Quadratic Variation vs Ordinary Variation
Both quantities take the same chopped up path and add up its movements. The first adds the sizes of those movements. The second adds their squares. On a path with a slope the first settles on a finite number and the second collapses to nothing. On a Brownian path the two answers swap places, and the reason is one exponent.
The swap is the whole of this guide, so it is worth saying plainly before anything is defined. Two totals follow, built from exactly the same list of numbers and differing in one operation. Nothing else about them differs: not the path, not the grid, not the order of the arithmetic. And yet on the path this subject actually runs on, one of them is a clean finite answer and the other has no value at all. Every ruler and every odometer produces the first kind of total. A reader who has only ever met that kind will find the swap backwards. The square root scalingThe rule that the size of a Brownian movement over a short span grows with the square root of that span rather than in proportion to it. of a Brownian movement is the single fact that makes it come out that way, and everything below is careful bookkeeping around that fact.
One note before the definitions. This comparison is often the reason a reader arrives at all. Both quantities are therefore built here from nothing rather than assumed. A reader who has already met the squared total elsewhere will find the second section quick; it is present so that a reader who has not met it is never asked to take one side of a comparison on trust.
What is ordinary variation, once it is defined from scratch?
The body already computes this quantity, and has done since the age of three. Start with a walk. A walker leaves the house, goes three hundred metres towards the market, remembers something and comes back a hundred and twenty, carries on two hundred and sixty past the market, then comes back a hundred and forty to a friend's gate and stops. Where is that walker, relative to the front door? Three hundred metres away. How far did the walk cover? Eight hundred and twenty metres.
Distance and displacement are two different questions, and they have two different answers from the same four legs. The first is displacementThe end position less the start position, which knows only about the two ends and nothing about the route between them.: end less start. Even a long walk can produce a small displacement. The second is total distance travelled, and it is the sum of the four leg lengths with their directions thrown away. Ordinary variation is the second question asked of a path: it is the total distance travelled, and it is what an odometer reads while a map reads the displacement.
Now make it a definition rather than a walk. Take an interval from zero to a horizon. Choose a partitionA division of the interval into a finite run of consecutive pieces, marked by the times where one piece ends and the next begins. of it, meaning a finite list of times starting at zero and ending at the horizon. Over each consecutive pair the path has a change. Take the size of each of those changes, throwing away whether it was up or down, and add them. The addition gives one number for that one grid. Then refine the grid and do it again, and keep going.
| \(V_T(X)\) | the ordinary variation of the path \(X\) over the interval from zero to \(T\) |
| \(\Pi\) | one particular grid, being times \(0=t_0<t_1<\cdots<t_n=T\) |
| \(\|\Pi\|\) | the meshThe length of the longest piece in a grid. Driving the mesh to zero forces every piece to shrink, not merely the average piece., being the length of the longest piece in that grid |
| \(|\cdot|\) | the size of a change with its direction discarded, so a fall of three counts the same as a rise of three |
| \(n\) | how many pieces the grid has, which grows as the mesh shrinks |
One property of this construction matters more than any other, and it will do a lot of work later. Splitting a piece into two can never lower the total. Say a piece carried a change of the size five and it is split into two changes of three and minus two. The old grid recorded five and the new grid records three plus two, five as well. If instead the two halves went four and one, the old grid recorded five and the new grid records five again. Whenever the two halves pull in opposite directions the new grid records more, and it never records less. Refining a grid can only raise an ordinary variation total or leave it alone. The limit under refinement is therefore also the largest value the total ever reaches.
What is quadratic variation, once it is defined from scratch?
Change one operation. Where the first construction took the size of each change, this one squares each change. Everything else is identical: the same interval, the same grid, the same list of changes, the same addition, the same refinement at the end.
| \([X]_T\) | the quadratic variation of the path \(X\) over the interval from zero to \(T\) |
| \(\Pi\) | the same object as before, one particular grid over the same interval |
| \(\|\Pi\|\) | the mesh again, and the same limit is being taken |
| \((\cdot)^2\) | squaring, which also discards direction, because a negative number squared is positive |
| \(T\) | the horizon, one year throughout this guide |
Squaring looks like a small change and it is not. Taking a size is a gentle operation: halve a movement and its size halves. Squaring is a violent one: halve a movement and its square quarters. Push that further and the asymmetry becomes obvious. A movement of 0.500000 has a size of 0.500000 and a square of 0.250000, so the square is half the size. A movement ten times smaller, 0.050000, has a size of 0.050000 and a square of 0.002500, so the square is now a twentieth of the size. Squaring shrinks a movement in proportion to how small that movement already was. The two constructions therefore treat a grid of many tiny pieces in completely different ways.
The shrinking rule is the bridge between the two quantities, and it settles both of the cases below at once. Writing it down properly is worth a line. If every movement on a grid is no larger than some biggest movement, then every square is no larger than that biggest movement multiplied by that movement's own size. Adding them all up leaves the squared total sitting under a product of two quantities that already have names.
| \(d_i\) | the change in the path across the grid piece that ends at \(t_i\) |
| \(d_i^{\,2}\) | that change squared, which is one term of the quadratic total |
| \(\max_i |d_i|\) | the size of the single largest movement anywhere on that grid |
| \(\sum_i |d_i|\) | the ordinary total on that same grid |
| \(n\) | the number of pieces in the grid |
Read the bridge one way. On a continuous path the largest single movement is squeezed to zero as the grid refines, and the ordinary total sits still. A path with a finite ordinary total must therefore have a squared total of nothing. Read it the other way, by dividing through, and it says that a path with a squared total that refuses to vanish must have an ordinary total that grows without bound, for exactly the same reason. One inequality, read in two directions, produces both halves of the comparison.
Do either of the two totals record the direction of the movements?
On a path with a slope everywhere, which of the two totals comes out as nothing?
What do the two totals read on a path that has a slope?
Take the tamest path this subject has to hand. Sitting inside the standard processThe single invented traded quantity this reading order runs on, starting at Rs 100/-, with a drift of 8 per cent and a volatility of 20 per cent a year, over one year. is a deterministic trend that lifts its logarithm at six per cent a year in a perfectly straight line. Strip the randomness away and that trend runs from Rs 100/- to Rs 106.18/- over the year without ever turning back. The trend is monotoneMoving in one direction only, so the path never turns back on itself over the interval., it has a slope everywhere, and it is exactly the kind of path an ordinary calculus was built for.
The ordinary total on it is easy, and it is easy in an instructive way. Every movement is upward, so taking the size changes nothing, and a sum of consecutive changes that all point the same way collapses to the end less the start. Cut the year into one piece and the total is Rs 6.183655/-. Cut it into twelve and the total is Rs 6.183655/-. Cut it into a million and it is Rs 6.183655/- still. The pieces telescope, so on a monotone path the ordinary total is the same number at every grid, and refinement has nothing left to find.
| \(f\) | a path that only ever rises on the interval, in place of a random one |
| \(f(t_i)-f(t_{i-1})\) | the change across one piece, which here is never negative |
| \(|\cdot|\) | the size, which changes nothing when every change already points the same way |
| \(f(T)-f(0)\) | the end less the start, with no reference to the grid at all |
| \(n\) | the number of pieces, which cancels out entirely |
The squared total on the same trend goes the other way. On the logarithm the whole year's movement is 0.060000, so a grid of one piece squares that to 0.003600. Two pieces gives two squares of 0.030000, for 0.001800 in total. Twelve pieces gives twelve squares of 0.005000, for 0.000300. Each doubling of the grid halves the answer, and the bridge above says exactly why: the ordinary total is pinned at 0.060000 while the largest single movement keeps shrinking, so their product keeps shrinking too. Nothing about that argument used randomness, and nothing about it used this particular trend.
What do the same two totals read on the locked path?
Now swap the path and change nothing else. The locked pathThe one constructed Brownian path this reading order draws everywhere, fixed at twelve nodes so that every worked figure reproduces exactly rather than approximately. is a Brownian path over the same year, published at twelve nodes. Its twelve movements are each a driving value multiplied by 0.288675, the square root of one twelfth. The table below runs both constructions across those twelve movements at once, in adjacent columns. Two totals can then be watched being built from one list.
| Month | Driving value | Movement | Its size | Its square |
|---|---|---|---|---|
| 1 | -0.5 | -0.144338 | 0.144338 | 0.020833 |
| 2 | 1.6 | 0.461880 | 0.461880 | 0.213333 |
| 3 | -1.3 | -0.375278 | 0.375278 | 0.140833 |
| 4 | -0.1 | -0.028868 | 0.028868 | 0.000833 |
| 5 | 0.1 | 0.028868 | 0.028868 | 0.000833 |
| 6 | 1.5 | 0.433013 | 0.433013 | 0.187500 |
| 7 | -1.3 | -0.375278 | 0.375278 | 0.140833 |
| 8 | -0.5 | -0.144338 | 0.144338 | 0.020833 |
| 9 | -1.4 | -0.404145 | 0.404145 | 0.163333 |
| 10 | 0.4 | 0.115470 | 0.115470 | 0.013333 |
| 11 | 0.9 | 0.259808 | 0.259808 | 0.067500 |
| 12 | 0.6 | 0.173205 | 0.173205 | 0.030000 |
| Total | 0.0 | 0.000000 | 2.944486 | 1.000000 |
Read the two right hand totals and the comparison is already visible. The same twelve numbers give an ordinary total of 2.944486 and a squared total of 1.000000 at this grid. The ordinary total can be checked mentally without the table: the twelve driving values have sizes adding to 10.2, and 10.2 multiplied by 0.288675 gives 2.944486. The squared total can be checked the same way: the twelve driving values have squares adding to exactly 12.0, and dividing by twelve gives 1.000000. Both of those construction facts are built in rather than lucky. The figures therefore reproduce to six places rather than nearly. The individual entries in the two right hand columns are rounded to six places for display. The totals are computed from the unrounded values, so adding the entries by eye lands a few units in the last place away.
Now the part that carries the comparison, and it has to be said carefully. Refine the grid. The squared total stays at the elapsed year however finely the interval is cut, and that is a theorem rather than an observation. The ordinary total grows past any figure that can be named as the cutting gets finer, and that is a theorem too. Neither of those statements is a statement about 2.944486 or about 1.000000. Nothing finite approximates a quantity that is unbounded. The reading at twelve nodes is not the limit of either quantity, and 2.944486 is not an approximation to anything. The locked path is defined at twelve nodes and at no finer grid, so no refined reading of it exists to be quoted.
At twelve nodes the two totals read 2.944486 and 1.000000. Is 2.944486 an approximation to the ordinary variation of the path?
Does adding more grid points always push the totals up?
Here is a question that looks like housekeeping and is not. The six readings of the squared total on the locked path are 0.000000, 0.281667, 0.031667, 1.345000, 1.018333 and 1.000000, in the order of one, two, three, four, six and twelve pieces. The list of squared readings falls as often as it rises. The ordinary total across the same six grids reads 0.000000, 0.750555, 0.288675, 1.962991, 2.424871 and 2.944486, a list that also falls once. Neither list is a march toward anything, and no honest drawing of them can pretend otherwise.
Two separate reasons sit behind that, and separating them is worth the paragraph. The first is that these six grids are not nested. Twelve pieces splits the year into months, so the three piece grid cuts at months four and eight while the two piece grid cuts at month six. Neither of those grids contains the other's cut point, so moving from two pieces to three is not a refinement at all: it is a different grid entirely, and there is no reason for a different grid to give a bigger answer.
Follow a genuinely nested chain instead and the ordinary total behaves exactly as the triangle inequality demands. One piece, then two, then four, then twelve: each grid contains every cut point of the one before it. Splitting a piece into two parts that pull in opposite directions can lower a sum of squares, though it can never lower a sum of sizes. The squared total on the same nested chain therefore still refuses to march.
| The nested chain | 1 piece | 2 pieces | 4 pieces | 12 pieces |
|---|---|---|---|---|
| Ordinary total | 0.000000 | 0.750555 | 1.962991 | 2.944486 |
| Squared total | 0.000000 | 0.281667 | 1.345000 | 1.000000 |
| Behaviour along the chain | ordinary total never falls | squared total falls at the last step | ||
The ordinary total is monotone under genuine refinement and the squared total is not. The two quantities differ in their structure, then, and not only in their limits. That difference is worth carrying: a rising sequence of ordinary readings proves nothing about the limit being reached, since a rising sequence can rise forever, while a squared reading that jumps about says nothing about the limit being missed.
Why does refining the grid drive the two totals apart?
One exponent, and here it is. On a Brownian path the movement across a piece of length delta-t has a size of the order of the square root of delta-t. Cut the year into n equal pieces and each piece has length one over n, so each movement is of the order of one over the square root of n. Now do the two sums.
The squared total adds n terms, each of size one over n, so the count and the shrinking cancel exactly and the total sits still. The ordinary total adds n terms, each of size one over the square root of n, so the count wins: n divided by the square root of n is the square root of n, and that climbs. Double the number of pieces and the ordinary total rises by roughly a factor of 1.414214, forever, with no ceiling anywhere. The same square root scaling that pins one total to the elapsed time is what sends the other one to infinity, so the two behaviours are not two facts but one fact seen twice.
| \(W_t\) | standard Brownian motion under the physical measure P |
| \(\mathbb{E}[\cdot]\) | the expectation, taken over every path the process could take |
| \(n\) | the number of equal pieces the horizon is cut into |
| \(T\) | the horizon, one year here |
| \(\sqrt{2/\pi}\) | 0.797885, the average size of a standard normal quantity once its sign is discarded |
The expression above is a statement about a Brownian path in general and about the average over all of them. The formula is emphatically not a reading of the locked path. The locked path exists at twelve nodes and nowhere else. Used as intended the expression gives 2.763953 for a general Brownian path cut into twelve pieces, 5.753627 at fifty two, 12.666025 at two hundred and fifty two and 40.053487 at two thousand five hundred and twenty. Those four figures are illustrative readings of a formula rather than measurements of anything, and their only job is to show the square root climbing.
A Brownian movement scales with the square root of the step. What happens to the sum of the sizes as the grid refines?
The grid is refined by the control below. Which of the two totals settles down, and which one climbs?
Cut the same year into fewer pieces and watch both totals at once
The locked path never changes. Only the grid does. The grey line is the path and the dark chords are what the grid actually measures. Below the path, each piece gets two bars: the bar above the line is the size of that movement and the bar below is its square. The two meters at the bottom carry the two totals on their own scales.
The error that gets made, and what it costs
Assuming that a path which stays in a narrow band, and which finishes near where it began, must have travelled a modest distance. For every path anyone has ever physically walked the assumption holds. The intuition is almost impossible to shake for that reason.
The locked path settles it. Over the year the standard process built on it never rises above Rs 111.08/- in June and never falls below Rs 93.74/- in September, so the whole year is contained in a band Rs 17.34/- wide. The twelve driving values were built to sum to zero, so the Brownian path underneath finishes the year at exactly where it started. And yet the distance travelled by that process, read at its twelve nodes, is already Rs 60.30/- in rupee terms, three and a half times the width of the band it never left. Refine the grid and that figure keeps climbing, with no number it is heading toward. Staying in a narrow band constrains the displacement and says nothing whatever about the distance travelled, and on a Brownian path the second of those is unboundedGrowing past any figure that can be stated as the grid refines, so having no limit at all rather than a large one..
The cost is a calculation resting on total distance travelled, a quantity every instinct says must exist. The distance travelled does not converge as the sampling is refined, so it returns a different answer every time somebody changes how often the path is read, and it returns a larger one every time they read it more often. Nothing in the arithmetic flags that. The figure comes back looking perfectly respectable, carries six decimal places, and is a property of the reading schedule rather than of the path. The quantity that does behave is the sum of squares, the one that looks as though it ought to be the more delicate of the two.
The locked path stays inside a band Rs 17.34/- wide all year. Is the distance it travelled finite?
What does each of the two totals actually measure?
Words worth holding, now that both have been computed. The ordinary total measures effort: how much movement happened, counted without regard to whether it got anywhere. The ordinary total is the pedometer, the odometer, the running tally of every step out and every step back. Its natural units are the units of the thing itself, so on the standard process it is quoted in rupees.
The squared total measures something with no everyday name, and the nearest honest description is accumulated spread. The squared total is what a variance budget spends. Its natural units are the units of the thing squared, so on the standard process it would be quoted in rupees squared. Nobody has an intuition for rupees squared, and that is exactly why the squared total gets read on the logarithm instead. On the Brownian path underneath the scale is dimensionless, and there the squared total comes out as the elapsed time. No answer is cleaner.
Here is the property both share, and it is the one readers most often forget. Neither total knows anything about direction. Taking a size discards the sign and squaring discards it too, so an upward movement and a downward movement of the same size contribute identically to both. Neither total records where the path ended up. The locked path proves as much without argument. Its twelve movements add to exactly 0.000000. The sizes of those same twelve movements add to 2.944486 and their squares add to 1.000000. Three totals from one list of twelve numbers, and only the one nobody was asking about knows that the path came home.
Why does the answer decide which calculus applies?
An ordinary calculus is built on one move made so often that it stops being visible: when a small change is broken into a first order piece and a second order piece, the second order piece is discarded. Every differentiation rule in ordinary calculus is that discarding, performed once and then given a name. The discarding is legitimate precisely when the second order pieces add up to nothing across the interval, and the sum of the second order pieces is the squared total.
So the comparison is not a curiosity about two ways of adding. The comparison is a gate. If a path has a finite ordinary total then the bridge inequality forces its squared total to be nothing, the discarding is legitimate, and ordinary calculus applies without a word of modification. The usual name for that condition is finite variationThe property of having an ordinary variation that stays bounded however finely the grid is refined.. If instead a path has a squared total that will not vanish, the bridge inequality forces its ordinary total to be unbounded, the discarding is illegitimate, and a different construction is needed. Which calculus applies is decided by whether the ordinary total survives refinement, and on a Brownian path it does not.
Ordinary calculus needs which of the two totals to be finite?
Which of the two has been handed over?
Variation figures arrive without labels. Somebody hands over a number and calls it how much the quantity moved over the year, and the two totals in this guide are both honest answers to that phrasing while being completely different objects. Three tests separate them, and none of them requires access to any data or any software.
- Check the units
An ordinary total carries the units of the thing itself, so on the standard process it is quoted in rupees. A squared total carries the units squared. If the figure is quoted in rupees it is a distance travelled, and if it is quoted in rupees squared, or as a plain dimensionless number on a logarithm, it is a squared total.
On the locked path the two read Rs 60.30/- and 1.000000, and the second has no rupee sign because it cannot have one.
- Change the reading frequency and look
Read the same path twice as often and see what the figure does. A distance travelled on a Brownian path rises by roughly a factor of 1.414214 for every doubling of the frequency, forever. A squared total has no such systematic drift, because the count and the shrinking cancel.
Any figure that grows every time the sampling gets more frequent is measuring the schedule, not the path.
- Compare it with the displacement
Work out end less start, which takes one subtraction. A number close to that is a displacement. A number far above it is one of the two variation totals, and the units test then says which.
The locked path has a displacement of Rs 6.18/- and a twelve node distance travelled of Rs 60.30/-, and the two are not confusable once both are stated side by side.
- Ask what grid produced it
Neither total means anything without its grid. A distance travelled without a stated reading frequency is not a figure that can be compared with any other figure, because it is a property of the frequency.
The locked path returns Rs 6.18/-, Rs 15.97/-, Rs 6.18/-, Rs 40.87/-, Rs 50.25/- and Rs 60.30/- across the six grids on the same unchanged path.
The household version of the same test is a walk to the market. A step counter and a map disagree about how far the walk went, and the disagreement is not an error in either device: they are answering different questions. A Brownian path does what no ordinary walk has ever done: its step counter reading has no final answer at all. A finer step counter raises the number. A finer one again raises it again. A figure that has no ceiling is not a measurement of the path, and asking which grid produced it reveals as much.
Somebody quotes a total movement figure for a Brownian path and doubles the reading frequency. What should the figure be expected to do?
Readers of this subject sometimes look for an authority. Here is a closing note on universality. No jurisdiction anywhere sets either definition, no regulator publishes a value for either total, and no market convention alters what a limit over refining grids equals. Both quantities are statements about paths rather than about anything traded, so no authority governs the arithmetic. The only thing a convention ever decides is how often somebody reads a path, and one of the two totals depends entirely on that while the other does not.
References
| Source | Document | Where |
|---|---|---|
| arXiv Quantitative Finance | Preprint repository for pathwise variation results and their use in pricing theory | arxiv.org |
| Social Science Research Network | Working paper repository for the same material | ssrn.com |
| Ito | The integral and the chain rule that carry his name, both made necessary by the failure of the ordinary total | named in the text only |
| Hull, Shreve and Wilmott | Standard texts setting out quadratic variation and the construction that rests on it | textbooks |
The standard process and the locked path are invented.
Educational material. Not advice on any investment, tax, budget or market position.
