# Generated Sigma Algebra - Help

1. Jul 5, 2012

### woundedtiger4

Consider S = {(H,H), (H,T), (T,H), (T,T)},
the repeated coin toss. The Sigma-Algebra generated by
C = {{(HH), (TT)}} is
σ(C) = { ∅ , S, {(HT), (TH)}, {(HH), (TT)} }

How does "(H,T), (T,H)" jumped in the σ(C) ? What is the difference between σ-algebra & generated σ-algebra? Why "(HT), (TH)" in σ(C) have been seperated from "(HH), (TT)" with curly brackets?

Last edited by a moderator: Jul 5, 2012
2. Jul 5, 2012

### mathman

{(H,T),(T,H)} is the complement of {(H,H),(T,T)}.

σ-algebra is the general term. Generated σ-algebra is a specific term, refering to the smallest σ-algebra containing the generator.

3. Jul 5, 2012

### theorem4.5.9

This is poorly worded. A sigma algebra is a sigma algebra, regardless of where it comes. I believe your confusion is with the generating set. $C$. $C$ is not a sigma algebra, but there is a unique minimal sigma algebra containing $C$, you've denoted it $\sigma (C)$. In this sense, the sigma algebra (that you have denoted by) $\sigma (C)$ is generated by $C$.

4. Jul 6, 2012

thanks

5. Jul 6, 2012

### woundedtiger4

what does it mean that it is "poorly worded"? I have just posted it from the text, & if you think my questions are stupid then I am trying to learn it & if I have some questions in my mind then shouldn't I clarify them? :(

6. Jul 6, 2012

### Staff: Mentor

One entity that you named was a sigma algebra, i.e. $\sigma (C)$. It's a sigma algebra because the set that $\sigma (C)$ represents satisfies the sigma algebra axioms. The second entity you named, $C$ is not a sigma algebra. It's a set with no special properties. However, you can generate a sigma algebra from it by finding the smallest possible sigma algebra that contains $C$. You chose to name this sigma algebra $\sigma (C)$ because it makes it clear that $C$ is the generating set.
You could also ask what sigma algebra is generated by $\sigma (C)$? You might as well call it $\sigma ( \sigma (C))$. The answer is of course the same set, $\sigma (C)$ since it was a sigma algebra to begin with, and so, in equations, $\sigma ( \sigma (C)) = \sigma (C)$