The Ito Integral vs the Riemann Integral: Where They Part

A Riemann integral multiplies the integrand by the width of each step, and it converges to the same number wherever inside the step the integrand is read. An Ito integral multiplies the integrand by the random increment of each step, and it converges to different numbers for different reading points. Every other difference between the two constructions follows from that one substitution.

Both of these objects are built the same way. An interval is cut into steps, the integrand is read once in each step, that reading is multiplied by something, the products are added up, and the question is what happens to the total as the steps get finer. Four moves, in the same order, in both constructions. The only thing that changes between them is what the reading gets multiplied by.

In one case it is multiplied by the width of the step, a quantity that is always positive, is the same at every step, and shrinks in exact proportion as the grid is refined. In the other it is multiplied by the change in a random path across the step, a quantity that is signed, is different at every step, and shrinks only like the square root of the step. The difference in exponent, one against one half, is where the two constructions part. Every other difference between them is a consequence of that one.

The worked instance runs on the locked path published earlier in this subject: twelve equal steps across one year, driven by the twelve values minus 0.5, 1.6, minus 1.3, minus 0.1, 0.1, 1.5, minus 1.3, minus 0.5, minus 1.4, 0.4, 0.9 and 0.6, each multiplied by the square root of one twelfth, or 0.288675. The path was constructed rather than sampled, so it reproduces on every reload and every figure below can be checked by hand. The standard process built on it starts at Rs 100/-, reaches a high of Rs 111.08/- at month six and a low of Rs 93.74/- at month nine, and finishes at Rs 106.18/-.

What is a Riemann integral, defined from scratch?

The construction starts with a function of time and an interval to integrate it over. The interval is cut into steps. In each step, any point at all is chosen and the function is read there. The reading multiplied by the width of the step gives a rectangle. All the rectangles are added up. The steps are then allowed to get narrower without limit, and the question is what the total settles down to. If it settles on one number, and if that same number appears whichever point inside each step was read at, that number is the Riemann integral.

The freedom in the middle of that description is the thing to hold on to. The function may be read anywhere inside the step, and the definition insists that the answer must not care which choice was made. Reading at the left edge of every step, at the right edge of every step, or at the middle of every step must all give the same limit. If they did not, there would be no single object to call the integral, and the construction would have failed before it started.

There is a household version of this that makes the freedom obvious. Suppose water is charged for by the rate the meter is running at, multiplied by how long it runs. Over a whole hour the rate wanders a little, so it matters somewhat whether the meter was read at the top of the hour or at the end of it. Billing by the minute instead, the rate still wanders, but each reading error is now multiplied by one minute rather than by one hour, and there are only sixty times as many of them. Shortening the billing window further, the choice of when in the window to read stops mattering at all. The argument works only because the multiplier is a length of time that shrinks along with the window.

Two words of vocabulary before the second definition. A partitionThe set of grid points that cuts an interval into steps. Refining a partition means adding more grid points so that every step gets shorter. is the set of cuts that produces the steps, and the integrandThe quantity being read inside each step and multiplied by something. The integrand is the thing under the integral sign. is whatever is being read inside them. Both words carry over unchanged into the second construction. The shared vocabulary is part of why the two look so alike when written down.

The whole of ordinary integration hangs on the last line of that figure. The three rectangles are genuinely different at any grid that can be drawn. The construction does not say they are equal. The construction says the total difference between the three sums goes to nothing as the grid refines, and the reason is arithmetic about sizes rather than anything subtle. A height gap times a width, added up over more and more steps, shrinks. The width falls faster than the number of steps grows. The arithmetic is worth holding on to. In the second construction the same arithmetic is done and it comes out the other way.

What is an Ito integral, defined from scratch?

Take the same interval and the same partition. Take a random path, Brownian motion, written W with a time subscript and starting at zero. In each step, read the integrand at the start of the step and nowhere else. Multiply that reading not by the width of the step but by the incrementThe change in a random path from one grid point to the next, taken as a quantity in its own right. An increment carries a sign and is different at every step. of the path across the step, which is the value of the path at the end of the step less its value at the start. Add up the products. Refine the grid and ask what the total settles down to. The limit, taken in a particular sense described below, is the Ito integral, and it carries the name of Ito, who built it.

Two things in that description are not free choices, and both of them are load bearing. The reading point is fixed at the start of the step rather than left open, and the limit is taken in a sense that is weaker than the one ordinary integration uses. Neither restriction is decoration. Each one exists because the construction falls apart without it, and the rest of this guide is largely an explanation of why.

Of the two, the reading point has the plainer meaning, so take it first. The multiplier for a step is the move the path makes during that step. If the integrand were read at the end of the step instead, that reading would already contain the move it is about to be multiplied by. The reading and the multiplier would be the same event seen twice. The overlap is not a small technical annoyance. Reading at the end changes the answer by a fixed amount that never goes away, and the size of that amount is computed exactly below.

The rule has a counterpart outside mathematics. A weighing scale that settles on a number and a scale that keeps flickering give the same reading if either is looked at before the object is placed on it. Once the object is on, the flickering scale can be read at a moment that happens to be high or a moment that happens to be low, and choosing the moment after the object lands is no longer measuring the object. Reading at the start of the step is the rule that forbids choosing after the fact.

Now the limit. In ordinary integration the sums approach the answer in the plain sense: with the grid fine enough, the sum is as close to the answer as was asked for. Here the sums are random quantities and the path is too rough for the plain argument to run, so that promise is not available. The available promise is mean square convergenceA sum approaches a limit in mean square when the average of the squared difference between them goes to zero. Mean square convergence is a weaker requirement than the sums approaching the limit on every single path., which means the average of the squared difference between the sum and the limit goes to zero. Mean square convergence is real, and it is enough to build everything on. It is still a different promise from the one ordinary integration makes, and the difference matters once the question becomes what kind of object the answer is.

The twelve increments are the whole of what makes this construction different. Two of their properties get used later, so name them now. The increments sum to exactly zero, and the path therefore returns to where it started at the end of the year. The twelve driving values were built to have a sum of squares of exactly 12.0, so the squares of the increments sum to exactly 1.000000, the elapsed year. The sum of those squares is the quadratic variationThe sum of the squared increments of a path over a partition. For Brownian motion it settles on the elapsed time as the grid refines, rather than on zero as it would for a smooth curve. of the path, and it is the quantity that will turn out to separate the two integrals.

Where exactly do the two constructions part?

Both definitions are now on the table, so the comparison can begin. Set them side by side and almost everything matches. Same interval, same partition, same integrand, same act of reading it once per step, same addition, same refinement of the grid. Six of the seven moves are identical. The seventh is that one sum multiplies by a width and the other multiplies by an increment, and that is the only place the two constructions touch different material.

The single point of divergence is that the Riemann limit is required to be the same for every reading point, and for the Ito sum that requirement cannot be met. It is not that the Ito construction chooses the start of the step for tidiness and could have chosen otherwise. The three choices give three different limits, so the definition has to name one, and the one it names is the start.

The algebra that shows it is short enough to check on the back of an envelope. Suppose the path between grid points is read along a straight line. The locked path is published at twelve nodes and nowhere in between, so a straight line is the interpolation rule in use. Reading the integrand a fraction of the way across each step adds, to each term, the move the path made over that fraction. In the Riemann sum that extra piece is multiplied by the width, and the extra pieces telescope: they add up to the fraction times the width times the total movement of the path across the whole year. In the Ito sum the extra piece is multiplied by the increment itself, and at the end of the step the extra piece is exactly the increment, so the products are squares.

Read the two right hand sides against each other. The contrast is total. On the left, a step width multiplies something. Halve the step and that term halves, whatever the path did. On the right, no step width appears at all. There is only a sum of squares, and every one of its terms is non-negative, so nothing in it can cancel anything else. Refining the grid makes each square smaller and makes there be more of them, and the two effects offset each other exactly.

Why does the reading point matter for one and not the other?

The answer is a comparison of sizes, and the vocabulary for it is the order of a termHow fast a quantity shrinks as the step shrinks. A term of order one half shrinks like the square root of the step; a term of order one shrinks in proportion to it; a term of order one and a half shrinks faster still.. The picture is worth fixing first. Over a step of width one twelfth of a year, a Brownian path typically moves by about the square root of that width, or 0.288675. Over a step of width one forty eighth it typically moves by about 0.144338. The width fell by a factor of four and the typical move fell only by a factor of two. The mismatch between those two rates drives everything that follows.

Now run it through both sums. In the Riemann sum, moving the reading a little way into the step changes the reading by something of the order of the square root of the step, and that change is then multiplied by the step. Square root times step is step to the power one and a half. There are as many such terms as there are steps, one over the step of them. Even in the worst case where none of them cancel, the total is of the order of the square root of the step. Take more steps and that total falls. At twelve steps the worst case is 0.288675 of the horizon; at forty eight it is 0.144338; at seven hundred and sixty eight it is 0.036084. The worst case is heading for nothing.

In the Ito sum the same change in the reading is again of the order of the square root of the step, and the increment it multiplies is of that order too. Square root times square root is the step itself. There are one over the step of them, and one over the step multiplied by the step is one. The total does not shrink at all, and it cannot. At the far end of the step every term is a squared increment, and a sum of squares has nothing in it that can cancel.

There is a coastline in this. A ragged shoreline measured with a one kilometre rule gives a length. The shorter rule follows wiggles the long one cut across, so the same shoreline measured with a hundred metre rule gives a longer answer. The measured length does not settle down as the rule is shortened, and no amount of refinement makes it settle. A sum of squared increments behaves the same way, and that is why the Ito spread will not go away. The Riemann sum, by contrast, is asking how much area sits under the shoreline rather than how long the shoreline is, and area settles down perfectly well however ragged the edge.

What do the two integrals actually read on the locked path?

Now put numbers on all of it. The integrand is the Brownian path itself, the interval is the year, and the partition is the twelve equal steps the locked path is published on. Two integrals get taken. The first is the path against time, an ordinary Riemann integral that asks how much area the path swept out. The second is the path against its own increments, an Ito integral that asks something the first one cannot express. Each is computed three times, reading at the start of every step, then at the middle of every step, then at the end of every step.

The Riemann integral returns minus 0.079386 all three times. Not close to it three times: identical, to as many decimal places as anyone cares to print. The reason is the identity two sections up. The end reading exceeds the start reading by one step width multiplied by the net movement of the path across the whole year, and the twelve driving values were built to sum to exactly zero, so the net movement is exactly zero and one twelfth of exactly zero is exactly zero. The middle reading is the average of the other two, so it agrees as well.

The Ito integral returns minus 0.500000 read at the start, 0.000000 read at the middle and plus 0.500000 read at the end. One construction gives one answer at every reading point, and the other gives three answers that are a full 1.000000 apart from end to end. That 1.000000 is not a coincidence and it is not the elapsed year by luck: the gap between the end reading and the start reading is exactly the sum of the squared increments, and on the locked path over twelve steps that sum is exactly the elapsed year of 1.000000. The middle reading sits exactly halfway between them, and the construction that reads at the middle carries the name of Stratonovich rather than of Ito.

One honest caveat before that picture is taken as a general law. The Riemann readings on the locked path agree not merely in the limit but at every partition that can be built from these twelve nodes, and that is a property of this path rather than of the Riemann integral. This particular path returns to exactly where it started. The net movement is therefore exactly zero, and the whole spread with it. On a path that ends somewhere other than where it began, the three Riemann readings differ at a coarse grid and then close up as the grid refines. The claim survives either way, but both versions are worth seeing.

So take the first half of the same locked path, from the start of the year to month six, where the path is up at 0.375278 rather than back at zero. Over that half year the three Riemann readings genuinely do differ, and the difference is the width of a step multiplied by 0.375278. Halve the step and the difference halves. Cut it to a sixth and the difference falls to a sixth. Meanwhile the Ito spread over the same half year does the opposite. It climbs from 0.140833 at one step to 0.564167 at six steps, heading toward the half year of 0.500000 and overshooting it on the way. A single path at a coarse partition is noisy.

What happens as the grid refines?

A single pair of numbers cannot carry a claim about a trend, so the control below recomputes both sets of readings at six partitions of the same locked path. Every figure it prints is computed from the twelve published driving values, never from a fresh random draw. The default therefore reproduces the worked example above to the last decimal place, and will do so on every reload. The upper panel shows which nodes of the path each partition actually looks at, and this is the part worth watching: at one step the sum sees only the two ends of the year, and the two ends of the year are the same number.

The one step setting is the sharpest thing in this subject, so take it first. Cut the year into a single step and the Ito spread is 0.000000. The only two nodes the coarsest possible grid looks at are the two ends of the year, and they are the same number. So the coarsest grid sees no variation at all in a path that travelled between Rs 93.74/- and Rs 111.08/- on the standard process built from it. Refine to two steps and the spread is 0.281667. Refine to three and it drops back to 0.031667. Refine to four and it jumps to 1.345000, overshooting the elapsed year by a third. Then 1.018333, then 1.000000.

The sequence is not a climb, and no honest drawing of it can be made to look like one. The mathematical claim is about what happens in the limit as the grid refines without bound, and a claim about a limit says nothing reliable about any particular coarse grid on any particular path. A reader who has been shown a tidy curve rising to one has been shown something the numbers do not do, and will then be surprised by real output. The reason the coarse readings jump about is easy to see: at three steps the partition happens to land on nodes where the path is close to its starting level, so the chords it draws are short and their squares are tiny.

What is each answer a property of?

Most readers skip the question of what each answer is a property of, and then get caught by it. A finished Riemann integral is a number, and that number belongs to the function that was integrated. The same function and the same interval, handed to somebody on the other side of the country, produce the same number. Nothing else in the calculation could differ.

A finished Ito integral is something else. The answer depended on the increments of one particular path, and a different path would have produced a different answer. The Ito construction yields a random variable rather than a number, a quantity whose value is settled only once the path is settled. The two constructions do not merely give different answers to the same question, they produce different kinds of object.

Approaching a random variable is why the Ito construction had to weaken its notion of a limit. When the thing being approached is a random variable, saying that the sums get close to it has to mean something, and mean square closeness is the meaning that works. Ordinary integration never faced this problem because it never had anything random to approach.

A lift makes the distinction concrete. The floor a lift is on is a number anybody in the building would agree with. How much total up and down travel the lift did today depends on the particular sequence of calls it received. The travel figure is a different kind of fact about a different kind of object, though both are measurements of the same lift on the same day.

Which rules of ordinary integration survive the move?

A reader arrives at this integral carrying a set of rules that have been correct in every context met so far, and the honest summary is that most of them need checking and some of them need repairing. Knowing which is which is a lookup rather than a derivation, and each row below carries its reason.

Two of those rows deserve a sentence more. The row about the size bound is the one people find strangest, so name what replaces it. Instead of an inequality bounding the answer, this construction offers an exact statement about the average of the squared answer, and that average equals the ordinary integral of the squared integrand against time. On the locked path that ordinary integral works out to 0.067986, and it is worth being clear that this is an equality about averages taken across many paths rather than a prediction for any one of them. On the single locked path the squared Ito answer is 0.250000. One path is not an average, so there is no contradiction.

The row about the missing rate of change is the deepest one. In ordinary calculus, integration and differentiation undo one another, and the rate at which the thing integrated against is changing can always be asked for. Here it cannot be asked for. Brownian motion has no rate of change at any instant. The absence is not an admission of difficulty. The missing rate is why the notation writes the increment rather than a rate multiplied by a width, and why the whole subject is built on integrals rather than on derivatives.

Tested against the worked instance, the rule closes, and no cheaper check exists. With both processes set equal to the Brownian path, the rule says the integral of the path against its own increments equals the final value squared, less zero, less the same integral again, less the covariation of the path with itself. The covariation of a path with itself is the quadratic variation, and here the quadratic variation is the elapsed year. Moving the repeated integral to the other side and halving everything gives half the final value squared less half the elapsed year. On the locked path the final value is zero and the elapsed year is one, so the answer is minus 0.500000. The twelve step computation printed exactly that.

How does somebody checking a calculation use this?

Rebuilding anybody's working is not necessary here. Three checks, all cheap, catch most of what goes wrong when these two constructions are confused, and every one of them can be run on a printed sheet of output with nothing else to hand.

1. Find where the integrand is read relative to the step Any routine that sums an integrand against the moves of a path has to read the integrand somewhere, and the reading point is usually one index in one line. If it reads the level at the end of the step rather than at the start, the answer is too high by the quadratic variation of the path over the interval.
   On the locked path over the year that shift is exactly 1.000000, which turns minus 0.500000 into plus 0.500000.
2. Ask whether a rule was carried across without its extra term Take the working and test it on the path itself. If the method returns half the final value squared for the integral of the path against its own increments, it is using the ordinary rule. If it returns that less half the elapsed time, it is using the right one.
   On the locked path those two are 0.000000 and minus 0.500000, which is as clean a test as could be asked for.
3. Ask what the reported number is a property of If a single number is reported for an integral taken against a path, ask which path. An integral against time needs no such answer and an integral against a path always does, and a report that cannot say which path has confused the two kinds of object.
   A quantity computed from one path is one draw, and the average across paths is a different statement, as the 0.250000 against 0.067986 comparison above shows.

All three checks are the same question asked three ways: has somebody treated a construction that depends on the reading point as though it did not? The question catches the error long after the algebra has gone, and it is worth carrying away even if every formula fades.

The standard process, its four parameters and the twelve driving values of the locked path are invented.
Educational material. Not advice on any investment, tax, budget or market position.