Chapter 01

Easy

Sample Spaces & Events

00 · Symbol Glossary

$\Omega$Omega — sample space

The set of every outcome the experiment could produce. Each run of the experiment yields exactly one element of $\Omega$ .

$\omega$omega — outcome

A single point of $\Omega$ : the complete description of what actually happened on one run, such as the pair of directions $UD$ for two trading days.

$A,\ B,\ C$Events

Subsets of $\Omega$ . An event $A$ occurs when the realized outcome $\omega$ lies in $A$ .

$A^c$A complement — not A

The outcomes in $\Omega$ that are not in $A$ : $A^c=\Omega\setminus A$ . The event " $A$ does not happen."

$A\cup B,\ A\cap B$Union and intersection — or / and

$A\cup B$ : at least one of $A$ , $B$ occurs. $A\cap B$ : both occur.

$A\setminus B$Set difference — A but not B

$A\setminus B = A\cap B^c$ : outcomes where $A$ happens and $B$ does not.

$\emptyset$Empty set — impossible event

The event containing no outcomes. It never occurs, whatever $\omega$ is realized.

$\Omega_1\times\Omega_2$Product space

All ordered pairs $(\omega_1,\omega_2)$ with $\omega_1\in\Omega_1$ and $\omega_2\in\Omega_2$ . The sample space of a two-stage experiment, such as two consecutive trading days. $\Omega^n$ is the $n$ -fold product.

$|A|$Cardinality of A

The number of outcomes in a finite set $A$ . For example, $|\{U,D\}^3|=8$ .

$\{B_i\}$A partition of Omega

Events that are pairwise disjoint and whose union is all of $\Omega$ . Every outcome lands in exactly one cell. Rating buckets and "number of up days" are partitions.

$\mathbf{1}_A$Indicator of the event A

The function on $\Omega$ that equals $1$ when the outcome lies in $A$ and $0$ otherwise. A cash-or-nothing digital pays a notional times $\mathbf{1}_A$ . Chapter 07 shows that the probability-weighted average of this payoff equals $\mathbb{P}(A)$ .

01 · Why Probability Starts With Sets

A risk statement such as "the probability the book loses money tomorrow is 45%" makes three separate choices. It fixes what can happen tomorrow, it fixes which of those possibilities count as losing money, and only then does it attach a number. The number is meaningless until the first two choices are precise.

This chapter makes the first two choices precise. The universe of possibilities is the sample space; the statements you care about ("exactly two defaults," "the option finishes in the money," "up three days running") are subsets of it. Chapter 02 attaches the numbers. Later chapters turn events into random variables (Chapters 05 and 06), averages (Chapter 07), joint behavior (Chapter 10), and tail risk (Chapter 15). Every one of those constructions still refers back to an event inside a sample space.

02 · Sample Spaces

Definition — Sample Space

A sample space $\Omega$ is a set whose elements, called outcomes, are mutually exclusive and exhaustive descriptions of the result of an experiment: every run of the experiment produces one and only one $\omega\in\Omega$ .

Plain language: write down a list of possible results so that, after the fact, exactly one entry on the list is true. Two conditions are doing the work. Exclusive means no two outcomes can both be true; exhaustive means nothing that could happen is missing.

Example — Finance-shaped sample spaces

One day's direction for an index: $\Omega=\{U,D\}$ (up or down).
The same day with flat closes allowed: $\Omega=\{U,F,D\}$ .
A BBB issuer's rating one year from now: $\Omega=\{\mathrm{AAA},\mathrm{AA},\mathrm{A},\mathrm{BBB},\mathrm{BB},\mathrm{B},\mathrm{CCC},\mathrm{D}\}$ , with $\mathrm{D}$ for default.
Number of defaults in a basket of five bonds: $\Omega=\{0,1,2,3,4,5\}$ .
Number of trading days until the first down day: $\Omega=\{1,2,3,\ldots\}$ . This space is infinite but countable — its outcomes can be listed in a sequence.

Granularity is a modeling choice

The same trading day can be modeled by $\{U,D\}$ , by $\{U,F,D\}$ , or by return buckets. None is "the" sample space. A finer space can express more questions — $\{U,D\}$ cannot even state "the close was unchanged." The working rule: choose $\Omega$ fine enough that every event you will need is a subset of it, and no finer than that.

❌ Outcomes that overlap or leave gaps

A desk lists tomorrow's outcomes as "up," "down," and "up more than 2%."

Why it breaks: a $+3\%$ day is both "up" and "up more than 2%," so the list is not exclusive. It also has no entry for an unchanged close, so it is not exhaustive. Either failure means a realized day does not map to exactly one outcome.

Consequence: once probabilities are attached (Chapter 02), overlapping outcomes double-count and missing outcomes leave probability unaccounted for, so the weights no longer total $1$ . A valid version is $\{\text{up more than } 2\%,\ \text{up at most } 2\%,\ \text{flat},\ \text{down}\}$ .

Outcomes must also be checkable. "The market moves a lot" is not an outcome: after the close you cannot say yes or no without first deciding what "a lot" means. Make the threshold explicit and it becomes a valid event on a suitable space.

Finite and countable spaces for now

The opening chapters use only finite or countable sample spaces, where every outcome can be listed. A price or a return measured on a continuum needs an uncountable space such as an interval of real numbers. That setting opens in Chapter 06, together with the extra care it requires about which subsets count as events.

03 · Events

Definition — Event

An event is a subset $A\subseteq\Omega$ . The event $A$ occurs on a given run when the realized outcome satisfies $\omega\in A$ .

Plain language: an event is a yes/no question about the outcome, identified with the set of outcomes for which the answer is yes. "Did the index rise on both days?" is the set of outcomes in which it did.

Three events have their own names:

$\Omega$ , the certain event — it occurs whatever happens.
$\emptyset$ , the impossible event — it never occurs.
A singleton $\{\omega\}$ , an elementary event — one specific outcome.

Example — Events on two trading days

Take $\Omega=\{UU,UD,DU,DD\}$ , where $UD$ means up on day 1 and down on day 2.

"Day 1 is up": $A=\{UU,UD\}$ .
"Exactly one up day": $B=\{UD,DU\}$ .
"Both days move the same way": $C=\{UU,DD\}$ .
"At least one up day": $E=\{UU,UD,DU\}$ .

If the realized outcome is $\omega=UD$ , then $A$ , $B$ , and $E$ all occur and $C$ does not. Many events occur at once; exactly one outcome does.

Outcome versus event

$UU$ is an outcome; $\{UU\}$ is the event that this outcome occurs. The distinction looks pedantic, but probabilities are assigned to events, so you will always write $\mathbb{P}(\{UU\})$ — usually shortened to $\mathbb{P}(UU)$ once the convention is clear.

Practice — Translate into sets

On $\Omega=\{UU,UD,DU,DD\}$ write each statement as an explicit set: (a) "day 2 is down"; (b) "the two days differ"; (c) "no up day"; (d) "day 1 is up and day 2 is down." Which of your four sets are elementary events?

04 · Combining Events

Logical words become set operations. That translation is the whole skill of this section.

Definition — Set Operations on Events

For events $A,B\subseteq\Omega$ :

A\cup B=\{\omega: \omega\in A\text{ or }\omega\in B\}, \qquad A\cap B=\{\omega: \omega\in A\text{ and }\omega\in B\}, \qquad A^c=\{\omega\in\Omega: \omega\notin A\}

The difference is $A\setminus B=A\cap B^c$ . $A$ and $B$ are disjoint (mutually exclusive) when $A\cap B=\emptyset$ . $A\subseteq B$ means every outcome in $A$ is also in $B$ .

Statement about the outcome	Set expression
$A$ or $B$ (or both)	$A\cup B$
$A$ and $B$	$A\cap B$
not $A$	$A^c$
$A$ but not $B$	$A\setminus B$
whenever $A$ happens, $B$ happens	$A\subseteq B$
$A$ and $B$ cannot both happen	$A\cap B=\emptyset$

"Or" in probability is always inclusive: $A\cup B$ contains the outcomes where both occur. "Exactly one of $A$ , $B$ " is a different event, $(A\setminus B)\cup(B\setminus A)$ .

Definition — De Morgan's Laws

For any events $A,B$ ,

(A\cup B)^c = A^c\cap B^c, \qquad (A\cap B)^c = A^c\cup B^c

More generally, for events $A_1,\ldots,A_n$ ,

\Big(\bigcup_{i=1}^n A_i\Big)^c=\bigcap_{i=1}^n A_i^c, \qquad \Big(\bigcap_{i=1}^n A_i\Big)^c=\bigcup_{i=1}^n A_i^c

Plain language: "not (A or B)" means "neither," which is "not A and not B." "Not (A and B)" means "not both," which is "at least one fails." Negation swaps "or" with "and."

Step-by-step — Proving a set identity by tracking one outcome

Pick an arbitrary outcome. To show two sets are equal, show every

\omega

in the left set is in the right set, and vice versa.

Left to right. If

\omega\in(A\cup B)^c

, then

\omega\notin A\cup B

, so

\omega

is in neither

A

nor

B

. Hence

\omega\in A^c

and

\omega\in B^c

, so

\omega\in A^c\cap B^c

Right to left. If

\omega\in A^c\cap B^c

, then

\omega\notin A

and

\omega\notin B

, so

\omega\notin A\cup B

, which means

\omega\in(A\cup B)^c

Conclude. Each set contains the other, so they are equal. The second law follows by applying the first to

A^c

and

B^c

and taking complements.

Example — Two-name credit events

Let $D_1$ be "firm 1 defaults this year" and $D_2$ be "firm 2 defaults this year."

"Neither defaults" is $(D_1\cup D_2)^c=D_1^c\cap D_2^c$ : both survive.
"Not both default" is $(D_1\cap D_2)^c=D_1^c\cup D_2^c$ : at least one survives.
"Exactly one defaults" is $(D_1\setminus D_2)\cup(D_2\setminus D_1)$ .

The second event contains the first. The outcomes in the second but not the first are exactly the outcomes in the third.

❌ Negating “and” into “and”

A protection contract is meant to pay unless both reference names default. It is written down as paying on "firm 1 survives and firm 2 survives."

Why it breaks: the complement of $D_1\cap D_2$ is $D_1^c\cup D_2^c$ , not $D_1^c\cap D_2^c$ . The written event is "neither defaults," which is strictly smaller than the intended "not both default."

Consequence: the contract fails to pay in every outcome where exactly one firm defaults — precisely the scenarios the buyer was most likely to care about. Write the event as a set and apply De Morgan before reading it back in words.

05 · Partitions

Definition — Partition

A collection of events $\{B_1,\ldots,B_n\}$ is a partition of $\Omega$ when the events are pairwise disjoint ( $B_i\cap B_j=\emptyset$ for $i\neq j$ ) and together cover the space:

\bigcup_{i=1}^n B_i=\Omega

Plain language: a partition sorts every outcome into exactly one bucket. No outcome is left out, and none lands in two buckets.

Example — Rating buckets

On the one-year rating space of Section 02, the three events

B_{\mathrm{IG}}=\{\mathrm{AAA},\mathrm{AA},\mathrm{A},\mathrm{BBB}\}, \qquad B_{\mathrm{HY}}=\{\mathrm{BB},\mathrm{B},\mathrm{CCC}\}, \qquad B_{\mathrm{D}}=\{\mathrm{D}\}

form a partition: investment grade, high yield, and default. The individual ratings form a finer partition. "Downgraded below BBB" is not a fourth cell — it is the union $B_{\mathrm{HY}}\cup B_{\mathrm{D}}$ of existing cells.

Every partition splits every event into disjoint pieces:

A=\bigcup_{i=1}^n (A\cap B_i), \qquad (A\cap B_i)\cap(A\cap B_j)=\emptyset \text{ for } i\neq j

This decomposition is the raw material for the law of total probability in Chapter 03: to analyze $A$ , analyze it separately inside each bucket and add the pieces back up.

Example — Partitioning by the number of up days

On $\Omega=\{U,D\}^3$ (three trading days), let $N_k$ be the event "exactly $k$ up days" for $k=0,1,2,3$ . Then $N_0=\{DDD\}$ , $N_1=\{UDD,DUD,DDU\}$ , $N_2=\{UUD,UDU,DUU\}$ , $N_3=\{UUU\}$ . These four events partition $\Omega$ , with $1+3+3+1=8$ outcomes in total. The event "at least one up day" is $N_1\cup N_2\cup N_3=N_0^c$ .

Sorting outcomes by a numerical summary — here, the count of up days — is the first glimpse of a random variable, the subject of Chapter 05.

06 · Product Spaces

Multi-period and multi-asset experiments are built by pairing up simpler ones.

Definition — Product Space

The product of sample spaces $\Omega_1$ and $\Omega_2$ is the set of ordered pairs

\Omega_1\times\Omega_2=\{(\omega_1,\omega_2): \omega_1\in\Omega_1,\ \omega_2\in\Omega_2\}

The $n$ -fold product $\Omega^n$ is the set of ordered $n$ -tuples from $\Omega$ . For finite spaces, $|\Omega_1\times\Omega_2|=|\Omega_1|\,|\Omega_2|$ and $|\Omega^n|=|\Omega|^n$ .

Plain language: an outcome of the combined experiment records the result of every stage, in order. $UD$ and $DU$ are different outcomes, because the order of the days matters.

Step-by-step — Building a multi-day space

Choose the one-day space. For example

\{U,D\}

, or

\{U,F,D\}

if flat days matter.

Take ordered tuples. Three days of

\{U,D\}

give

\Omega=\{U,D\}^3

, with outcomes such as

UDU

Count.

|\{U,D\}^3|=2^3=8

; with flat days,

|\{U,F,D\}^3|=3^3=27

. Counts grow geometrically with the horizon.

Define events through coordinates. "Day 2 is up" fixes the second coordinate and leaves the others free:

\{UUU,UUD,DUU,DUD\}

, which has

2^2=4

outcomes.

Example — Two assets on one day

An equity index moves up or down, and a corporate bond either defaults ( $1$ ) or does not ( $0$ ). The joint space is

\{U,D\}\times\{0,1\}=\{(U,0),(U,1),(D,0),(D,1)\}

"Equity down and bond defaults" is the elementary event $\{(D,1)\}$ . "Bond defaults" is $\{(U,1),(D,1)\}$ — the equity coordinate is left free.

A product space says nothing about dependence

Writing two days as $\{U,D\}^2$ only lists the joint outcomes. It does not claim that day 2 is unaffected by day 1. Whether the stages are related is a statement about probabilities, made in Chapter 04 (independence) and later Chapter 10 (joint distributions). The same product space can carry independent days or strongly trending ones.

07 · Events a Contract Can Name

Once the space is a product, most contracts do not pay on a single outcome. They pay on a set of paths. The payoff of a cash-or-nothing digital is the indicator $\mathbf{1}_A$ : one unit of cash if the realized path lands in $A$ , and nothing otherwise. Two contracts differ when their events differ, even if both can be described with the word "up".

Example — Terminal digital versus a barrier, on three days

Take $\Omega=\{U,D\}^3$ , eight paths. Write $U_i$ for "day $i$ is up".

"Up at expiry" (day 3 up): $A=\{UUU,UDU,DUU,DDU\}$ . Four paths. A digital on the terminal close pays on $A$ .
"Up on every day": $C=\{UUU\}$ . One path. This is $U_1\cap U_2\cap U_3$ .
"Down on at least one day": $C^c$ , the other seven paths. By De Morgan this is $U_1^c\cup U_2^c\cup U_3^c$ .
"Up at expiry, but not in a straight line": $A\setminus C=\{UDU,DUU,DDU\}$ .

$A$ and $C$ are different events. A term sheet that says only "the index finishes up" names $A$ . A term sheet that says "the index never trades down" names $C$ .

❌ Pricing the path with the terminal event

A three-day note pays if the index is up at expiry. The pricing sheet instead counts only the single path that is up every day.

Why it breaks: "up at expiry" is a set of four paths on $\{U,D\}^3$ . "Up every day" is one of those four. The second event is a subset of the first, so any probability attached later (Chapter 02) is smaller: under equal weights, $1/8$ against $4/8$ .

Consequence: the note is priced like a much stricter barrier. Holders who finish up after a down day — a path inside $A\setminus C$ — are owed a payment the model never reserved. Name the event as a set of paths before assigning it a number.

The indicator is the payoff, not yet a random variable

$\mathbf{1}_A(\omega)$ is a complete description of a digital's payoff on each outcome. Calling that function a random variable, and calling its probability-weighted average an expectation, is the work of Chapters 05 and 07. Nothing here requires that language: the contract is the event.

08 · Exercises

EXERCISE 1.1

List the $8$ outcomes of $\{U,D\}^3$ , sort them by how many $U$ they contain, then use the partition $N_0,\ldots,N_3$ of Section 05.

"At least two up days" is $N_2\cup N_3=\{UUD,UDU,DUU,UUU\}$ , which has $4$ outcomes. Its complement is "at most one up day," $N_0\cup N_1=\{DDD,UDD,DUD,DDU\}$ , also $4$ outcomes. The two sets are disjoint and their union is all $8$ outcomes, so $\{N_2\cup N_3,\ N_0\cup N_1\}$ is a two-cell partition of $\Omega$ — as $\{A,A^c\}$ always is.

For three trading days with $\Omega=\{U,D\}^3$ , write the event "at least two up days" as an explicit set of outcomes. Write its complement both in words and as a set, and check that the two together partition $\Omega$ .

EXERCISE 1.2

Translate each phrase using the table in Section 04, then use De Morgan to rewrite any complement of a union or intersection.

(a) "At least one defaults" is $D_1\cup D_2\cup D_3$ . (b) "None defaults" is $(D_1\cup D_2\cup D_3)^c=D_1^c\cap D_2^c\cap D_3^c$ by De Morgan — all three survive. (c) "Not all default" is $(D_1\cap D_2\cap D_3)^c=D_1^c\cup D_2^c\cup D_3^c$ — at least one survives. (d) (b) is a subset of (c): if all three survive, then certainly at least one does. The reverse fails — for example, the outcome where only firm 1 defaults lies in (c) but not (b).

A basket holds three bonds; $D_i$ is the event "bond $i$ defaults this year." Express in terms of $D_1,D_2,D_3$ : (a) at least one default; (b) no defaults; (c) not every bond defaults. (d) Which of (b) and (c) contains the other, and why are they not equal?

EXERCISE 1.3

Check the two partition conditions one at a time: does any rating appear in two buckets, and does every rating appear in some bucket?

It is not a partition. $\mathrm{CCC}$ lies in both "high yield" and "distressed," so the buckets are not pairwise disjoint. (Every rating does appear somewhere, so the cover condition holds.) One fix is high yield $=\{\mathrm{BB},\mathrm{B}\}$ and distressed $=\{\mathrm{CCC},\mathrm{D}\}$ , keeping investment grade $=\{\mathrm{AAA},\mathrm{AA},\mathrm{A},\mathrm{BBB}\}$ . Under the fixed partition, the event $A=\{\mathrm{A},\mathrm{BBB},\mathrm{BB},\mathrm{B},\mathrm{CCC},\mathrm{D}\}$ ("rated A or below") splits as $A\cap\mathrm{IG}=\{\mathrm{A},\mathrm{BBB}\}$ , $A\cap\mathrm{HY}=\{\mathrm{BB},\mathrm{B}\}$ , and $A\cap\mathrm{Distressed}=\{\mathrm{CCC},\mathrm{D}\}$ , three disjoint pieces whose union is $A$ .

On the rating space $\Omega=\{\mathrm{AAA},\mathrm{AA},\mathrm{A},\mathrm{BBB},\mathrm{BB},\mathrm{B},\mathrm{CCC},\mathrm{D}\}$ , a report uses the buckets investment grade $=\{\mathrm{AAA},\mathrm{AA},\mathrm{A},\mathrm{BBB}\}$ , high yield $=\{\mathrm{BB},\mathrm{B},\mathrm{CCC}\}$ , and distressed $=\{\mathrm{CCC},\mathrm{D}\}$ . Is this a partition? If not, fix it, and then split the event "rated A or below" into its pieces across your corrected buckets.

EXERCISE 1.4

List all eight paths. "Up at expiry" fixes only the last coordinate. "A down day occurs" is the complement of the single path $UUU$ . Their overlap is the paths that end in $U$ and are not $UUU$ .

(a) Up at expiry: $\{UUU,UDU,DUU,DDU\}$ , four paths. A down day occurs: every path except $UUU$ , seven paths. (b) The intersection is $\{UDU,DUU,DDU\}$ . (c) Let $U_i$ be "day $i$ is up". No down day is $U_1\cap U_2\cap U_3=\{UUU\}$ , so a down day occurs is $(U_1\cap U_2\cap U_3)^c=U_1^c\cup U_2^c\cup U_3^c$ . (d) The digital pays on "up at expiry", the four-path event, which properly contains the one-path event "up every day".

For $\Omega=\{U,D\}^3$ , (a) write "up at expiry" and "a down day occurs" as sets of paths. (b) Write the event that both happen. (c) Express "a down day occurs" using De Morgan, starting from the three events "day $i$ is up". (d) A cash-or-nothing digital pays if the index is up at expiry. Which of the events in (a) is that contract, and why is "up every day" a different contract?

09 · Chapter Summary

Concept	Meaning
Sample space $\Omega$	exclusive, exhaustive list of outcomes; exactly one occurs
Event	subset of $\Omega$ ; occurs when $\omega$ lands in it
$\cup$ , $\cap$ , $A^c$ , $A\setminus B$	or (inclusive), and, not, but not
De Morgan	negation swaps union and intersection
Partition	disjoint events that cover $\Omega$ ; every outcome in one cell
Product space	ordered tuples of stage outcomes; sizes multiply
Indicator $\mathbf{1}_A$	digital payoff: $1$ on the event, $0$ off it
Path versus terminal	"up at expiry" and "up every day" are different subsets of the same product space

Next: Chapter 02 — Probability Axioms, which attaches numbers to events so that they behave consistently under unions, complements, and partitions.