soft number and a very good error estimate. Arithmetic knows how to compute the spread of an average. Grow the number of processes and the count of nodes explodes with them, so the lattice becomes unusable. The sampling method barely notices. One outcome is drawn per run whether that outcome has one component or forty. Let the payoff depend on the whole path and the lattice struggles again. A recombining lattice has thrown away which route it took to a node. The sampling method still has the route in hand.

Try it out

Which of the four chops nothing?

How is any of them checked?

An approximation that is not measured against something is an assertion. Most treatments reach a point where a reader is told that a method converges and is expected to take it on trust, and that is exactly the place where trust should not be extended. The discipline used throughout is simple: never claim an error, measure one.

Measuring one needs a known answer, and a known answer is what a benchmarkA known answer to check an approximation against. Its job is not to be the answer wanted, but to be the answer already in hand. is. Here it is the closed form. Rs 10.450584/- appears not because anyone needs the price of the at-the-money contract, but because it lets every number set against it be a measurement rather than a claim. The closed form is the calibration weight in the drawer under the shop scale. Nobody wants to buy the weight. The weight sits there so that when the scale reads 1.004 kilograms for a known kilogram, the shopkeeper knows the scale is four grams heavy rather than wondering whether the packet is.

And that is how anyone who has to trust a computed number actually works, whatever the field. The machinery is not tested on the problem that cannot be solved. The machinery is tested on the nearest problem that can be solved, how far off it was and in which direction is recorded, and only then is it pointed at the problem actually at hand. A method that has never been run against a known answer has no known accuracy, however impressive its output looks. Everything from a kitchen scale to a national census works this way, and so does every one of the four methods here.

The closed form is not here to be the answer. It is here to be the ruler. WITHOUT A KNOWN ANSWER Rs 10.253409/- ? Rs 10.351260/- ? Rs 10.394588/- ? Three assertions. No way to rank them, and no way to say how far off any of them is. AGAINST Rs 10.450584/- minus Rs 0.197175/- 3rd minus Rs 0.099323/- 2nd minus Rs 0.055996/- 1st Three measurements. Ranked, signed, and all three on the same side of the truth. Same three outputs on both sides. Only the right-hand panel contains any information about accuracy. Educational illustration on invented parameters. The rank shown is the rank at these particular settings, not a ranking of the methods.
The locked contract is priced by formula so that every approximation set against it can be measured rather than trusted, which turns three bare outputs into three signed and ranked errors.

How to Compare a Tree Model With a Closed-Form Result

The comparison itself has a shape, and doing it in the wrong order is how people talk themselves into believing a method has converged when it has not. Four moves, in this order.

First, the closed form is evaluated once and written down to more decimals than seem necessary. Six is the working standard here. Second, the tree is run at a ladder of step counts rather than at one. A single reading says where the tree is and nothing about where it is going. Third, what matters is the signed difference at each count, never the size alone. The direction of the error is half of what the ladder has to teach, and throwing it away hides the most interesting behaviour a tree has. Fourth, include both odd and even step counts in the ladder. The fourth move sounds fussy and matters most of the four, for a reason the next section is entirely about.

The ladder is then read with the ratio test: the error at one count divided by the error at double that count. If the answer is near two, the error is falling in proportion to the step size and the method is first order. If it is near four, the error falls with the square of the step size and the method is second order. The ratio test says what a refinement will buy before it is paid for.

The ratio test, and what it says about each method here
$$ \frac{e(h)}{e(h/2)} \approx 2^{\,p} \qquad\text{binomial: } \frac{-0.197175}{-0.099323} = 1.985177, \quad \frac{-0.039892}{-0.019972} = 1.997407 $$
\(e(h)\)the signed error at chopping size \(h\), measured against Rs 10.450584/-
\(p\)the order, recovered as the base-two logarithm of the ratio
1.985177the binomial error at ten steps divided by the error at twenty
1.997407the same test at fifty and a hundred steps, closer to two still
4.249071the grid, at space steps of 0.10 and 0.05, so near four rather than two
What it says in wordsDoubling the step count of the binomial roughly halves its error, which places the method at first order in the step size, and the ratio moves closer to two as the count rises because the shape only holds in the limit. Halving the space step of the grid divides its error by roughly four instead, which places the grid at second order in that step, so the same amount of extra refinement buys the grid far more accuracy than it buys the tree.
Try it out

Why is the locked contract priced by formula throughout this reading order?

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How far apart are the four in practice?

Now the worked instance, and every figure in it is arithmetic on the invented parameters set out at the top. One contract, one process, four methods, and a known answer to hold them all against. The effort has been set at a level that is small enough to see: ten steps for each of the two trees, a space step of 0.05 with a thousand time steps for the grid, and ten thousand draws for the sampling method.

MethodWhat it choppedSettingResultSigned error
Binomial latticeTime only10 stepsRs 10.253409/-minus 0.197175
Trinomial latticeTime only, finer10 stepsRs 10.351260/-minus 0.099323
GridTime and the level0.05 and 1,000Rs 10.394588/-minus 0.055996
Sampling methodNothing10,000 drawsa band, not a figurespread 0.147194
Closed formNothing to chopevaluated onceRs 10.450584/-nought by construction

Read the fourth row again. The fourth row is the one that does not fit the table. The other three have a result and an error. The sampling method has neither. What it has is a centre that moves on every run and a spread that does not, and the honest way to report it is as a range: the answer lies within 0.147194 of the truth about two times in three, and within 0.294388 about nineteen times in twenty. Three of the four methods answer the question with a number, and the fourth answers it with a width. That difference is the whole reason the Monte Carlo method is treated on its own.

Three deterministic methods. Three different wrong numbers. One right one. 10.20 10.48 CLOSED FORM Rs 10.450584/- BINOMIAL, 10 STEPS Rs 10.253409/- minus 0.197175 0.197175 short TRINOMIAL, 10 STEPS Rs 10.351260/- minus 0.099323 GRID, STEP 0.05 Rs 10.394588/- minus 0.055996 All three sit to the left of the truth at these particular settings, and none of them is on it.
Against a closed form of Rs 10.450584/- the three deterministic methods return errors of minus Rs 0.197175/-, minus Rs 0.099323/- and minus Rs 0.055996/-, so three different wrong numbers stand against one right one.

One more thing the table hides, and it matters when deciding what to run. The three deterministic methods did not do the same amount of work. The binomial at ten steps evaluates 66 node values in total. The trinomial at ten steps evaluates 121. The grid at a space step of 0.05 spans 33 levels and runs a thousand time steps, so it evaluates 33,000. The accuracy ranking and the effort ranking are the same ranking here, the ordinary case rather than a coincidence. The sampling method at ten thousand draws is doing more work than any of them and is still returning a width rather than a number.

Turned the other way round, the comparison becomes the sharpest thing in this guide. How many draws would the sampling method need before one spread was as small as each deterministic error? To match the binomial at ten steps it needs about 5,573 draws. To match the trinomial it needs about 21,963. To match the grid it needs about 69,099. And even then the match is not a match. The deterministic methods are wrong by that amount; the sampling method is merely within that amount most of the time.

And the second row deserves a caution. The trinomial looks like a free upgrade and is not one. The trinomial used here is built by taking two half-length binomial moves and folding them into one step, so its up move, its middle move and its down move are just the four outcomes of two coin tosses collected together. Work through the arithmetic and the trinomial at ten steps returns Rs 10.351260/- to twelve decimal places, the same figure the binomial returns at twenty steps. The trinomial is not a better description of the process; at the same step count it has simply resolved time twice as finely and paid for it in nodes. Both are first order in the step count and both carry the same behaviour underneath.

Try it out

At these settings the trinomial lattice is nearer the truth than the binomial. What is the honest reason?

Which side of the truth does a lattice sit on?

Every number in the worked instance above came out below the closed form, and a reader who stops there will carry away something false. The step counts in that table are all even. Run the same binomial at odd counts and every single reading sits above the closed form instead.

Check the first twelve counts. One step reads Rs 12.162285/-, above the closed form. Two steps reads Rs 9.540501/-, below. Three steps Rs 11.043871/-, above. Four steps Rs 9.970523/-, below. Five above, six below, seven above, eight below, and so on without a single exception at any count up to a hundred. The binomial lattice does not approach its limit from one side; it straddles it by parity, with odd counts above and even counts below at every count checked. Which side appears is decided entirely by which counts are looked at.

The everyday version is a set of steel measuring rods and a doorway. Lay an odd number of rods across the doorway and the last one always overhangs; lay an even number and there is always a gap left. Neither run is wrong and neither is converging from one side. The two runs are interleaved sequences closing on the same width from opposite directions, and reporting only one of them describes a march that is not happening.

The ratio test earns its keep here, and here is also where it can mislead. Take the errors at ten and twenty steps, both even, and the ratio is 1.985177. Take fifty and a hundred, both even, and it is 1.997407. Both are proper first-order readings. But take twenty-five and fifty, one odd and one even, and the ratio comes out at minus 1.764314. Minus 1.764314 is not an order at all; it is the sign flip showing up as a negative number. A ladder that mixes parities without saying so produces exactly that kind of nonsense.

One more detail worth carrying away. From nine steps onward the odd count is nearer the truth than the even count immediately above it, at every pair up to a hundred. Nine steps is out by 0.196576 and ten steps by 0.197175. Ninety-nine steps is out by 0.017722 and a hundred by 0.019972. The two parities are converging at the same rate with different constants, so the sawtooth never closes up.

Odd above, even below. Every count, without exception. Rs 10.450584/- the closed form ODD COUNTS SIT ABOVE EVEN COUNTS SIT BELOW 0.196576 over 0.197175 under 3 4 5 6 7 8 9 10 11 12 steps One step and two steps are off this panel, at plus 1.711701 and minus 0.910082. Educational illustration on invented parameters.
The signed error of a binomial lattice alternates side by side as the step count rises, with odd counts sitting above the closed form and even counts below it, so a ladder of even counts alone shows a one-sided approach that is not happening.
Try it out

The step count is about to be raised from one to a hundred. Before it moves: does the binomial price approach the closed form from below?

Play with it

Move the step count and watch two ladders close on one line

Held fixed: the standard process at Rs 100/-, the strike at Rs 100/-, the rate at 5 per cent, the volatility at 20 per cent and the horizon at one year. The only thing that moves is the number of steps, from one to a hundred. The top panel draws both lattices against the closed form, and the closed form does not move. The lower panel draws the same readings as an error, and the two buttons under it switch between the signed error and the size of that error on a log scale. Every reading is computed from the lattice recursions on each move of the control, and nothing anywhere in it is sampled.

1 step10 steps100 steps
Move the count. Both ladders close in. Neither one lands. BOTH LADDERS, IN RUPEES 9.5 10.0 10.5 11.0 11.5 12.0 binomial lattice trinomial lattice closed form, Rs 10.450584/- 1 3 10 20 50 100 steps THE SIGNED ERROR, IN RUPEES 1.5 1.0 0.5 minus 0.5 minus 1.0 0 The binomial line crosses the zero line at every single step. That is the parity straddle.
Binomial price
10.253409
Binomial error
minus 0.197175
Trinomial price
10.351260
Trinomial error
minus 0.099323

At 10 steps the binomial lattice reads Rs 10.253409/- and the trinomial reads Rs 10.351260/-, against a closed form of Rs 10.450584/-. 10 is an even count, so the binomial sits below the closed form, out by 0.197175, and the trinomial sits below it, out by 0.099323. Neither of them lands on the line.

Educational illustration. Every reading is computed from the lattice recursions on each move of the control rather than sampled, so the default of ten steps reproduces the worked instance above exactly. The binomial ladder reads 12.162285, 9.540501, 11.043871, 9.970523, 10.805934, 10.253409, 10.351260, 10.410692 and 10.430612 at one, two, three, four, five, ten, twenty, fifty and a hundred steps. The trinomial ladder reads 9.540501, 9.970523, 10.125573, 10.205099, 10.253409, 10.351260, 10.400751, 10.430612 and 10.440591 at the same counts. Both approach Rs 10.450584/- and neither reaches it. The trinomial ladder never shows the straddle: because this trinomial equals the binomial at twice the count, it only ever reads even counts, and a run of even counts is one-sided by construction rather than by convergence. Starting value Rs 100/-, strike Rs 100/-, rate 5 per cent, volatility 20 per cent, one year, all invented for teaching.
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What do all four have in common?

All four can be made to converge to the wrong thing, and the full account of it is set out under numerical error and stability. The mechanism is always the same. A method has more than one source of error. One source is refined, the answer stops moving, and the conclusion drawn is that it has converged. Converged it has, on a number that is not the truth. The source left alone is still sitting there at its original size.

Here it is with figures, on the grid. Hold the space step at 0.05, refine the time step alone, and watch the answer.

Time stepsSpace stepGrid resultSigned errorWhat it is doing
2500.05Rs 10.397813/-minus 0.052771closest of all of them
5000.05Rs 10.395663/-minus 0.054921moving away
1,0000.05Rs 10.394588/-minus 0.055996moving away
4,0000.05Rs 10.393782/-minus 0.056802moving away
8,0000.05Rs 10.393648/-minus 0.056936almost stopped
1,28,0000.05Rs 10.393522/-minus 0.057062settled, and settled wrong

Read that table slowly. It inverts the thing everyone believes about refinement. Every row is a finer chopping of time than the row above it. Every row is further from the truth than the row above it. And by the bottom row the answer has stopped moving in the fifth decimal, exactly the behaviour a reader takes as proof of convergence. The grid has converged perfectly well. It has converged on Rs 10.3935/- and change, wrong by about Rs 0.057/-, and the space step of 0.05 was never touched.

Every one of the four does a version of this. The lattice has its own second source in how the strike sits relative to the nodes. The path scheme has one error from chopping time and another from how the random packet is built. The grid has time, level and the edges of its own mesh. The sampling method has the spread of the draws, and also whatever bias the path construction underneath it carries, and refining the count of draws does nothing at all to the second one. The general rule is that refining one source while another is untouched buys a very precise answer to a question nobody asked.

Try it out

What can all four methods be made to do?

The failure: treating stability as evidence of accuracy

Three of the four methods here return the same answer every time they are run. Repeatability feels like accuracy and is nothing of the kind. Repeatability is a property of the arithmetic. A grid at a space step of 0.05 and a thousand time steps returns Rs 10.394588/- on the first run, on the second run, and on the ten thousandth, and it is wrong by Rs 0.055996/- on every one of them.

The everyday version is the clock on the wall that stopped last Tuesday. Look at it on Monday and it says ten past four. Look again on Friday and it still says ten past four. The stopped clock is the most consistent clock in the building. A reader who runs a method twice, gets the same answer and concludes it has converged has established that the method is deterministic and nothing else at all.

The failure is expensive rather than merely embarrassing because of the direction it pushes in. A method with a visible spread announces its own uncertainty; a sampling answer of Rs 10.39/- plus or minus 0.15 cannot be read without noticing that it is uncertain. A deterministic method returns six decimal places and no warning label. The cost is a wrong figure carrying a reassuring consistency. A right figure with a visible spread invites a second look, and a wrong figure with six decimals invites none.

The repair is the one thing this guide has been circling throughout. The question to put to a method is not whether it agrees with itself, but whether it agrees with something known, at a ladder of settings, with the sign kept. Repeatability is free and says nothing. A measured error costs a benchmark and says everything.

Three runs. Three identical answers. Three identical errors. RUN 1 Rs 10.394588/- space step 0.05, 1,000 steps RUN 2 Rs 10.394588/- space step 0.05, 1,000 steps RUN 3 Rs 10.394588/- space step 0.05, 1,000 steps WRONG BY Rs 0.055996/- ON ALL THREE The agreement between the runs contains no information about the agreement with the truth. Repeatability is a property of the arithmetic. Accuracy is a property of the answer. They are unrelated.
The grid returns Rs 10.394588/- on every single run and is wrong by Rs 0.055996/- on every single run, so running it twice and getting the same answer establishes only that the method is deterministic.
Try it out

A method returns the same answer on two runs. What has that established?

Every numerical answer carries an error nobody quotes. See what the four share.

What does none of them do?

None of them produces an exact answer, and bluntness about that is owed before any of the four is opened in turn. Not one of the four is ever right. Every reading in every table in this guide is an approximation, and the useful question is never whether a method is exact but how far off it is, in which direction, and what a refinement would cost to shrink that distance.

None of them reports its own error either. The second omission is the quiet one, and it catches good readers. A method that could report how wrong it was would have to know the truth, and if it knew the truth it would not need to be running. What a method can report is how much its answer moved when the chopping was refined. Movement is a different quantity, and only sometimes a good guide to the error. The grid table above is the counterexample: the answer barely moved between four thousand steps and eight thousand, and the error was seven hundredths of a rupee the whole time.

And none of them removes the modelling question underneath. Every figure in this guide depends on a volatility of 20 per cent that was assumed rather than discovered, and no amount of refinement in any of the four methods touches that. A numerical method answers the question it was handed, to whatever accuracy was paid for, and takes no view at all on whether it was the right question. Fitting the parameters to observed prices is set out under calibration.

Everywhere

Where this holds

No jurisdiction owns the arithmetic of approximation. It reads the same wherever it is written down: a lattice straddles by parity in every country, and a spread falls with the square root of the count of draws in every country. Contract conventions, quotation practice and the way any instrument is actually settled do vary by place, and those belong to the subject area on instruments rather than to the mathematics.

The Monte Carlo method, the discretisation schemes that step a path forward, tree methods and the finite difference method each carry their own construction, failure modes and error behaviour in full. Variance reduction, which is how to need fewer draws for the same width, comes last. The way small errors grow rather than shrink, and the condition that decides which happens, is set out under numerical error and stability. What any contract pays, and why anyone would hold one, is settled in the subject area on instruments. Fitting a parameter to observed prices belongs to calibration.

References

SourceDocumentWhere
arXiv, Quantitative FinancePreprints on lattice convergence, oscillation by parity and finite difference schemes for pricing equationsarxiv.org
Social Science Research NetworkWorking papers on the error behaviour of simulation and lattice methodsssrn.com
Phelim P. Boyle, 1977Options: A Monte Carlo Approach, Journal of Financial Economics, the origin of the simulation approach and of the trinomial lattice used heresciencedirect.com
John C. HullOptions, Futures, and Other Derivatives, the chapters on binomial trees, Monte Carlo simulation and finite difference methodspearson.com

The standard process used throughout and the at-the-money contract struck on it are invented.
Educational material. Not advice on any investment, tax, budget or market position.

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