Say I have a set $\{E_1, ... ,E_n\}$ of independent events of a sample space $\Omega$. I want to show independence of any partition of $\{E_1, ... ,E_n\}$, by which I mean a set $\{P_1,...,P_m\}, 2\leq m \leq n $, with $P_j = \bigcup\limits_{i \in I_k} E_i , I_k \subset \{1,...,n\} $ and no two $E_i$ appear in the same union and no $E_i$ appears in more than one (a.k.a each $E_i$ appears in exactly one union, or $I_k \cap I_j = \emptyset,\cup I_l = \{1,...,n\}$). Now I want to show that the $P_j$ are independent events.

The intuition here is that if knowledge of any event in $\{E_1, ... ,E_n\}$ or any intersection thereof happening won't tell me anything about the probability of any other event in $\{E_1, ... ,E_n\}$ happening, then likewise knowledge of any union of events in $\{E_1, ... ,E_n\}$ or intersections of these unions won't tell me anything about the probability of another union of $\{E_1, ... ,E_n\}$.

My best guess is to show that the complements of the $P_j$ are independent. That's not hard to show, in fact, since $P_j^c = (\cup E_i)^c= \cap E_i^c$ and the $\cap E_i^c$ are independent. Am I on the right track here?

  • $\begingroup$ You are on a good track. Do you want to show that the $P_{j}$'s are independent for every pair or that they are jointly independent? $\endgroup$
    – madprob
    Commented Nov 7, 2017 at 19:42
  • $\begingroup$ jointly independent, not just pairwise $\endgroup$
    – ghthorpe
    Commented Nov 8, 2017 at 5:58

1 Answer 1


We want to show that, for any choice of $ P_i = E_{i_1} \cup \cdots \cup E_{i_n} $ and $P_j = E_{j_1} \cup \cdots \cup E_{j_m}$, that

$\mathbb{P}(P_i \cap P_j) = \mathbb{P}((E_{i_1} \cup \cdots \cup E_{i_n}) \cap (E_{j_1} \cup \cdots \cup E_{j_m})) = \mathbb{P}(P_i)\mathbb{P}(P_j) $

From set theory,

$$ (E_{i_1} \cup \cdots \cup E_{i_n}) \cap (E_{j_1} \cup \cdots \cup E_{j_m}) = (E_{i_1} \cap E_{j_1}) \cup (E_{i_1} \cap E_{j_2}) \cup \cdots \cup (E_{i_n} \cap E_{j_m})$$ $$ = \bigcup_{k=1}^n \bigcup_{a=1}^m (E_{i_k} \cap E_{j_a}) $$

$$ \mathbb{P}(P_i \cap P_j) = \mathbb{P}\left( \bigcup_{k=1}^n \bigcup_{a=1}^m (E_{i_k} \cap E_{j_a}) \right) $$

Now use the fact that $ \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B) $, and further that $ \mathbb{P}(A \cup B \cup C) = \mathbb{P}(A) + \mathbb{P}(B) + \mathbb{P}(C) - \mathbb{P}(A \cap B) - \mathbb{P}(A \cap C) - \mathbb{P}(B \cap C) + \mathbb{P}(A \cap B \cap C)$

This gives $ \mathbb{P}(P_i \cap P_j) $ as a double sum subtract two triple sums plus a quadruple sum.

Observe that $$ \mathbb{P}(P_i) = \mathbb{P}(E_{i_1} \cup \cdots \cup E_{i_n}) = \sum_{k=1}^n \mathbb{P}(E_{i_k}) - \sum_{k=1}^n \sum_{l=1}^{k-1} \mathbb{P}(E_{i_k})\mathbb{P}(E_{i_l}) $$

(using the independence of the $E$ events) and taking the product of two of these (with different indices) gets exactly the expression for $ \mathbb{P}(P_i \cap P_j) $.

  • $\begingroup$ This is independence of two events. The question asks for the independence of the whole family of events $(P_m)$. $\endgroup$
    – Did
    Commented Nov 7, 2017 at 22:06
  • $\begingroup$ could it be used as an induction base case for the full proof though? $\endgroup$
    – ghthorpe
    Commented Nov 8, 2017 at 6:00

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