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?

Definition of an event space


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

  • $\begingroup$ 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. $\endgroup$ – xAly Jan 14 at 21:52
  • $\begingroup$ 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 $\mathcal{F}$. $\endgroup$ – xAly Jan 14 at 22:03

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