# Independent of random variables.

When reading Shiryaev's Probability. In the chapter 1, section 4. problem 11:

Show that the random variables $\xi_1,\cdots,\xi_n$ are independent if and only if $$F_{\xi_1\cdots\xi_n}(x_1,\cdots,x_n)=F_{\xi_1}(x_1)\cdots F_{\xi_n}(x_n)$$ for all $x_1,\cdots, x_n$,where $F_{\xi_1\cdots\xi_n}(x_1,\cdots,x_n)=\Bbb{P}\{\xi_1\leq x_1,\cdots,\xi_n\leq x_n\}$

But in the text, there is a definition 3:

Definition 3 The random variables $\xi_1,\cdots,\xi_n$ are said to be independent(collectively independent) if $$\Bbb{P}\{\xi_1=x_1,\cdots,\xi_n=x_n\}=\Bbb{P}\{\xi_1=x_1\}\cdots\Bbb{P}\{\xi_n=x_n\}$$ for all $x_1,\cdots,x_n \in X$ or equivalently,if$$\Bbb{P}\{\xi_1\in B_1,\cdots,\xi_n\in B_n\}=\Bbb{P}\{\xi_1 \in B_1\}\cdots\Bbb{P}\{\xi_n \in B_n\}$$ for all $B_1,\cdots,B_n$ is in a algebra of subsets of $X$

I think this problem is the same as the definition because $\Bbb{P}\{\xi \leq x\} = \Bbb{P}\{\xi \in B\}$. so it doesn't need to prove. But I think author will not do this. So what's wrong with me?

• It's true that one direction is obvious by taking $B_i=(-\infty,x_i]$ for $1\leq i\leq n$. But for the other direction you need to work a bit. – Stefan Hansen Jul 2 '14 at 8:58
• @StefanHansen thanks very much. – Laura Jul 29 '14 at 8:48