Strong and Weak Solutions Compared: What Each Delivers
Solving one of these equations does not yield a number or a formula. Solving yields a process. A strong solution is built on randomness handed to it, so it can say which path went with which driving values. A weak solution builds its own randomness alongside, so it fixes the distribution and nothing finer. Almost every question here needs only the second.
The familiar idea of what it means to solve an equation is about to work against anybody who carries it in. The familiar idea has been reliable since school, it has never once failed in an ordinary setting, and it does not survive the crossing into this one. So before either kind of solution can be named, the word itself has to be taken apart and rebuilt.
Start with something concrete. Two people have been asked to describe the same set of five hundred parcels. The first offers a numbered list: parcel one weighed 1.4 kilograms, parcel two weighed 2.9 kilograms, and so on to parcel five hundred. The second offers a chart of how many parcels fell in each weight band, and nothing else. Asked for the average weight, both answer with the same number. Asked how many parcels weighed more than three kilograms, they agree again, exactly, not approximately. Asked what parcel one hundred and seven weighed, only the first has anything to say at all. The second has not given a worse answer to that question. The second never recorded which parcel was which, so the second cannot express the question at all.
The two parcel descriptions are the whole distinction in one image. The numbered list is the stronger notion of a solution and the chart is the weaker one, and the interesting fact, the one worth carrying away, is how many of the questions actually worth asking are answered identically by both. Three questions put to both notions, once each has been defined completely, show precisely where the two part.
What does it mean to solve one of these equations at all?
The existing idea of a solution comes in two shapes. Both are worth naming, and both are about to be set aside. The first shape is a number. Solving three x plus six equals zero gives minus two, a single value that can be substituted back to check. The second shape is a formula. Solving the ordinary growth equation that says a quantity grows at 8 per cent of its own level every year, starting from Rs 100/-, gives an expression that returns the level at any time named: at half a year it reads Rs 104.081077/-, at one year Rs 108.328707/-, and it will do that all day. Both shapes share a property. The answer can be written down on one line and evaluated.
Now put a random term into the equation and ask the same question. Which object could possibly satisfy it? Not a number. The equation describes something that moves through a year rather than sitting at a value. Not a curve either. The equation is compatible with an enormous set of paths and the randomness decides which of them happens, so there is no single curve to hand back. A solution has to be a whole random object, one that assigns a value to every instant of the year for every way the randomness could have come out.
| \(S_t\) | the standard process at time \(t\), an invented traded quantity used throughout |
| \(S_0\) | its starting value, Rs 100/- exactly |
| \(\mu\) | the drift, 0.08 a year under the physical measure \(\mathbb{P}\) |
| \(\sigma\) | the volatility, 0.20 a year |
| \(W_u\) | standard Brownian motion at time \(u\) under the physical measure \(\mathbb{P}\) |
| \(\mathbb{P}\) | the physical measure, the rule that weights the outcomes here |
| \(T\) | the horizon, one year |
The shift that statement asks for is larger than it looks. Read it twice. The thing on the left is not a number at each time; it is a random value at each time. The equality has to hold at every instant, for essentially every way the randomness turns out. A solution here is a process, and the two notions compared here are two different answers to a question the integral form leaves completely open: where did the randomness on the right hand side come from?
Two questions sit underneath everything that follows. The first is existenceWhether any process satisfies the equation at all.. Existence asks whether any process satisfies a given equation. The second is uniquenessWhether more than one process can satisfy the same equation.. Uniqueness asks whether more than one can. The general conditions that settle either belong in a mathematics text and are covered separately. Both questions have two versions, a strong one and a weak one, and the two versions do not always give the same verdict.
Is the solution of one of these equations a formula that could be evaluated?
What is a strong solution?
Begin with the demanding version. A complete probability setup arrives already built, with a Brownian path W running on it. The path is not open to choice. The path arrives finished, every wiggle already decided, and the task is to build a process S out of that particular randomness so that the equation holds. Such a process is a strong solutionA process built on a Brownian path that was given, so the correspondence is kept..
For the standard process, the object to be handed back can be written down in one line. The standard process is therefore the easiest possible case to look at, and most equations are not so obliging: having a closed expression is luck rather than part of the definition.
| \(\mu - \tfrac{1}{2}\sigma^{2}\) | 0.06 exactly, being 0.08 less half of the variance rate 0.04 |
| \(\tfrac{1}{2}\sigma^{2}\) | 0.02 exactly, the half variance correction |
| \(W_t\) | the handed-over Brownian path at time \(t\), which was not chosen |
| \(\exp\) | the exponential function |
| \(t\) | elapsed time in years, running from 0 to the horizon \(T=1\) |
The second half of the definition is the part readers skate over, and it is the part that does all the work. Satisfying the equation is not enough. The process being built has to be adapted to the given pathUsing only the handed-over randomness, which is what strong requires.. At every time, its value must be computable from the handed-over randomness up to that time and from nothing else. No extra coin flips from any other source. No peeking ahead at where the path goes later.
| \(\mathcal{F}^{W}_{t}\) | the information generated by the handed-over path up to time \(t\), and in prose the information available at that time |
| measurable | computable from that information, with no further randomness needed |
| \([0,T]\) | every time from the start of the year to the horizon |
Think of a lift that only knows which floor it is currently on. The lift cannot act on a button that has not been pressed yet, and it cannot invent a floor of its own. A strong solution is under exactly that discipline. A strong solution is a machine that takes the handed-over path as input and returns the process as output, and it is allowed no other input at all.
Now feed it something concrete. The locked path publishes twelve driving values for the year, 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 0.288675 to give that month's Brownian increment. Hand those twelve values to the formula above and it returns readings, month by month, with nothing left to choose.
| Month | Driving value | Brownian level | Strong solution reads |
|---|---|---|---|
| 1 | -0.5 | -0.144338 | Rs 97.641506/- |
| 2 | 1.6 | 0.317543 | Rs 107.627772/- |
| 3 | -1.3 | -0.057735 | Rs 100.345896/- |
| 4 | -0.1 | -0.086603 | Rs 100.268308/- |
| 5 | 0.1 | -0.057735 | Rs 101.354389/- |
| 6 | 1.5 | 0.375278 | Rs 111.077230/- |
| 7 | -1.3 | 0.000000 | Rs 103.561971/- |
| 8 | -0.5 | -0.144338 | Rs 101.119468/- |
| 9 | -1.4 | -0.548483 | Rs 93.735186/- |
| 10 | 0.4 | -0.433013 | Rs 96.405923/- |
| 11 | 0.9 | -0.173205 | Rs 102.056764/- |
| 12 | 0.6 | 0.000000 | Rs 106.183655/- |
The table above is what a strong solution delivers and no other notion of solution can produce it. Given that randomness, this path, reading for reading. The twelve driving values were constructed to sum to zero exactly, so the Brownian level returns to zero at the horizon and the process lands at Rs 100/- multiplied by the exponential of 0.06, which is Rs 106.183655/-. Both of those facts are construction rather than coincidence, and they are what allow a path to be published that a reader can check.
One more property belongs to this notion and it is stronger than it sounds. If two people both build a strong solution on the same handed-over path, their two processes agree at every time, on essentially every outcome, not merely on average and not merely in shape. Agreement of that kind is called pathwiseHolding path by path rather than only in distribution. uniqueness, and it would let one person check the other's twelve readings line by line and find no difference anywhere.
Which requirement does a strong solution impose that a weak one does not?
What is a weak solution?
Weakening the demand relaxes exactly one part. Nothing is handed over. Only the equation is given, with the task of producing a pair: a process and a Brownian path, on a probability setup freely chosen, such that the equation holds between them. The randomness and the process are built together, as one construction. Such a pair is a weak solutionA process and a Brownian path constructed together, fixing only the distribution..
The freedom to choose the setup sounds like a technicality. The choice is the whole difference. Because the Brownian path was built by the constructor, it has no relationship to anybody else's Brownian path, and in particular it has no relationship to the twelve locked driving values published above. One such construction is a valid weak solution and so is a completely different one built by somebody else, and the two need not agree on any single outcome.
| \(\widehat{S}_t\) | the process the weak solution constructs, written with a hat to keep it distinct from a process built on a handed-over path |
| \(\widehat{W}_t\) | the Brownian path the weak solution constructs for itself, alongside the process, and not the risk-neutral \(\tilde{W}\) |
| \(\ln\) | the natural logarithm |
| \(\mathcal{N}\) | the normal distribution, given by its mean and its variance |
| \(4.665170\) | the mean of the logarithm at the horizon, being the logarithm of 100 plus 0.06 |
| \(0.040000\) | the variance of the logarithm at the horizon, being 0.20 multiplied by itself |
The distribution statement still contains a great deal. From it the mean at the horizon can be read, Rs 108.328707/-, being Rs 100/- multiplied by the exponential of 0.08. The median can be read, Rs 106.183655/-, being Rs 100/- multiplied by the exponential of 0.06. The gap between them can be read, Rs 2.145052/-, the half variance correction made visible in rupees. Any probability, any quantile, any variance, any average of any function of the value can be read. A weak solution is not a partial answer to the same question; it is a complete answer to a different and slightly smaller question, namely what the whole set of outcomes looks like and how they are weighted.
A weak solution has thrown away the correspondenceThe link between a particular set of driving values and a particular path., the link between a particular set of driving values and a particular path. Consider a queue that forgets who joined it. The queue can still report, accurately and completely, how many people waited under five minutes and what the average wait was. The queue cannot report what happened to the person who arrived at ten past three. The record that would connect them was never kept.
Uniqueness has a weak version too, and it matches. Two weak solutions to the same equation are considered the same when they agree in distributionAgreeing about the whole set of outcomes and their weights, and nothing finer.. Agreeing in distribution means assigning the same weights to the same sets of paths. Two weak solutions are allowed to disagree about every single outcome, provided the bookkeeping over all outcomes matches. Agreement in distribution is a real notion of sameness and it is exactly the right one when the questions being asked are questions about weights.
Which notion keeps the link between a particular set of driving values and a particular path?
What can each one answer, side by side?
Definitions settle nothing on their own. The way to see what separates two notions is to put the same questions to both and watch where the answers stop matching. Three questions do it, and the order matters: the first two are there to establish how much agreement there is before the third finds the seam.
Question one. On average, what is the value of the standard process at the horizon? The strong solution has the whole process, so it can take the average over all outcomes and it gets Rs 108.328707/-. An average is computed from the distribution at the horizon, and the weak solution has exactly that, so it too gets Rs 108.328707/-. Identical, to every decimal place shown and to every decimal place beyond.
Question two. How large a fraction of paths finishes above Rs 100/-? This is a probability, which is a statement about weights over sets of paths. Both notions have the weights.
| \(K\) | the level being asked about, Rs 100/- here, which is also the starting value |
| \(N\) | the standard normal distribution function |
| \(\sigma\sqrt{T}\) | 0.20 exactly, since the horizon is one year |
| \(0.30\) | 0.06 divided by 0.20, the number of standard deviations by which the median logarithm clears the level |
| \(\mathbb{P}\) | the physical measure, under which the drift is 0.08 |
Both answer 0.617911. Not close, not to within a rounding, but the same number arrived at by the same arithmetic. Two thirds of the way through the comparison, the two notions have not disagreed about anything at all, and that is the finding rather than the setup for one.
Question three. Given the twelve locked driving values, what is the value at the horizon? Here the two part, and they part in a way that is easy to describe wrongly. The strong solution answers immediately. The strong solution was handed those twelve values, is a function of them, and returns Rs 106.183655/- exactly.
| \(z_k\) | the driving value for month \(k\), from the twelve published numbers |
| \(0.288675\) | the square root of one twelfth, the length of one step |
| \(\sum z_k\) | 0.000000 exactly, a construction choice of the locked path |
| \(\bigm|\) | read as given, meaning the driving values are known rather than averaged over |
The weak solution does not return a different number. The weak solution returns nothing. The weak solution never received those twelve values and built its own randomness instead, so the symbols on the left of that expression have no meaning inside its construction. Asking it this question is like asking the weight chart what parcel one hundred and seven weighed. The gap between the two notions is not a gap in accuracy, it is a gap in vocabulary, and that is why it never shows up as a disagreement about a number.
Before the calculator below is run: asked what fraction of paths finish above Rs 100/-, do the two notions agree?
Put each of the three questions to both notions and read the two columns
Selecting a question recomputes both columns from the formulas above, never by sampling, so the readings are the same on every visit. The default is question one, where the two notions agree.
| Question | Strong solution | Weak solution |
|---|---|---|
| The mean at the horizon | Rs 108.328707/- | Rs 108.328707/- |
| The fraction finishing above Rs 100/- | 0.617911 | 0.617911 |
| The value given the twelve locked driving values | Rs 106.183655/- | no answer at all |
Which of the three questions separates the two notions?
Why is the weaker notion enough for almost every question here?
Now count what is actually asked of a model like this one. An average. A probability that the value clears a level. A variance, a standard deviation, a quantile, the average of some function of the value at the horizon. How the answer moves when a parameter moves. Every single item on that list is computed from the distribution and from nothing else.
The pattern in that list is not a coincidence. The list is a statement about what this subject area is for. The mathematics beneath a price is used to attach numbers to sets of outcomes, and a set of outcomes with a weight attached is precisely what a distribution is. A weak solution delivers exactly the object that almost every question in this subject area is a question about. The weaker notion is not a compromise but the natural fit.
There is a one-line test that settles which notion a question needs, and it needs no mathematics at all. The test is to read the question aloud and ask whether it names a particular set of driving values, a particular realisation, a specific path that was handed over. If it does not, the weaker notion has everything the question requires. If it does, the stronger one is needed. The test works because the only thing the stronger notion adds is the correspondence, so the only questions it can answer that the weaker cannot are questions that mention one.
There is a second reason, and it is the one that turns a preference into a rule. The weaker notion is easier to satisfy. Conditions that are too rough to give a strong solution can still give a weak one, so there are equations for which the weaker notion delivers a complete description of the distribution while the stronger notion delivers nothing at all. The general conditions belong to a mathematics text and are covered separately, but the direction of the asymmetry is worth carrying: weak existence is the more forgiving requirement, and weak uniqueness is the more forgiving notion of sameness.
The question mentions no particular set of driving values. Which notion is needed?
Before reading on: what kind of question would need the stronger notion?
Which question needs the stronger notion, and why?
Three situations need it. Each one has a shape that can be recognised in somebody else's working, so naming the three concretely is more useful than a general principle.
The first is the one the simulation demonstrated. A question that conditions on a realisation. Not what happens on average, but what happened this time, given these driving values. On the locked path that question has the exact answer Rs 106.183655/-, and it has that answer because the twelve values sum to zero, so the whole Brownian contribution to the exponent vanishes and only the steady 0.06 for the year survives. A weak solution cannot be asked this. The phrase "these driving values" refers to nothing inside a weak solution.
The second is comparison on shared randomness, and it is the one most likely to catch a working modeller out. Suppose the question is what a change in the drift is worth, holding everything else fixed. Running the same twelve driving values through the same equation with a drift of 0.05 instead of 0.08 ends the process at Rs 103.045453/-, against Rs 106.183655/- on the original drift. The randomness was identical in both runs, so the drift alone accounts for that Rs 3.138201/- gap. The same exercise with two weak solutions fails. Each builds its own randomness, so the two endings differ for two reasons at once and the two reasons cannot be separated. On this path the spread between the high of Rs 111.077230/- and the low of Rs 93.735186/- is Rs 17.342044/-, more than five times the effect being measured. Reusing the same handed-over randomness across two runs is what stops the noise swamping the difference actually being asked about, and only the stronger notion permits that reuse.
The third is any statement that has to hold path by path rather than on average. The month-end readings of the process built on this handed path never fell below Rs 93.735186/-, the highest of them was Rs 111.077230/- in month six, and the year ended Rs 6.183655/- above where it began. Each of those claims is about the given path rather than about the weights over all paths, and the weaker notion has no way to express any of them.
How does someone checking a model actually use this?
Nothing needs to be built to get value out of this distinction. Three checks, all of them readable off a specification or a written-up calculation, catch the two mistakes that matter: reaching for the stronger notion when the weaker one would have done, and quietly relying on the stronger notion without having established it.
- Read what the question conditions on
Find the sentence that states what is being asked. If it conditions on a particular realisation of the randomness, or on a specific handed-over path, the stronger notion is in use whether or not anybody said so. If it conditions on nothing, the weaker one is sufficient and any extra work spent establishing the stronger one bought nothing.
Of the three worked questions above, only one conditions on anything, and it is the only one the weaker notion cannot answer.
- Check whether two outputs are being compared on the same randomness
Wherever two variants of a model are set against each other, ask whether both were run on identical driving values. If they were, the correspondence is doing real work and the stronger notion is load bearing. If they were not, the difference reported is a mixture of the model change and the noise.
A drift change from 0.08 to 0.05 moves the horizon reading by Rs 3.138201/- on the shared path, against a path range of Rs 17.342044/- if the randomness is redrawn.
- Ask what would break if the randomness were redrawn
Take the claim being made and imagine a different but equally valid set of driving values. If the claim survives unchanged, it was a claim about the distribution and the weaker notion supports it. If the claim becomes meaningless, it depended on the correspondence, and the working needs the stronger notion for it to stand.
The reading 0.617911 survives any redraw. The reading Rs 106.183655/- does not survive it, because it belongs to one particular set of twelve values.
All three checks are the same check asked three ways. Does this claim mention a particular realisation of the randomness, or does it not? That single question is worth carrying long after the notation has faded, and it also survives being asked about equations never seen before, which is more than a memorised definition can manage.
The error that gets made, and what it costs
Treating a weak solution as a second best answer, and concluding that the stronger notion should always be sought where it can be had. The words invite it. Strong sounds like more and weak sounds like less, and a reader who has just met both will reasonably assume that anything the weaker one says, the stronger one says better.
It does not. On every question about the distribution the two agree exactly, so the extra work of establishing the stronger notion has bought a correspondence the question never asked for. The extra work is a waste rather than an error, and a modest one.
The real cost arrives when the equation at hand is one of those for which the weaker notion exists and the stronger one does not. There are such equations, and the standard textbook illustration carries the name of Tanaka. Its diffusion coefficient is essentially the sign of the process itself, and no process built on a handed-over path satisfies it. A construction that builds the process and the path together does satisfy it. Insisting on the stronger notion there does not give a worse answer, it gives no model at all, and the distributional question originally posed goes unanswered.
The check that prevents it is the one-line test above. The question decides the notion, and not the other way round. Whatever the question conditions on settles which notion is wanted.
Is a weak solution a lesser answer?
References
| Source | Document | Where |
|---|---|---|
| arXiv Quantitative Finance | Preprint repository for work on stochastic differential equations and their solution notions | arxiv.org |
| Social Science Research Network | Working paper repository for the same material | ssrn.com |
| Hull, Shreve and Wilmott | Standard textbook treatments of stochastic differential equations and both notions of solution | Textbooks under the authors' names |
| Kiyosi Ito | The stochastic integral and the change of variable rule that the integral form above rests on | Textbooks under the author's name |
| Hiroshi Tanaka | The equation with a weak solution and no strong one, named above | Textbooks under the author's name |
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.
