$\displaystyle X_1$, $\displaystyle X_2$ are nonempty sets. Let $\displaystyle \mathcal{F}_1 \subset 2^{X_1}$ and $\displaystyle \mathcal{F}_2 \subset 2^{X_2}$ be $\displaystyle \sigma$-algebras.

Let $\displaystyle \mathcal{R} = \{A \times B: A \in \mathcal{F}_1, B \in \mathcal{F}_2 \}$

And $\displaystyle \mathcal{R}_1 = \left\{\bigcup_{i=1}^n R_i: n \in \mathbb{N}, \ \{R_i\}_{i=1}^n \subset \mathcal{R} \right\}$

Prove that $\displaystyle \mathcal{R}_1$ is an algebra.

It was not hard to prove that it is colsed with respect to finite unions. However I failed to prove it is closed with respect to complement!

2. And this relates to Probability and Statistics, how?

-Dan

3. Originally Posted by topsquark
And this relates to Probability and Statistics, how?

-Dan
Because axiomatic probability theory involves measures and $\displaystyle \sigma$-algebra.

4. Frankly speaking I was working on some probability theory theorem when this problem occured. I'm 99% sure $\displaystyle R_1$ is algebra but the full proof seems bit too long.... If somebody knows a short proof, please post it.

PS I think I can prove that $\displaystyle R_1$ is closed under finite intersections. Than I take complement of set that belongs to $\displaystyle R_1$ and I get finite intersection - the only problem is to find a simple proof that the "intersected" sets belong to $\displaystyle R_1$ .

5. Originally Posted by albi
Frankly speaking I was working on some probability theory theorem when this problem occured. I'm 99% sure $\displaystyle R_1$ is algebra but the full proof seems bit too long.... If somebody knows a short proof, please post it.

PS I think I can prove that $\displaystyle R_1$ is closed under finite intersections. Than I take complement of set that belongs to $\displaystyle R_1$ and I get finite intersection - the only problem is to find a simple proof that the "intersected" sets belong to $\displaystyle R_1$ .
these two trivial identities should be enough for you to complete the proof:

$\displaystyle 1) \ \ (A_1 \times B_1) \cap (A_2 \times B_2) = (A_1 \cap A_2) \times (B_1 \cap B_2).$

$\displaystyle 2) \ \ (A \times B)^c = (A^c \times B) \cup (A \times B^c) \cup (A^c \times B^c),$ where $\displaystyle Y^c$ is the complement of $\displaystyle Y.$

6. The last one is not true. I belive it should be:
$\displaystyle (A \times B)^c = (A^c \times B) \cup (A \times B^c) \cup (A^c \times B^c),$

However that will help.

7. Originally Posted by albi
The last one is not true. I belive it should be:
$\displaystyle (A \times B)^c = (A^c \times B) \cup (A \times B^c) \cup (A^c \times B^c),$
of course! thanks! i fixed it. i've already done too much latex for today ... i guess i'm getting really tired!