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:

	\(\displaystyle \)	\(\displaystyle \Pr \left({A \setminus B}\right)\)	\(=\)	\(\displaystyle \Pr \left({A}\right) - \Pr \left({A \cap B}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({A}\right) - \Pr \left({A}\right) \Pr \left({B}\right)\)	\(\displaystyle \)	as $A$ and $B$ are independent
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({A}\right) \left({1 - \Pr \left({B}\right)}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({A}\right) \Pr \left({\Omega \setminus B}\right)\)	\(\displaystyle \)	Elementary Properties of Probability Measure

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.

$\blacksquare$

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.

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \Pr \left({\bigcap_{i=1}^{k+1} A_i}\right)\)	\(=\)	\(\displaystyle \Pr \left({\bigcap_{i=1}^k A_i \cap A_{k+1} }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \prod_{i=1}^k \Pr \left({A_i}\right) \times \Pr \left({A_{k+1} }\right)\)	\(\displaystyle \)	as all of the $A_1, A_2, \ldots, A_k, A_{k+1}$ are independent
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({\bigcap_{i=1}^k A_i}\right) \times \Pr \left({A_{k+1} }\right)\)	\(\displaystyle \)

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

So $\displaystyle \bigcap_{i=1}^k A_i$ 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.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 1.7$: Exercises $22, 23$