If $A_1,A_2,…,A_n$ are independent then $A_1^c,A_2^c,…,A_n^c$ are also independent.

Question

Let $A_1,A_2,…,A_n$ be events in a probability space $(\Omega,\Sigma,P)$.

If $A_1,A_2,…,A_n$ are independent then $A_1^c,A_2^c,…,A_n^c$ are also independent, (where $A^c = \Omega \setminus A$).

I have found a proof by induction for this exercise, however, I have not been able to understand the conclusion of the proof, which I have marked in red. That is, why can it be immediately concluded that $A_1^c , A_2^c ,..., A_{k+1}^c$ are independent? I would really appreciate if someone can give me a clear explanation of what happens in that conclusion.

Proof by induction.

Basis for the Induction.

If $A_1$ and $A_2$ are independent then $A_1^c$ and $A_2^c$ are independent.

Assume $A_1$ and $A_2$ are independent. Then \begin{align*} P(A_1^c \cap A_2^c) &= 1 - P(A_1 \cup A_2) \\ &= 1 - P(A_1) - P(A_2) + P(A_1 \cap A_2) \\ &= 1 - P(A_1) - P(A_2) + P(A_1)P(A_2) \\ &= (1-P(A_1))(1-P(A_2)) \\ &= P(A_1^c)P(A_2^c). \end{align*}

Induction Hypothesis.

This is our induction hypothesis:

If $A_1,A_2,…,A_k$ are independent then $A_1^c,A_2^c,…,A_k^c$ are independent.

Then we need to show:

If $A_1,A_2,…,A_{k+1}$ are independent then $A_1^c,A_2^c,…,A_{k+1}^c$ are independent.

Induction Step.

This is our induction step.

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

Then: \begin{align} P\left( {\bigcap_{i = 1}^{k + 1} A_i}\right) &= P\left( \bigcap_{i=1}^{k}A_i \cap A_{k+1} \right) \\ &= \prod_{i=1}^{k}P(A_i) \cdot P(A_{k+1})\\ &= P\left(\bigcap_{i=1}^{k}A_i\right) \cdot P(A_{k+1}) \end{align}

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

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

$\color{red}{\text{So, from the above results, we can see that} A_1^c,A_2^c,…,A_{k+1}^c \text{are independent}}.$

Answer 1

Partial attempt:

Let $B_1 := \bigcap_{i=1}^k A_i^c$ and $B_2 := A_{k+1}^c$.

\begin{align} P\left(\bigcap_{i=1}^{k+1} A_i^c\right) &= P(B_1 \cap B_2) \\ &= 1 - P(B_1^c \cup B_2^c) \\ &= 1 - P(B_1^c) - P(B_2^c) + P(B_1^c \cap B_2^c) \\ &\overset{*}{=} 1 - P(B_1^c) - P(B_2^c) + P(B_1^c) P(B_2^c) \\ &= P(B_1) P(B_2) \\ &= P\left(\bigcap_{i=1}^{k} A_i^c\right) P(A_{k+1}^c) \\ &= \prod_{i=1}^{k+1} P(A_i^c). \end{align}

It remains to verify the starred equality $P(B_1^c \cap B_2^c) = P(B_1^c) P(B_2^c)$.

\begin{align} P(B_1^c \cap B_2^c) &= P\left( \left(\bigcap_{i=1}^k A_i^c\right)^c \cap A_{k+1} \right) \\ &= P\left( \left(\bigcup_{i=1}^k A_i\right) \cap A_{k+1} \right) \\ &\overset{?}{=} P\left(\bigcup_{i=1}^k A_i \right) P(A_{k+1}) \\ &= P(B_1^c) P(B_2^c). \end{align}

For the "?" equality, I think you can show this by writing $\bigcup_{i=1}^k A_i$ as the disjoint union of intersections of $A_1, \ldots, A_k, A_1^c, \ldots, A_k^c$.

Answer 2

Hi Inquirer: My apologies for the delay. I've had a lot going on so I just got back to this today.

I figured I'd put it in an answer since you get more space that way.

So, you agree that you proved the statement for $i = 2$ case, right. You showed that if $A_1$ and $A_2$ are independent, then $A^{c}_{1}$ and $A^{c}_{2}$ are independent. So, what induction says is that, if we have proven the statement for $i = 2$ and then we can show that, it being true for $i = 2$, implies that it is also true for the $i = 3$ case, then we are done with the proof by an induction argument.

So, let us assume that $A_{1}, A_{2}$ and $A_3$ are independent. We want to show that, given that the statement is true for $A_{1}$ and $A_{2}$, then this implies that $A^{c}_{1}, A^{c}_{2}$ and $A^{c}_{3}$ are independent.

So, what we can do for clarity is let $k = 2$ and then use the same proof that you used in your question.

\begin{align} P\left( {\bigcap_{i = 1}^{2 + 1} A_i}\right) &= P\left( \bigcap_{i=1}^{2}A_i \cap A_{2+1} \right) \\ &= \prod_{i=1}^{2}P(A_i) \cdot P(A_{2+1})\\ &= P\left(\bigcap_{i=1}^{2}A_i\right) \cdot P(A_{2+1}) \\ &= P\left(\bigcap_{i=1}^{2}A_i\right) \cdot P(A_{3}) \end{align}

Next we can use the trick of letting $A_{1}^{*} = \bigcap_{i=1}^{2}A_i$ and letting $A_{2}^{*} = A_{3}$.

So, what we have shown above is that that $A^{*}_1$ and $A^{*}_2$ are independent. But we then know that (because it's already been proven for $i = 2$) that $A^{c*}_{1}$ and $A^{c*}_{2}$ are independent. But $A^{c*}_{1} = \bigcap_{i=1}^{2}A^{c}_i$ and $A^{*c}_{2} = A^{c}_{3}$ so this means that $A^{c}_{1}, A^{c}_{2}$ and $A^{c}_{3}$ are independent. This is because $\left(\bigcap_{i=1}^{2}A^{c}_i\right) \bigcap A^{c}_{3} = \bigcap_{i=1}^{3}A^{c}_i$.

So, what we have shown is that the $i = 2$ case being true implies that the $i = 3$ case is also true so, since this argument holds for $k = 2$, it implies that it also holds for any value of $k$. So we have proved the statement using induction. Does that make sense ? If not, let me know what part doesn't and I'll try to explain it more clearly.