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I am confused on how to determine a Sigma-algebra.

The following partitions of a set are given:
$$ A1 = \{1,3\} $$ $$ A2 = \{2,4,6,8\} $$ $$ A3 = \{5,7,9\}$$ And Omega is $$ \Omega = \{1,2,3,4,5,6,7,8,9\} $$

I know that at least Omega and the empty set has to be part of the Sigma-algebra as well as the single partitions and their complements: $$ \sigma(A1,A2,A3) = \{\emptyset,\Omega,\{1,3\},\{2,4,5,6,7,8,9\},\{2,4,6,8\},\{1,3,5,7,9\},\{5,7,9\},\{1,2,3,4,6,8\}\} $$

Additionally, there has to be the union of the elements inside and this is the point I am not sure about. Are the following elements part of the Sigma-algebra as well? $$ A1 \cup A2 = \{1,2,3,4,6,8\}$$ $$ A2 \cup A3 = \{2,4,5,6,7,8,9\}$$ $$ A1 \cup A3 = \{1,3,5,7,9\} $$ $$ A1 \cup A2 \cup A3 = \Omega $$

I think I can ignore all of them since they are already in my Sigma-algebra.

Is my Sigma-algebra fullfinished and where my thoughts about creating the algebra correct? I know that there have to be 2^n elements inside the algebra and due to the fact of there are 8 elements this might be a correct solution.

Another question:

In case a union (e.g. A1 U A2) would not have been in the algebra. Do I just have to add A1 U A2 to the algebra or also its complement?

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You're right on target. The ($\sigma$-)algebra on $\Omega$ generated by the partition $\{A_1,A_2,A_3\}$ is precisely $$\{\emptyset,A_1,A_2,A_3,A_1\cup A_2,A_1\cup A_3,A_2\cup A_3,\Omega\}.$$ More generally, if you're given a partition of any set $\Omega$ into $n$ subsets, the ($\sigma$-)algebra on $\Omega$ generated by that partition will consist of $2^n$ subsets, consisting of all finite unions of the partition elements (including the empty union).

When you aren't dealing with partitions, it is less simple. At that point, you'll need to worry about complements. Roughly speaking, take all the countable unions, take all the complements, and repeat until you're done.

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  • $\begingroup$ Thanks! Don't I have to unite all the partitions as well? $$ A1 \cup A2 \cup A3 $$ In my script there's something like $$ \sigma(A1,...An) = \{A1^{\epsilon1} \cup A2^{\epsilon2} \cup ... \cup An^{\epsilon_n} : \epsilon_i \in \{0,1\}, i=1,...,n\} $$ and this has confused me. I thought I have to unite all elements as well. $\endgroup$
    – Drudge
    Commented Oct 26, 2013 at 13:16
  • $\begingroup$ Yes, but $A_1\cup A_2\cup A_3=\Omega,$ so that's in there. As for the other thing, I suspect it is intended that $A^0=\emptyset$ and $A^1=A$, so (for example) $A_1\cup A_2=A_1^1\cup A_2^1\cup A_3^0$ has the desired form, as does $\emptyset=A_1^0\cup A_2^0\cup A_3^0,$ as does $\Omega=A_1^1\cup A_2^1\cup A_3^1.$ $\endgroup$ Commented Oct 26, 2013 at 13:23
  • $\begingroup$ Ah okay, I think I got it! If I take a look at the superiored numbers, I think I just have to unite each combination of them 000, 001, 010, 011, 100, 101, 110, 111 which are 2^3 possibilities, right? And 000 is the empty set an 111 the omega. Thats why there are always 2^n possibilities. I got it! the only question left is when I have to determine the complements. In this case I didn't really need them. Or can I say that e.g. 001 is the complement of 110 and e.g. 101 the complement of 010? I didn't calculated it but I think that is correct? $\endgroup$
    – Drudge
    Commented Oct 26, 2013 at 13:28
  • $\begingroup$ Bingo! That's exactly right. $\endgroup$ Commented Oct 26, 2013 at 13:30
  • $\begingroup$ Thanks! Sorry, I still have a last but short question to the topic. In case there are doubled or empty partitions, can I just ignore them? $\endgroup$
    – Drudge
    Commented Oct 26, 2013 at 13:42

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