# Struggling with event spaces in probability

My textbook gives the following definition of an event space (see attached). Omega here is the sample space (i.e. all possible outcomes of an experiment)

Based on this definition I have been trying to solve the following exercises, but haven't really gotten anywhere and would like some advice on how to continue: (see attached)

For 1.11: I tried to define a set $$K$$ where exactly $$k$$ of the points in $$\Omega$$ belong to $$K$$, however I don't know how to do this concisely/what notation to use.

$$K = (A_1\setminus A_2... A_k)\cup (A_2\setminus A_1...A_k)\cup...(A_k\setminus A_1...A_{k-1})$$ does't work I think.

For 1.12: Isn't this just the power set hence the answer is $$2^n$$. I feel like this isn't right though. Won't an event space always be just the power set of omega?

• I think you're right about 1.12 – Zubin Mukerjee Jan 14 at 20:04
• for 1.12 yes it's true for the power set and trivial set but you can construct the smallest sigma algebra containing any $A_i$. – user29418 Jan 14 at 20:13

Hint for 1.11: one way to write "points that belong to only $$A_1, A_3, A_5$$ and no other $$A_i$$" is $$A_1 \cap A_2^c \cap A_3 \cap A_4^c \cap A_5 \cap A_6^c \cap \cdots \cap A_m^c,$$ (if you are unfamiliar with this complement notation, I denote the complement of $$B$$ by $$B^c := \Omega \setminus B$$). Using something like the above as a building block can help you finish this question.
1.12: The English is a little misleading in the definition. $$\mathcal{F}$$ is some collection of subsets of $$\Omega$$ that satisfies the three conditions. It is not necessarily the power set (all subsets). Note that the power set always satisfies the three conditions.
Hint: is there a natural way to group the elements of $$\mathcal{F}$$ into pairs?
• What is the best way to choose the $A_i$'s that contain the set of points? I defined a set $K = \{A'_1,A'_2,...,A'_k\}$ with $k$ members as well as the set $M = \{A''_1,A''_2,...,A''_{m-k}\}$ with $m-k$ members. Then I used your hint to complete the question. – xAly Jan 14 at 21:52
• For 1.12: Let $\mathcal{F} = \{A_1,A_2,...,A_m\}$, so $\mathcal{F}$ has $m$ members. Now since $A_1 \in \mathcal{F}$,$A^c_1 \in \mathcal{F}$, and so we can pair each $A_i$ with $A^c_i$, so there must be an even number of elements in $\mathcal{F}$. – xAly Jan 14 at 22:03