# Independence of events question

1. Feb 7, 2016

### geoffrey159

Hello,

I am studying independence of events and I came across a formula that I don't understand. It is rather technical, not very interesting, but I feel stuck and it stays in my mind. Could you explain the following :

If $(A_i)_{i\in I}$ is a sequence of independent events on $(\Omega,{\cal A}, P)$, then, given the partition $I_1 \cup I_2$ of $I$, the sequence of events $(B_i)_{i\in I}$ defined by $B_i = A_i$ if $i\in I_1$ and $B_i = A_i^c$ if $i\in I_2$ is also independent.

To prove this, one must show that for any finite set $J\subset I$ : $P(\bigcap_{j\in J} B_j) = \prod_{j\in J} P(B_j)$

Since $J = J_1 \cup J_2$, where $J_i = I_i \cap J$, we have $\bigcap_{j\in J} B_j = (\bigcap_{j\in J_1} A_j) \bigcap (\bigcap_{j\in J_2} A_j^c) =(\bigcap_{j\in J_1} A_j) - (\bigcup_{j\in J_2} A_j)$

so that
$P(\bigcap_{j\in J} B_j) = P(\bigcap_{j\in J_1} A_j) - P(\bigcup_{j\in J_2} A_j)$

With the inclusion-exclusion principle
$P(\bigcup_{j\in J_2} A_j) = \sum_{ K\subset J_2 \ |K| = 1} P(\bigcap_{j\in K} A_j ) - \sum_{ K\subset J_2 \ |K| = 2} P(\bigcap_{j\in K} A_j ) + ... + (-1)^{|J_2|-1} P(\bigcap_{j\in J_2} A_j)$

But I don't see how to finish this once $P(\bigcap_j A_j )$ is developped into a product !

Last edited: Feb 7, 2016
2. Feb 7, 2016

### geoffrey159

Oh I've made a mistake I get it now :

$\bigcap_{j\in J} B_j = A\cap B^c = A - A\cap B$

with $A = \bigcap_{j\in J_1} A_j$ and $B = \bigcup_{j\in J_2} A_j$

we have $A \cap B = \bigcup_{j\in J_2} \bigcap_{k\in J_1} (A_k \cap A_j)$

And with the principle of inclusion exclusion and the independence of $(A_i)_{i\in I}$ we find $P( A \cap B) = P(A) P(B)$

Last edited: Feb 7, 2016