# coin/ sigma-algebra

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.

edit:

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:

• $$\varnothing$$
• $$\{(h,h)\}$$
• $$\{(t,t)\}$$
• $$\{(h,t),(t,h)\}$$
• $$\{(h,h),(t,t)\}$$
• $$\{(h,h),(h,t),(t,h)\}$$
• $$\{(t,t),(h,t),(t,h)\}$$
• $$\{(h,h),(t,t),(h,t),(t,h)\}$$
• When I don't conisder the order, then I have only 8, haven't I? – Steven33 Apr 22 at 12:01
• $8=2^3$ 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. – drhab Apr 22 at 12:03
• 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. – Steven33 Apr 22 at 12:06
• I dont' get to 8 elements: $A \cap B= \emptyset$ , $A \cap \bar{B}=A =(H,H)$, $\bar{A} \cap B=B =(T,T)$ $\bar{A} \cap \bar{B}=(Z,K),(K,Z)$ – Steven33 Apr 22 at 12:09
• $A^{\complement}\cap B^{\complement}=\{(h,t),(t,h)\}\neq A$ (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 $2^3=8$. – 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 $$F=\{\emptyset, A, B ,A\cup B, \overset {-} A,\overset {-} B,\overset {-} A\cap \overset {-} B,\{((h,h),(t,t),(h,t),(t,h)\}$$.

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