Independent Events are Independent of Complement

Theorem
Let $$A$$ and $$B$$ be events in a probability space $$\left({\Omega, \Sigma, \Pr}\right)$$.

Then $$A$$ and $$B$$ are independent iff $$A$$ and $$\Omega \setminus B$$ are independent.

Corollary
$$A$$ and $$B$$ are independent iff $$\Omega \setminus A$$ and $$\Omega \setminus B$$ are independent.

General Theorem
Let $$A_1, A_2, \ldots, A_m$$ be events in a probability space $$\left({\Omega, \Sigma, \Pr}\right)$$.

Then $$A_1, A_2, \ldots, A_m$$ are independent events (in the general sense) iff $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$$ are also independent (in the same sense).

Proof
For $$A$$ and $$B$$ to be independent:
 * $$\Pr \left({A \cap B}\right) = \Pr \left({A}\right) \Pr \left({B}\right)$$

We need to show that:
 * $$\Pr \left({A \cap \left({\Omega \setminus B}\right)}\right) = \Pr \left({A}\right) \Pr \left({\Omega \setminus B}\right)$$

First note that $$\Omega \setminus B \equiv \mathcal C_{\Omega} \left({B}\right)$$ where $$\mathcal C_{\Omega}$$ denotes the relative complement.

From Set Difference Relative Complement, we have then that $$A \cap \left({\Omega \setminus B}\right) = A \setminus B$$.

From Set Difference and Intersection form Partition, we have that:
 * $$\left({A \setminus B}\right) \cup \left({A \cap B}\right) = A$$;
 * $$\left({A \setminus B}\right) \cap \left({A \cap B}\right) = \varnothing$$.

So from the Kolmogorov axioms, we have that:
 * $$\Pr \left({A}\right) = \Pr \left({A \setminus B}\right) + \Pr \left({A \cap B}\right)$$

Hence:

$$ $$ $$ $$

But as $$A \setminus B = A \cap \left({\Omega \setminus B}\right)$$ we have:
 * $$\Pr \left({A \cap \left({\Omega \setminus B}\right)}\right) = \Pr \left({A}\right) \Pr \left({\Omega \setminus B}\right)$$

which is what we wanted to show.

Now, suppose $$A$$ and $$\Omega \setminus B$$ are independent.

From the above, we have that $$A$$ and $$\Omega \setminus \left({\Omega \setminus B}\right)$$ are independent.

But $$\Omega \setminus \left({\Omega \setminus B}\right) = B$$ from Relative Complement of Relative Complement hence the result.

Proof of Corollary
Let $$A$$ and $$B$$ be independent.

Then from the main result, $$A$$ and $$\Omega \setminus B$$ are independent.

Setting $$A' = \Omega \setminus B$$ and $$B' = A$$, we see clearly that $$A'$$ and $$B'$$ are independent.

So from the main result, $$A'$$ and $$\Omega \setminus B'$$ are independent.

That is, $$\Omega \setminus B$$ and $$\Omega \setminus A$$ are independent.

The "only if" part of the result follows directly from Relative Complement of Relative Complement and another application of this result.

Proof of General Theorem
Proof by induction:

For all $$n \in \N: n \ge 2$$, let $$P \left({n}\right)$$ be the proposition:
 * $$A_1, A_2, \ldots, A_n$$ are independent iff $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$$ are independent.

Basis for the Induction

 * $$P(2)$$ is the case:
 * $$A_1$$ and $$A_2$$ are independent iff $$\Omega \setminus A_1$$ and $$\Omega \setminus A_2$$ are independent.

This has been proved above, as the corollary.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$A_1, A_2, \ldots, A_k$$ are independent iff $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_k$$ are independent.

Then we need to show:
 * $$A_1, A_2, \ldots, A_{k+1}$$ are independent iff $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k+1}$$ are independent.

Induction Step
This is our induction step.

Suppose $$A_1, A_2, \ldots, A_{k+1}$$ are independent.

Then:

$$ $$ $$

So we see that $$\bigcap_{i=1}^{k} A_i$$ and $$A_{k+1}$$ are independent.

So $$\bigcap_{i=1}^{k} A_i$$ and $$\Omega \setminus A_{k+1}$$ are independent.

So, from the above results, we can see that $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k+1}$$ are independent.

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

The reverse implication follows directly.

Therefore:
 * $$A_1, A_2, \ldots, A_n$$ are independent iff $$\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$$ are independent.