Sample Space and Events: Every Outcome, and Which Sets Count
A sample space is the set of every outcome a model admits, and an event is a subset of that set. Choosing the sample space decides which questions can be posed at all. A set holding only the finishing value cannot say whether a level was reached along the way, and no mathematics added later recovers what the set left out.
Everything downstream sits on top of that set. Probabilities are attached to its subsets, expectations are sums or integrals across it, and every later object in this subject area is defined relative to it. So a question the set cannot express is a question the model cannot answer at any price. The choice of sample space is the one modelling decision that cannot be repaired later, so it is made first and made explicitly rather than by accident.
What is a sample space, and how is one chosen?
A sample spaceThe set of every outcome the model admits, with nothing left out. is written with a capital omega, and it is a set in the ordinary sense: a collection of things, with no repeats and no order. Each thing inside it is an outcomeOne single element of the sample space, a complete description of one way things went., and an outcome is a complete description of one way the world went. Not a partial description. Not a summary. Complete, in the sense that once the element that occurred is named, there is nothing further the model could say.
Complete is doing all the work, and complete is where the decision hides. Complete relative to what? A model of a coin toss can take its outcomes to be heads and tails, or it can take them to be the full physical trajectory of the coin through the air. Both are legitimate sample spaces. The two sample spaces differ in how much detail one element carries, and therefore in what can be asked of them afterwards.
Here is the everyday version. Think of a lift in a building that only reports which floor it is standing on. Its outcome set is the list of floors, and that is genuinely complete for the questions the display was built for. Now ask how long the lift spent between the third floor and the seventh. The quantity was never in the set of things the display records, so the display has no answer and no amount of staring at it produces one. The lift certainly did spend a length of time in transit. The recording just does not contain it.
So choosing a sample space runs in three steps. List what could happen. Decide how much detail one element carries. Then check that list against every question the work will eventually have to answer. The first two steps feel like the whole job, and it is the third one that gets skipped.
Which step of choosing a sample space is the one usually skipped?
What is an event, and why is it a set rather than a sentence?
An eventA subset of the sample space, so a set rather than a sentence. is a subset of the sample space. Nothing more exotic than that. In ordinary speech an event sounds like a happening, something with a verb in it. In this subject it is a region: the collection of outcomes on which some description is true.
| \(\Omega\) | the sample space, the set of every outcome the model admits |
| \(\omega\) | one outcome, a single element of that set |
| \(A\) | an event, meaning any subset of the sample space |
Translating a sentence into a set is the move that makes the mathematics work. Take the standard process, the single invented traded quantity written S with a time subscript that this subject area uses throughout, starting at Rs 100/- and observed over one year. The sentence the process finished above Rs 110/- is not itself a mathematical object. The set of outcomes on which that sentence is true is one, and it can be intersected, complemented, measured and integrated over.
Once a description becomes a set, every operation on sets becomes an operation on descriptions, and that is the whole reason the formal treatment is worth the trouble. English sentences do not compose reliably and regions do, so reasoning about sentences gives way to reasoning about regions.
One more distinction matters before moving on. Being a subset is not the same as being allowed to carry a probability. Which outcomes exist is settled by the sample space. Which subsets may be assigned a number is settled by a second object entirely, and that object is the sigma-algebra, set out under sigma-algebras. Every subset mentioned here is one that can safely be measured, and the general question of which ones qualify belongs to sigma-algebras.
Can one process carry more than one sample space?
One process can carry several, and the abstraction earns its keep exactly here. Take the standard process, unchanged: the same starting value of Rs 100/-, the same drift of 8 per cent a year, the same volatility of 20 per cent a year, the same one year horizon. Now describe it three different ways.
The first description admits four outcomes. Split the year into two halves and let the process move up or down in each, giving four sequences of moves. The four sequences are the whole sample space. The model admits no fifth thing that could happen.
The second description admits every continuous path the process could trace over the year. Here one outcome is an entire path, from the first instant to the last, and the set of them is uncountableToo large to list even endlessly, as a set of continuous paths is.. Listing them forever would not exhaust them.
The third description admits only the finishing value. One outcome is a single number in rupees. The finishing value set is by far the smallest, it is easy to work with, and it is completely adequate for a great many questions.
Now put three questions to all three. How likely is a finish above Rs 110/-? How likely is a finish of exactly Rs 100/-? And was Rs 110/- ever reached at any point during the year? The first two are answered by all three descriptions, though not with the same numbers. The third is answered by the first two descriptions and is not answered by the third, and the reason is not that the sum is hard. There is nothing there to sum.
A model records only where the process finished. Which of these can it answer?
What do the four sequences and the continuous set actually give?
Take the four sequence description first. All of it is visible at once. The process starts at Rs 100/-, moves once over the first half year and once over the second. An up move multiplies by 1.151910 and a down move multiplies by its reciprocal, 0.868123, so an up followed by a down returns the process exactly to where it started. The weight attached to an up move under the risk-neutral measure Q is 0.553908.
Read the event that the process finishes at exactly Rs 100/- off that picture. The event is the subset holding two of the four sequences, up then down and down then up. Its weight is 0.494188, being twice 0.553908 multiplied by 0.446092. A weight of 0.494188 is a perfectly ordinary positive number, and on the four sequence sample space finishing at exactly Rs 100/- is a thoroughly likely thing.
| \(\Omega\) | the four sequences of two half year moves, the whole sample space here |
| \(A\) | the event that the process finishes at exactly Rs 100/- |
| \(q\) | the weight on an up move under the risk-neutral measure Q, 0.553908 |
| \(\mathbb{Q}\) | the risk-neutral measure, the second of the two measures used in this subject area |
Now switch to the continuous description, where the finishing value can be any positive number. Under the physical measure P the logarithm of the finishing value is normally distributed with mean 4.665170 and standard deviation 0.20, so the median finish is Rs 106.18/- and the mean finish is Rs 108.33/-. On this set the event that the process finishes at exactly Rs 100/- is a single point in an uncountable set, and it carries probability zero.
On the four sequence sample space, finishing at exactly Rs 100/- carries 0.494188. On the continuous one it carries zero. Which is right?
How do the set operations turn into and, or and not?
Because events are regions rather than sentences, the three connective words of ordinary language become three operations on regions, and the translation is exact rather than approximate. The unionThe set of outcomes lying in one set or the other or both, carrying the word or. of two events holds every outcome lying in one or the other or both, and it carries the word or. The intersectionThe set of outcomes lying in both sets, carrying the word and. holds every outcome lying in both, and it carries the word and. The complementThe set of outcomes lying outside the given set, carrying the word not. of an event holds everything in the sample space outside it, and it carries the word not.
| \(A\cup B\) | the union, every outcome in A or in B or in both |
| \(A\cap B\) | the intersection, every outcome lying in both A and B |
| \(A^{c}\) | the complement, every outcome of the sample space lying outside A |
Two events with no outcome in common are called disjointTwo sets with no outcome in common, so they cannot both happen., and their intersection is the empty set. No single sequence does both, so on the four sequence description finishing at Rs 132.69/- and finishing at Rs 75.36/- are disjoint. Holding those three pairings together is what makes every probability rule that follows readable rather than memorised. The rules are statements about regions, and the words are just the labels on the regions.
Which set operation carries the word and?
What is the difference between an outcome and a one outcome event?
This looks like pedantry and it is not. An outcome is an element of the sample space. The event containing only that outcome is a subset of the sample space. Written out, one is omega and the other is the set holding omega, and they sit at different levels: one is a thing, the other is a bag containing exactly that thing.
The everyday version is the difference between a weight and what a scale reads. The object has a weight whatever any instrument says. A scale reports a reading, and a reading is a statement about a range. Confusing the two is harmless until the range shrinks to nothing, and then the two come apart completely.
On a continuous sample space every one element subset carries probability zero, and yet exactly one of those outcomes occurs on every single run. An event of probability zero happening is not a paradox and not a rounding artefact. Probability zero and impossible are two different statements, and only the second one rules something out. Every path the standard process could trace over the year had probability zero before the year began, and one of them happened anyway.
On the continuous sample space, one particular path carries probability zero. Can it still happen?
What happens to a single value when the set cannot be counted?
Here is the thing readers accept in words and quietly disbelieve, so it is worth computing rather than asserting. On the continuous description the probability of finishing inside a range around Rs 100/- is a genuine positive number, and it shrinks in step with the range.
| \(S_T\) | the standard process at the horizon, in rupees |
| \(S_0\) | its starting value, Rs 100/- |
| \(K\) | the value the range is centred on, here Rs 100/-, also the strike of the at-the-money contract |
| \(w\) | the half width of the range, in rupees |
| \(\mu\) | the drift under the physical measure P, 0.08 |
| \(\sigma\) | the volatility, 0.20, so the variance rate is 0.04 and half of it is 0.02 |
| \(T\) | the horizon, one year |
| \(N\) | the standard normal distribution function |
Put the locked parameters in. The drift less half the variance rate is 0.06 exactly, and the volatility multiplied by the square root of the horizon is 0.20 exactly. The readings come out clean for that reason. Within Rs 10/- of Rs 100/- the probability is 0.365895. Within Rs 5/- it is 0.188705. Within Rs 1/- it is 0.038123. Within 10 paise it is 0.003814. Within one paisa it is 0.000381.
Look at the last three. Each time the range narrows by a factor of ten, the probability falls by a factor of ten. The tenfold fall is not a coincidence and not an approximation being forced. Over a range that narrow the density hardly changes across the range, so the area under it is very nearly the density multiplied by the width.
| \(\phi\) | the standard normal density |
| \(d\) | the standardised logarithm of the ratio of K to the starting value, minus 0.30 here |
| \(w\) | the half width of the range, in rupees |
| \(K\) | the centre of the range, Rs 100/- |
At the locked parameters that density factor comes to 0.019069 per rupee of width. Multiplied by two rupees of width it gives 0.038139, against the exact 0.038123. Multiplied by two paise of width it gives 0.000381, against the exact 0.000381. The probability of a single value is not asserted to be zero by decree; it is the end of a fall that can be watched happening, one factor of ten at a time.
The range around Rs 100/- is about to narrow from Rs 1/- to 10 paise. Before the control moves: does the probability fall by a factor of about ten, or by much less?
Narrow the range and watch the probability go
One control: the half width of a range around Rs 100/-, running from Rs 20/- down to one paisa on a logarithmic track. The shaded band on the distribution narrows, the linear bar collapses to nothing, and the marker on the logarithmic track keeps sliding left long after the bar has disappeared. Every reading is computed from the formula above, never sampled, so the calculator reproduces the same readings on every reload.
A range of Rs 1.00/- either side of Rs 100/- carries a probability of 0.038123, which works out at 0.019061 per rupee of width.
| Half width around Rs 100/- | Probability of finishing inside it | Per rupee of width |
|---|---|---|
| Rs 10/- | 0.365895 | 0.018295 |
| Rs 5/- | 0.188705 | 0.018871 |
| Rs 1/- | 0.038123 | 0.019061 |
| 10 paise | 0.003814 | 0.019070 |
| One paisa | 0.000381 | 0.019074 |
Which subsets of a sample space are allowed to carry a probability at all?
How does anyone use this before writing a line of mathematics?
Someone reviewing a model does not begin with the equations. The review begins with two lists side by side: the outcomes the model admits, and the questions the model will be asked over its life. Then the reviewer checks the second list against the first, item by item, and marks every question the outcome set cannot express. Every item the review catches would otherwise be caught much later by somebody who needed an answer and could not get one. The review takes an afternoon, and it is the cheapest afternoon in the whole exercise.
The list of questions is where the discipline lives. A researcher writing down a set of continuous paths has bought the ability to ask about maxima, minima, first crossings, time spent above a level and the order in which things happened. A researcher writing down a set of finishing values has bought a much smaller and much faster object, and has sold all of those abilities in exchange. Neither choice is wrong; the mistake is making the trade without knowing a trade was being made.
There is a second habit worth copying. When the outcome set is written down, one question it cannot answer should be written beside it, together with the reason that is acceptable. A sentence saying this set carries no path information, and that is acceptable because nothing in the work depends on a level being reached, is a sentence that either survives review or fails it loudly. A sentence that never gets written fails quietly, months later.
The locked path finished at Rs 106.18/-. Before reading on: what was the highest value it reached during the year?
What goes wrong when the set cannot express the question?
One path is kept fixed across this subject area so that the same illustration can be drawn each time. The locked path takes the standard process through twelve monthly steps over the year, and it is constructed rather than sampled, so it reproduces identically every time. The locked path starts at Rs 100/-, drops to Rs 97.64/- after a month, climbs to Rs 111.08/- at month six, falls as low as Rs 93.74/- at month nine, and finishes at Rs 106.18/-.
The error that gets made, and what it costs
Choosing an outcome set that cannot express a question the work will eventually need. The set of finishing values is smaller, faster and entirely adequate right up until somebody asks whether a level was reached along the way. At that point the set is not inadequate. Inadequate would at least be visible. The set is silent. The path was discarded at the very first step, so there is nothing left to compute from. The set does not return a wrong answer that a check could catch; it has no answer at all.
On the locked path the finish of Rs 106.18/- and the high of Rs 111.08/- differ by Rs 4.90/-, and a recording built on finishing values carries the first and has never heard of the second. The cost is a model rebuilt from the outcome set upward, and every object standing on that set rebuilt with it.
The uncomfortable part is the timing. The decision that caused it was taken before any mathematics began, by someone who was being sensible: the smaller set was simpler, it answered every question anyone had asked so far, and nothing in the arithmetic that followed was ever wrong. The third step of choosing a sample space, the one that checks the set against the questions, is worth the afternoon it takes for exactly that reason.
References
| Source | Document | Where |
|---|---|---|
| arXiv, Quantitative Finance | Preprints on probability spaces and continuous time models in finance | arxiv.org |
| Social Science Research Network | Working papers on model specification and outcome sets | ssrn.com |
| Shreve, Hull and Wilmott | Standard texts setting out probability spaces, events and continuous time models in finance | Published editions |
The standard process, the locked path, the four sequence description and the at-the-money contract at Rs 100/- are invented.
Educational material. Not advice on any investment, tax, budget or market position.
