Independent Events are Independent of Complement/General Result

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Theorem

Let $A_1, A_2, \ldots, A_m$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.

Then $A_1, A_2, \ldots, A_m$ are independent if and only if $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ are also independent.


Proof

Proof by induction:

For all $n \in \N: n \ge 2$, let $\map P n$ be the proposition:

$A_1, A_2, \ldots, A_n$ are independent if and only if $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent.


Basis for the Induction

$\map P 2$ is the case:

$A_1$ and $A_2$ are independent if and only if $\Omega \setminus A_1$ and $\Omega \setminus A_2$ are independent.

This is proved in Independent Events are Independent of Complement: Corollary.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$A_1, A_2, \ldots, A_k$ are independent if and only if $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_k$ are independent.


Then we need to show:

$A_1, A_2, \ldots, A_{k + 1}$ are independent if and only if $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k + 1}$ are independent.


Induction Step

This is our induction step.

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

Then:

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

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

So $\displaystyle \bigcap_{i \mathop = 1}^k A_i$ 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 $\map P k \implies \map P {k + 1}$ 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 if and only if $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent.

$\blacksquare$


Sources