Now that we have an understanding of what sets are, how to relate sets to other sets, and how to do some operations on sets, the next thing we will want to do is organize sets based off of what we think is important about them. What we think is important will be dependent on the context that we are in. For example, if we are playing a game that involves rolling dice, the order that the dice are rolled may or may not be important to the rules.

Definition Let $\mathcal{S}$ be a set. An event space $\mathcal{A}$ on $\mathcal{S}$ is a collection of subsets of $\mathcal{S}$ satisfying

1. $\mathcal{S} \in \mathcal{A}$
2. If $E\in \mathcal{A}$, then $E^c \in \mathcal{A}$
3. If $E_1\in\mathcal{A}$ and $E_2\in\mathcal{A}$, then $E_1\cup E_2\in \mathcal{A}$

We call the elements of $S$ outcomes, and the subsets that are in $\mathcal{A}$ events.

Example: Suppose we are playing a game in which you roll two dice. We can establish what the outcomes of this game are without needing to define an event space. They are the possibile results (order included) of rolling the two dice. We can represent them as the elements of the set $$ \mathcal{S} = \{(1,1),(1,2),(1,3),...,(2,1),(2,2),...,(6,5),(6,6)\}.$$ The set $\mathcal{S}$ will always be an event, which we interpret as "the event that you have rolled both dice".

Defining an event space on $\mathcal{S}$ establishes which events are important to the game. Suppose the game is simply: I win if the sum of both dice is seven. An important event (the important event) is the subset of $\mathcal{S}$ whose pairs sum to 7. I.e., $$E_7 = \{(1,6),(2,5),(3,4),(4,3),(5,2),(1,6)\}.$$ We can interpret this event as a "win". Another important event is the subset of $\mathcal{S}$ whose pairs don't sum to seven $$E_7^c = \{(1,1),(1,2),...,(6,6)\}.$$ We can interpret this event as a "loss". In this very simple case, the event space is $$\mathcal{A} = \{\mathcal{S},E_7,E_7^c,\varnothing\}.$$ Note that we have to include the empty set $\varnothing$ here, since $\mathcal{S}^c=\varnothing$.

Exercise: Consider a simple game where you roll a single six-sided die. The rules 1) you win a candy if you roll a one. 2) you win a chocolate if you roll a two. 3) you win nothing if you roll anything else.

1. What are the outcomes of this game?
2. What is the complete event space of this game? Interpret each event within the rules of the game.

Solution: (click to show)

We can take a more algorithmic approach to constructing event spaces if we think about what specific things events we want in the space. For example, if we want $E_1$ and $E_2$ to be events in the event space $\mathcal{A}$, then we need to add $E_1^c$ and $E_2^c$ to $\mathcal{A}$ as well as $E_1\cup E_2$ and $(E_1\cup E_2)^c$. We can then look to see if there are other unions or complements missing, if so, we add them to the event space as well. If we started with two events $E_1$ and $E_2$ we call the resulting event space the event space generated by $E_1$ and $E_2$.

Example: Let $\mathcal{S}= \{1,2,3,4,5\}$. Generate an event space that contains all the subsets of $\mathcal{S}$ with one outcome.

• Since the event space must contain $\mathcal{S}$, it must also contain $\varnothing$
• Since the event space contains $\{1\}$, we have to add $\{1\}^c = \{2,3,4,5\}$. And therefore every subset with four outcomes.
• Since the event space contains $\{1\}$ and $\{2\}$ we must include $\{1\}\cup\{2\} = \{1,2\}$, and therefore all subets with two outcomes.
• Since the event space contains $\{1,2\}$ we must include $\{1,2\}^c = \{3,4,5\}$, and therefore all subsets with three elements.
• At this point, we have included all of the possible subsets of $\mathcal{S}$, so there is nothing left to add and our event space is complete.

Exercises

• Suppose $\mathcal{A}$ is an event space and $E_1$ and $E_2$ are events:

  1. Is $E_3 = E_1^c \cap E_2^c$ an event?
  2. Is $E_4 = E_1 \cap E_2$ an event?
• Let $\mathcal{S} = \{1,2,3,4\}$. Find the event space generated by $E_1= \{1\}$ and $E_{23} = \{2,3\}$.
• Let $\mathcal{S} = (0,1).$ Find the event space generated by $E_1 = (0,\tfrac{3}{4})$ and $E_2 = (\tfrac{1}{4},1)$.