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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
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4Stochastic Differential Equations
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5Pricing Theory and No-Arbitrage
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6Option Pricing Theory
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10Calibration and Model Risk
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Euler and Milstein Schemes Compared: One Extra Term

A discretisation scheme turns a continuous equation into a rule for taking finite steps. The Euler scheme takes the drift and the randomness at the start of each step and applies both across it. The Milstein scheme adds one further term, built from the square of that step's randomness, and that single addition is the entire difference between them.

One extra term is the whole comparison. The extra term improves the rate at which the error shrinks as the steps are made shorter, and it makes no claim whatever about any particular path at any particular number of steps. A better rate and silence about single paths are not in tension. On the one path worked through below, the better scheme is beaten at the finest partition it is tested on, and holding both claims at once is the only way to read that table without drawing the wrong conclusion from it.

The one term that separates the two schemes is set out below, symbol by symbol, and then both schemes are run along a single path that can be checked by hand, at six different step counts.

What does a discretisation scheme actually do?

A continuous equation describes what happens over an interval of no length at all. A machine cannot take a step of no length. Something has to bridge the two, and a discretisation schemeA rule turning a continuous equation into finite steps a machine can take. is that bridge: a rule saying what to do over a step of real, positive length, given what was known when the step began.

Here is the everyday version. A weighing scale that reports once a second cannot say what a moving object weighed at every instant between readings. The scale gives a reading, and then a rule fills in the gap: hold the last reading until the next one arrives, draw a straight line between them, or something cleverer. Each of those fill-in rules is a discretisation scheme, and they differ only in what they assume happened between the readings. The scale is the same scale in every case. Only the rule changes.

The choices compared here are exactly of that kind. Both schemes read the process at the start of a step. Both apply something across the step. The two schemes differ only in what they apply.

The horizon here is one year, and the partitionThe set of times at which an interval is cut into steps, and the lengths those cuts produce. divides it into a whole number of equal steps. Twelve divides by one, two, three, four and six. So the step counts used here are one, two, three, four, six and twelve, and the same underlying path can be read at every one of those coarsenesses without the path itself being changed.

Six partitions of one year. Same path underneath every one of them. Each row cuts the same year at different places. Nothing about the path changes from row to row. 1 step 1.0000 2 steps 0.5000 3 steps 0.3333 4 steps 0.2500 6 steps 0.1667 12 steps 0.0833 Step length in years is shown on the right of each row. Twelve divides by every count above it, so one path serves all six rows.
The same one year interval is cut into one, two, three, four, six and twelve equal steps, and because twelve divides by every one of those counts the identical underlying path can be read at all six coarsenesses without being changed.
Try it out

What is a discretisation scheme for?

What is the Euler scheme, in one line?

The Euler schemeTaking the drift and the randomness at the start of each step and applying both across the whole of it. reads the drift and the volatility at the moment the step begins, treats both as fixed for the length of the step, and pushes the process forward by the drift multiplied by the step length plus the volatility multiplied by the step's random increment. Nothing is re-read part way through. Whatever the process did inside the step is invisible to the rule.

The standard process, an invented single traded quantity used throughout this subject area, starts at Rs 100/- with a drift of 8 per cent and a volatility of 20 per cent over one year. Written for that process, the rule is one line.

The Euler scheme on the standard process
$$ S_{k+1} \;=\; S_k\Bigl(1 \;+\; \mu\,\Delta t \;+\; \sigma\,\Delta W_k\Bigr) $$
\(S_k\)the standard process at the start of step \(k\), in rupees
\(\mu\)the drift, 8 per cent a year, decimal 0.08, an invented parameter
\(\sigma\)the volatility, 20 per cent a year, decimal 0.20, an invented parameter
\(\Delta t\)the length of one step, in years, being one over the number of steps
\(\Delta W_k\)the increment of the Brownian path across step \(k\), read from the locked path
What it says in wordsThe value at the end of a step is the value at the start multiplied by one plus the drift times the step length plus the volatility times the step's random increment, with the drift and the volatility both held at whatever they were when the step began.

That one line is the whole scheme. The rule carries the name of Euler from the deterministic case and of Maruyama from its stochastic form. It is nothing more than reading the equation literally and replacing the infinitesimals with finite quantities, so it is the rule that any first attempt at stepping an equation forward lands on unaided.

The assumption inside it is that the coefficients do not move within a step, and that assumption is exactly what the error will be made of. Over a step of one year the process moves a very long way, so the coefficients read at the start are badly out of date by the end. Over a step of one month they are only slightly out of date. Staleness of the coefficients is the intuition for why shortening the steps helps, and for what a better scheme has to correct.

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What one term does the Milstein scheme add?

The Milstein schemeThe Euler scheme with one further term added, built from the square of the step's randomness less the step length., published by Milstein in 1975, keeps every part of the line above and adds one more. The extra termHalf the volatility squared, multiplied by the square of the step's random increment less the length of the step. is half the volatility squared, multiplied by the square of the step's random increment less the step length.

The Milstein scheme on the standard process
$$ S_{k+1} \;=\; S_k\Bigl(1 + \mu\,\Delta t + \sigma\,\Delta W_k \;+\; \underbrace{\tfrac{1}{2}\,\sigma^{2}\bigl[(\Delta W_k)^{2} - \Delta t\bigr]}_{\text{the one added term}}\Bigr) $$
\((\Delta W_k)^2\)the square of the step's random increment, never negative
\(\Delta t\)the length of the step, subtracted so the bracket is centred on zero
\(\tfrac{1}{2}\sigma^{2}\)half the variance rate, here half of 0.04, which is 0.02 exactly
What it says in wordsThe Milstein rule is the Euler rule with one further quantity added inside the bracket: half the variance rate multiplied by the amount by which the square of the step's random increment exceeded the length of the step.

The bracket is the whole of the difference, and it repays slow reading. The square of the random increment is a number that is never negative. The step length is subtracted from it. So the added quantity is positive when the step's randomness came out larger than typical for a step of that length, and negative when it came out smaller. The added quantity is not a correction for the direction the process moved; it is a correction for how much the process moved, whichever way it went.

The added term exists to close a gap in the plain rule. The plain rule assumes the volatility contribution scales with the increment alone. The contribution does not scale that way. Over a step of real length the squared increment carries an effect of its own, and a rule ignoring that effect is systematically mis-stating every step. The added term is the leading part of what was being ignored.

Two rules, aligned row by row. Only the third row differs. THE EULER SCHEME drift times step length read at the start of the step volatility times the increment read at the start of the step nothing in this row THE MILSTEIN SCHEME drift times step length identical to the left panel volatility times the increment identical to the left panel half the variance rate, times squared increment less step length Two rows match exactly. One row is empty on the left and filled on the right. There is no other difference between the two schemes anywhere.
The Euler and Milstein schemes match exactly on the drift row and on the randomness row, and differ in one row only, which is empty for the Euler scheme and holds half the variance rate times the squared increment less the step length for the Milstein scheme.
Try it out

What is the Milstein scheme's extra term built from?

What does the extra term buy?

Two things are true of the added bracket at once. The first is that the bracket averages to exactly nothing. The square of a random increment over a step has an average equal to the length of that step, so subtracting the step length centres the whole quantity on zero. Over many steps and many paths it adds nothing at all to the answer.

The second is that it is not nothing on any actual step. Averaging to zero and being zero are different properties, and confusing them is the commonest way to misread this term. Think of a queue at a counter where the average wait is four minutes. Subtract four minutes from every wait and the adjusted figures average to zero. A zero average does not mean nobody waited. The average wait is unchanged by the adjustment, and an unchanged average is a far weaker claim than no waiting.

The locked path used here divides the year into twelve steps with twelve driving values whose squares sum to exactly twelve. Squares summing to exactly twelve are what make the bracket sum to exactly zero across the year, so the claim can be shown rather than asserted.

The added bracket, month by month. Five up, seven down, summing to exactly nothing. Squared increment less step length, at the twelve step partition of the locked path. 0 1 2 3 4 5 6 7 8 9 10 11 12 step number above zero: the step moved more than typical below zero: the step moved less than typical The twelve bars sum to exactly zero, by construction and not by rounding. No bar is zero, and that is the whole difference.
Across the twelve step partition of the locked path the added bracket sits above zero five times and below zero seven times and sums to exactly zero, yet no individual bar is zero, which is why the term shifts nothing on average while correcting every single step.

So what is bought is not a shift in the answer but a reduction in the spread of the answer around the truth. On steps where the randomness came out unusually large or unusually small, the plain rule is most wrong and the term is largest. Where the plain rule was nearly right, the term is near zero. A correction that averages to nothing but tracks the error it is correcting reduces error without introducing a bias, and that is the entire case for carrying the extra term.

Stated formally, what improves is the order of convergenceThe power of the step length at which the error shrinks as the steps are made shorter. in the strong convergenceConvergence of the approximate path itself toward the true path, rather than convergence of an average taken over many paths. sense. Strong convergence is convergence of the path itself rather than of an average over paths.

Strong order, the two schemes
$$ \mathbb{E}\bigl|\,\hat{S}_T - S_T\,\bigr| \;\le\; C\,(\Delta t)^{\,p}, \qquad p_{\text{Euler}} = \tfrac{1}{2}, \qquad p_{\text{Milstein}} = 1 $$
\(\hat{S}_T\)the value the scheme produced at the horizon, in rupees
\(S_T\)the exact value at the horizon on the same path, in rupees
\(\mathbb{E}\)the average taken over paths under the physical measure P
\(C\)a constant that does not depend on the step length
\(p\)the strong order, one half for the Euler scheme and one for the Milstein scheme
What it says in wordsThe average size of the gap between the approximated path and the true path is bounded by a constant multiplied by the step length raised to a power, and that power is one half for the Euler scheme and one for the Milstein scheme, which is what the added term buys.

Read the exponents rather than the constant. Halving the step length cuts the Euler bound by a factor of about 1.41 and the Milstein bound by a factor of 2. The gap compounds: dividing the step length by a hundred cuts one bound by ten and the other by a hundred. Every word of that sentence carries an average over paths inside it, and none of it is a statement about any single path that might be run.

Try it out

The extra term averages to nothing. How can it improve anything?

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What do the two schemes give on a path that can be checked?

Everything above is a claim about averages. Here is one path, worked all the way through, so the claim can be tested against something concrete rather than accepted.

The path is the locked twelve step path, an invented construction, used throughout this subject area. The path divides the year into twelve equal steps and carries twelve driving values that sum to exactly zero and whose squares sum to exactly twelve. Because the driving values sum to zero, the Brownian path returns to where it began at the horizon. The exact answer at the horizon is therefore the starting value of Rs 100/- multiplied by e to the 0.06, or Rs 106.183655/- exactly, and every error below is measured against that one number.

Both schemes are run along that identical path at six different step counts. Nothing about the path changes between rows. The only thing that changes is how many steps the year is cut into and which rule is applied across each step.

StepsEuler valueEuler errorMilstein valueMilstein errorCloser
1Rs 108.000000/-plus 1.816345Rs 106.000000/-minus 0.183655Milstein
2Rs 107.596667/-plus 1.413012Rs 106.107693/-minus 0.075961Milstein
3Rs 108.150785/-plus 1.967130Rs 106.121796/-minus 0.061859Milstein
4Rs 105.299068/-minus 0.884587Rs 106.273222/-plus 0.089567Milstein
6Rs 106.132004/-minus 0.051651Rs 106.179971/-minus 0.003684Milstein
12Rs 106.184491/-plus 0.000836Rs 106.177782/-minus 0.005872Euler

Start at the top row, the crudest thing anybody could do: cut the year into a single step and take it whole. With a Brownian increment of zero over the whole year the only thing left is the drift, so the plain rule gives Rs 108.000000/-, one hundred multiplied by one plus 0.08 being one hundred and eight. The corrected rule subtracts half the variance rate, 0.02, from that same bracket, giving exactly Rs 106.000000/- and landing within Rs 0.183655/- of the truth on a single step.

One step, one added term. The error falls by exactly Rs 2.000000/-. Error against the exact Rs 106.183655/-, at the one step partition of the locked path. 0 plus 1.816345 Euler error minus 2.000000 the added term half of 0.04, times zero less one, times Rs 100/- minus 0.183655 Milstein error a factor of nearly ten, on the crudest partition 1.816345 divided by 0.183655 is 9.89 Bars measured against the exact answer. Note the error also changed sign, from too high to slightly too low.
At one step the added term contributes exactly minus Rs 2.000000/-, which carries the error from plus Rs 1.816345/- down to minus Rs 0.183655/-, a reduction in size by a factor of 9.89 on the crudest partition available.

The two error columns settle the rest. The corrected scheme is closer at one step, at two, at three, at four and at six, and the margin is not small: a factor of 9.89, then 18.60, then 31.80, then 9.88, then 14.02. Five rows, five wins, and by an order of magnitude or better in every one of them. Then look at the last row, where the corrected scheme has an error of minus Rs 0.005872/- against the plain scheme's plus Rs 0.000836/-, and loses by a factor of seven.

Try it out

At one step, what does adding the extra term do to the error?

Try it out

The Milstein scheme has beaten the Euler scheme at one, two, three, four and six steps. Which wins at twelve?

Play with it

Step through all six partitions and watch the ranking reverse once

Held fixed: the starting value at Rs 100/-, the drift at 8 per cent, the volatility at 20 per cent, the horizon at one year, and the locked twelve step path underneath every setting. The only thing that moves is how many steps the year is cut into. Both staircases are recomputed from the locked driving values, never sampled, so the reading at any setting is identical every time it is opened. At twelve steps the two staircases finish 0.006709 apart and sit on top of each other on the price scale. The strip underneath therefore measures the two errors on a logarithmic scale instead.

1 step12 steps12 steps
Two staircases over one year, against the exact answer. 95 100 105 110 exact answer at the horizon Rs 106.183655/- start of year one year Euler staircase Milstein staircase Size of the error, on a logarithmic scale. Whichever marker sits further left is the closer of the two to the truth. 0.001 0.01 0.10 1.00 closer to the truth further from the truth Euler Milstein
Euler finish
106.184491
Euler error
plus 0.000836
Milstein finish
106.177782
Milstein error
minus 0.005872
At twelve steps the Euler scheme finishes at Rs 106.184491/- for an error of plus Rs 0.000836/-, and the Milstein scheme finishes at Rs 106.177782/- for an error of minus Rs 0.005872/-, so on this one path the plain scheme is closer by a factor of 7.02, which is the reversal this guide is about.
Educational illustration. Starting value Rs 100/-, drift 8 per cent a year, volatility 20 per cent a year, horizon one year, exact answer Rs 106.183655/-. Every figure is computed from the locked driving values by formula rather than drawn at random, so the reading at every setting is fixed and reproducible.
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What does the extra term not promise?

The extra term does not promise to win. Not on any particular path, not at any particular step count, not on any single run made to check it. The term promises a rate, and a rate is a statement about a limit taken over an average across many paths. One path is not an average, and six readings of one path are still not an average.

Look at what the two error columns actually do as the steps shorten. The plain scheme starts too high, stays too high, goes higher still at three steps, crosses to too low at four, stays too low at six, and crosses back to too high at twelve. The plain scheme's error makes a sign changeThe point at which an error crosses zero, from too high to too low or the other way round. twice over. The corrected scheme starts too low, stays too low, stays too low again, crosses to too high at four, and crosses back to too low at six. Also twice.

Neither sequence declines cleanly either. The plain scheme's error grows from 1.413012 at two steps to 1.967130 at three, so a refinement of the grid made the answer worse. The corrected scheme's error grows from 0.061859 at three steps to 0.089567 at four, and again from 0.003684 at six steps to 0.005872 at twelve. Refining a partition on a single path is not obliged to help, and on this path it repeatedly does not.

Size of error against number of steps. Both scales are logarithmic. Lower on the chart means closer to the exact Rs 106.183655/-. The lines cross once, at twelve steps. 1 0.1 0.01 0.001 at twelve steps the lines cross over plain scheme below, so closer Euler Milstein 1 2 3 4 6 12 number of steps in the year Euler plus plus plus minus minus plus Milstein minus minus minus plus minus minus The two rows above give the sign of each error at each step count. Each row changes sign twice, and neither is a clean decline anywhere.
Plotting the size of each error against the number of steps shows the Milstein line sitting below the Euler line at one, two, three, four and six steps and rising above it at twelve, while the sign rows underneath show each scheme crossing zero twice rather than settling.
Try it out

Does either scheme's error fall cleanly as the steps shorten on this path?

The error that gets made, and what it costs

The wrong reading is that the better scheme is the one that always wins, and that a test on a path is therefore a test of the schemes. Somebody implements both rules, runs them on one path at a handful of step counts, sees the corrected rule beaten at the finest partition, and concludes the correction is not worth carrying. Everything in that sequence was done carefully. The measurement is real, the arithmetic is right, and the conclusion is wrong.

On the locked path the Milstein scheme beats the Euler scheme at five of six step counts, often by a factor of ten or more, and loses at twelve steps by minus Rs 0.005872/- against plus Rs 0.000836/-. The improvement the added term offers is in the rate at which error falls when averaged over many paths, and one path is not a rate. A reader who takes a single path as a test will sometimes conclude the better scheme is worse, and will be looking at a perfectly real number that supports no such conclusion.

The cost is a correct measurement used to answer a question it cannot answer. In practice it costs either a correction quietly dropped from a working implementation because a one path check made it look useless, or the mirror of that, a correction kept for the wrong reason and then trusted on a single run where it happens to be worse. Both errors come from the same place: reading a rate as though it were a promise.

How many times smaller the corrected error is. Five wins and one loss. Euler error divided by Milstein error, in size. Above the line means the corrected scheme is closer. 1.0 9.89 18.60 31.80 9.88 14.02 0.14 seven times worse 1 step 2 steps 3 steps 4 steps 6 steps 12 steps above the line: corrected scheme closer below the line: plain scheme closer Every bar is a real, correctly computed measurement on one path. None of them, and not all six together, measures a rate of convergence.
Dividing the Euler error by the Milstein error in size gives 9.89, 18.60, 31.80, 9.88 and 14.02 at the first five step counts and 0.14 at twelve steps, so the single bar hanging below the break even line is the whole of what one path is entitled to do.
Try it out

Someone tests two schemes on one path and finds the better one loses. What have they established?

The extra term sharpens each path, not the average. See what it never promises.

When is the extra term worth its cost?

The added term costs one squaring and one subtraction per step. One squaring and one subtraction are close to free on any machine built in the last forty years, so the cost question is rarely about arithmetic time. The cost question is about whether the thing being improved is the thing being measured.

The added term improves how faithfully the whole approximated path tracks the true path. So the test is simple: does the quantity being computed read the whole path, or only where the path finished?

If the quantity reads the whole path, a better path approximation feeds straight through, and the correction earns its keep. A scheme can wander badly in the middle and still arrive close to the right place, so if the quantity reads only the endpoint, the improvement in path fidelity may buy very little. There is a separate notion of convergence covering exactly that case, and it treats the two schemes far more evenly than the path-fidelity notion does.

One question decides it. What does the quantity being computed actually read? Does the quantity read the whole path, or only where the path finished? the whole path only the finish THE CORRECTION EARNS ITS COST The reading depends on where the path went, not only on where it ended, so a closer path is a closer answer, step by step. Carry the extra term. OFTEN IT DOES NOT The reading is indifferent to the middle of the path, so a scheme can wander and still arrive close to the right place. Measure before carrying it. The added term improves the path. Whether that improves the answer depends entirely on whether the answer was reading the path in the first place.
The extra term is worth carrying when the quantity being computed reads the whole path, and often is not when the quantity reads only where the path finished, because a scheme can wander in the middle and still arrive near the right endpoint.
Try it out

The quantity being computed depends only on where the path finished. Does the extra term earn its cost?

How does somebody checking a computation rather than building one use this?

Most people who meet these two schemes are not writing them. The reader is instead handed a number somebody else produced, and has to decide how much of it to believe. Three questions get most of the way, and none of them needs the code.

The first is which rule was used and at what step length. A figure quoted without its step length is not a figure; it is one of a range of figures, and the range here runs from Rs 105.299068/- to Rs 108.150785/- for the plain rule on one path. The step length is part of the answer, not a footnote to it.

The second is what convergence evidence exists. A single run at a single step length is a number. A run repeated at halving step lengths, with the errors shrinking at roughly the promised power, is evidence. The promised power is a statement about an average, and an average needs many paths before it means anything. Ask for the repeated run, and ask how many paths went into it.

The third is whether the quantity being computed reads the whole path or only the finish. The whole-path question decides whether the corrected scheme was worth carrying, and somebody who cannot answer it about their own computation has not yet decided what they are measuring.

A household example makes the shape of it plain. A bill added up once, giving a total that looks fine, is one run at one setting. Adding it up a second time, a different way, and getting the same total is not proof either. The second reading is evidence about the method rather than about the arithmetic, so the two readings together are worth much more than twice one of them. Convergence evidence is exactly that second reading, done systematically.

Does any of this depend on jurisdiction?

No. There is no regulator of discretisation, no jurisdiction in which the squared increment behaves differently, and no local convention that changes the strong order of either scheme. Milstein published in 1975 and the result has held everywhere since. The strong order is a statement about the mathematics, not about any market, any exchange or any set of conduct rules.

Permission to use a number produced this way does vary by place, and permission belongs to conduct and disclosure rules rather than to numerical analysis.

The sampling method that generates paths in the first place is covered separately, alongside how its accuracy scales with the number of paths. Lattice methods and grid methods are each treated on their own. Fitting a parameter to an observation and pricing a contract are separate subjects again: the standard process here is stepped forward and checked against its own exact solution, nothing more.

References

SourceDocumentWhere
arXiv, Quantitative FinancePreprints on discretisation schemes and strong convergence for stochastic differential equationsarxiv.org
Social Science Research NetworkWorking papers comparing scheme accuracy in derivative pricing computationsssrn.com
Hull, Shreve and WilmottStandard texts covering discretisation schemes and strong convergence for stochastic differential equationsPublished editions

The standard process and the locked twelve step path are invented.
Educational material. Not advice on any investment, tax, budget or market position.

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