# Fully independent events and their complements

Suppose events $A_1, ..., A_n$ are fully independent, i.e., $P(A_1 \cap ... \cap A_k) = P(A_1)...P(A_k)$ for all $k$ between 2 and $n$. Does this mean that the complementary events are also fully independent: $P(A_1^c \cap ... \cap A_k^c) = P(A_1^c)...P(A_k^c)$ for all k?

I know this holds if $k = 2$, but I want to know in general.

I've tried to prove it by induction but it looks like hard work...

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What is wrong with hard work? Just remember that $P(A^C)=1-P(A)$ and everything should be straightforward, and just a bit tedious. – deinst Oct 15 '11 at 19:29
The definition that you give for "fully independent" is not quite the right one, we want the probability of the intersection of any finite subcollection to be the appropriate product. The definition given gets us into trouble if $P(A_i)=0$ for some early $i$. Then it is not hard to construct a counterexample for the complements question. But with the right definition the answer to the complements question is positive, by induction or inclusion/exclusion. – André Nicolas Oct 15 '11 at 19:42
Thanks for responses. – Court Oct 15 '11 at 22:21
If there is no formal answer by tomorrow, I will produce one. – André Nicolas Oct 16 '11 at 0:40
As noted by @AndréNicolas, your definition of fully independent is incorrect. But there is an alternative definition for independence of a finite number of events: $A_i, 1 \leq i \leq n$ are independent events if and only if the following $2^n$ equalities hold:$$P(A_1^*A_2^*\cdots A_n^*) = P(A_1^*)P(A_2)^*\cdots P(A_n^*)$$ where $A_i^*$ denotes either $A_i$ or $A_i^c$ (the same on both sides of the equation). The $2^n$ equations thus correspond to the $2^n$ choices for $*$. The standard definition follows upon adding sets of these equations and using $P(A) + P(A^c) = 1$ to simplify. – Dilip Sarwate Oct 16 '11 at 2:42

We show that under the non-standard definition of fully independent events given in the post, the desired result is not true. We then give a standard definition of fully independent events, and show that under this definition the desired result is true.

A counterexample: We toss a fair coin. Assume that the possible events are $A_1$, the coin rolls around forever (probability $0$), $A_2$, we get a head (probability $1/2$) and $A_3$, we get a tail (probability $1/2$). It is easy to verify that under the definition of fully independent given in the post, the sequence $A_1, A_2, A_3$ is fully independent. But $A_2^c$ and $A_3^c$ are not independent, for $P(A_2^c\cap A_3^c)=0$, but $P(A_2^c)P(A_3^c)=1/4$. It is also easy to verify that the sequence $A_1^c, A_2^c, A_3^c$ is not fully independent.

A proof: We first give a standard definition of full independence. The events $A_1,A_2,\dots, A_n$ are fully independent if, whenever $B_1, B_2, \dots B_k$ are distinct $A_i$, $$P(B_1\cap B_2 \cap \cdots \cap B_k)=P(B_1)P(B_2)\cdots P(B_k).$$ We show that if $A_1, A_2, \dots, A_n$ are fully independent, then so are $A_1^c,A_2^c,\dots, A_n^c$.

There is a not difficult proof by induction. However, we prefer to avoid formal induction, in order to get a proof that has more symmetry. We need to prove that if $B_1, B_2, \dots B_k$ are distinct $A_i$, then $$P(B_1^c\cap B_2^c \cap \cdots \cap B_k^c)=P(B_1^c)P(B_2^c)\cdots P(B_k^c).$$

To save space, let $b_i=P(B_i)$. So we want to prove that $$P(B_1^c\cap B_2^c \cap \cdots \cap B_k^c)=(1-b_1)(1-b_2)\cdots (1-b_k).$$

Let $p$ be the probability on the left. Then $$1-p=P(B_1\cup B_2 \cup \cdots \cup B_k).$$ Thus, by the Principle of Inclusion/Exclusion, $$1-p=\sum_{i=1}^k b_i -\sum_{1 \le i <j}b_ib_j+\sum_{1 \le i <j<k}b_ib_jb_k-\cdots$$ and therefore $$p=1 -\sum_{i=1}^k b_i +\sum_{1 \le i <j}b_ib_j-\sum_{1 \le i <j<k}b_ib_jb_k+\cdots.$$ The right-hand side is just $(1-b_1)(1-b_2)\cdots (1-b_k)$. This completes the proof.

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thanks very much for this. I recognise the "Inclusion/Exclusion" principle as the Bonferonni inequalities. – Court Oct 17 '11 at 18:13
@André Nicolas: How can you say on the last sentence, that the RHS is just that product? How does one prove it? – An old man in the sea. Jun 24 '14 at 11:33

As noted by others, what you write is NOT the definition of the independence of $(A_1,A_2,\ldots,A_n)$. You should ask that $\mathrm P(A_{k_1}\cap A_{k_2}\cap\cdots \cap A_{k_i})=\mathrm P(A_{k_1})\mathrm P(A_{k_2})\cdots \mathrm P(A_{k_i})$ for every choice of all distinct indices $k_j$ in $\{1,2,\ldots,n\}$.

This point set aside, let me mention that a strategy which might help avoid some tediousness in such a context is to translate everything in terms of random variables. I know, random variables are supposed to be always more complicated than events but in fact, the opposite holds quite often (did somebody just say linearity?), and the present question is a good example of the phenomenon.

We first make the, seemingly odd, general remark that, for every event $A$, $$\mathrm P(A)=\int_\Omega\mathbf 1_A\mathrm dP=\mathrm E(\mathbf 1_A),$$ where $\mathbf 1_A$ denotes the indicator function of $A$ defined by $\mathbf 1_A(\omega)=1$ if $\omega\in A$ and $\mathbf 1_A(\omega)=0$ if $\omega\in\Omega\setminus A$.

Turning to the question, let us choose $k$ events from the $n$ events $A_1$, $A_2$, ..., $A_n$, all different, rename them as $B_1$, $B_2$, ..., $B_k$, and introduce $B=\bigcap\limits_{i=1}^k(B_i)^c$. One knows that the indicator function of a complement is $1$ minus the original indicator function and that the indicatior function of an intersection is the product of the indicator functions, hence $$\mathbf 1_B=\prod\limits_{i=1}^k(1-\mathbf 1_{B_i})=Q_k(\mathbf 1_{B_1},\mathbf 1_{B_2},\ldots,\mathbf 1_{B_k}),$$ where $Q_k$ denotes the polynomial $$Q_k(x_1,x_2,\ldots,x_k)=\prod\limits_{i=1}^k(1-x_i).$$ Like every polynomial, $Q_k(x_1,x_2,\ldots,x_k)$ may be expanded into a sum of monomials in the unknowns $x_1$, $x_2$, ..., $x_k$. Since the (partial) degree of $Q_k$ in each $x_i$ is $1$ the expansion of $Q_k$ involves only monomials of the form $x_{i_1}x_{i_2}\cdots x_{i_\ell}$ for some distinct indices $i_j$. In other words, $$Q_k(x_1,x_2,\ldots,x_k)=\sum_Iq_I\prod_{i\in I}x_i,$$ where the sum runs over the $2^k$ subsets $I$ of $\{1,2,\ldots,k\}$, for some coefficients $q_I$ whose values will not be relevant. (The interested reader might note however that $q_\varnothing=1$ and $q_{\{1,2,\ldots,k\}}=(-1)^k$, and the motivated one might show that $q_I=(-1)^{|I|}$ for every $I\subseteq\{1,2,\ldots,k\}$.)

We stress that this relation holds between polynomials, hence every choice of the variables $x_i$ yields an equality, whether these are numbers or functions. In particular, evaluating both sides at the functions $\mathbf 1_{B_i}$ yields $$\mathbf 1_B=\sum\limits_Iq_I\prod\limits_{i\in I}\mathbf 1_{B_i}.$$ For every $I$, note that $$\prod\limits_{i\in I}\mathbf 1_{B_i}=\mathbf 1_{B_I},\quad \mbox{where}\ B_I=\bigcap\limits_{i\in I}B_i,$$ and that the independence of the events $(B_i)_{i\in I}$ yields $$\mathrm E(1_{B_I})=\mathrm P(B_I)=\prod\limits_{i\in I}\mathrm P(B_i).$$ Summing this over every $I$ yields $$\mathrm P(B)=\mathrm E(\mathbf 1_B)=\sum\limits_Iq_I\mathrm P(B_I)=\sum\limits_Iq_I\prod\limits_{i\in I}\mathrm P(B_i)=Q_k(\mathrm P(B_1),\mathrm P(B_2),\ldots,\mathrm P(B_k)),$$ where the last equality stems from the very definition of $Q_k$ evaluated at the real numbers $\mathrm P(B_i)$. But one knows the value of $Q_k$ at every point, in particular at $(\mathrm P(B_1),\mathrm P(B_2),\ldots,\mathrm P(B_k))$, which is $$\mathrm P(B)=\prod\limits_{i=1}^k(1-\mathrm P(B_i))=\prod\limits_{i=1}^k\mathrm P((B_i)^c),$$ and the proof is over. To conclude, note once again that we used the polynomial $Q_k$ twice, once for functions and the other one for real numbers.

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thanks for this proof. It's interesting and pretty different to what I was thinking. – Court Oct 17 '11 at 18:14