You flip a coin two times. You consider two events:

$$A=\{ " it \ lands\ heads \ up \ two \ times"\}$$

$$B=\{ " it \ lands\ tails\ up \ two \ times"\}$$

Which events do I have to add to get an sigma-algebra F? Firstly, it should be $ \bar{A} \in F$ and $ \bar{B} \in F$ So I have to add $$\bar{A}= \{ (t,h),(h,t),(t,t) \}, \ \bar{B}= \{ (t,h),(h,t),(h,h) \}$$ The last ones are already in F. Is this enough, when I consider the order?


To find the smallest $\sigma$-algebra $F$ satisfying $A,B\in F$ we first construct the collection: $$\mathcal V=\{A\cap B,A\cap B^{\complement},A^{\complement}\cap B,A^{\complement}\cap B^{\complement}\}$$It is evident that every element of $\mathcal V$ is an element of $F$.

Further the elements of $\mathcal V$ are mutually disjoint and cover the whole space.

Then: $$F=\{\cup\mathcal A\mid\mathcal A\subseteq\mathcal V\}$$

Or in words: elements of $F$ are exactly the sets that can be written as a union of elements of $\mathcal V$.

Usually (not always) $\mathcal V$ has $4$ distinct non-empty elements and consequently $F$ has $2^4=16$ elements.


with order:

  • $A\cap B=\varnothing$
  • $A\cap B^{\complement}=\{(h,h)\}$
  • $A^{\complement}\cap B=\{(t,t)\}$
  • $A^{\complement}\cap B^{\complement}=\{(h,t),(t,h)\}$

Then we find the following unions:

  • $\varnothing$
  • $\{(h,h)\}$
  • $\{(t,t)\}$
  • $\{(h,t),(t,h)\}$
  • $\{(h,h),(t,t)\}$
  • $\{(h,h),(h,t),(t,h)\}$
  • $\{(t,t),(h,t),(t,h)\}$
  • $\{(h,h),(t,t),(h,t),(t,h)\}$
  • $\begingroup$ When I don't conisder the order, then I have only 8, haven't I? $\endgroup$ – Steven33 Apr 22 at 12:01
  • $\begingroup$ $8=2^3$ is correct here. It finds its cause in the fact that $A\cap B=\varnothing$ in your case so that $\mathcal V$ contains $3$ disjoint non-empty sets (and next to that also the empty set). Then there are $2^3$ possible unions. Order is not relevant here. $\endgroup$ – drhab Apr 22 at 12:03
  • $\begingroup$ Ok. The tasks says, that I have to consider the order and don't consider the order. In one case I have 8 elements in F. In the other case 16. $\endgroup$ – Steven33 Apr 22 at 12:06
  • $\begingroup$ I dont' get to 8 elements: $A \cap B= \emptyset$ , $ A \cap \bar{B}=A =(H,H)$, $ \bar{A} \cap B=B =(T,T)$ $ \bar{A} \cap \bar{B}=(Z,K),(K,Z)$ $\endgroup$ – Steven33 Apr 22 at 12:09
  • $\begingroup$ $A^{\complement}\cap B^{\complement}=\{(h,t),(t,h)\}\neq A$ (with order). There are $3$ disjoint non-empty sets in $\mathcal V$. This also if order is disregarded (i.e. $A^{\complement}\cap B^{\complement}=\{\{h,t\},\{t,h\}\}=\{\{t,h\}\}$). In both cases the answer is $2^3=8$. $\endgroup$ – drhab Apr 22 at 12:16

You have to add one more event, namely $\overset {-} A \cap \overset {-} B=\{(h,t),(t,h)\}$. The sigma algebra is $F=\{\emptyset, A, B ,A\cup B, \overset {-} A,\overset {-} B,\overset {-} A\cap \overset {-} B,\{((h,h),(t,t),(h,t),(t,h)\}$.

  • $\begingroup$ Ok then I have $F= \{ \emptyset, \bar{A}, \bar{B}, \bar{A} \cap \bar{B}\}$ But there have to be 8 elements in this set? $\endgroup$ – Steven33 Apr 22 at 11:51
  • $\begingroup$ $F=\{\emptyset, A, B , \overset {-} A,\overset {-} B,\overset {-} A\cap \overset {-} B,\{((h,h),(t,t),(h,t),(t,h)\}$. $\endgroup$ – Kavi Rama Murthy Apr 22 at 11:54
  • $\begingroup$ But this are only 7 elements in F? $\endgroup$ – Steven33 Apr 22 at 11:57
  • $\begingroup$ @drhab Thanks for pointing out. $\endgroup$ – Kavi Rama Murthy Apr 22 at 13:02

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