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I was asked to prove the following:

If $\{ E_1,E_2,\ldots,E_n\}$ is a fully independent set of events, prove that:

$$ P\left(\bigcup_{i = 1}^n E_i \right) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] $$

(Hint: use complements)

I got the following:

Let $\alpha = E_1\cup E_2 \cup \cdots \cup E_n = \{E^C_1 \cap E^C_2 \cap \cdots\cap E_n^C\}^C$. Then, $\alpha^C = \{ E^C_1 \cap E^C_2 \cap \cdots \cap E^C\} $.

Assuming that the set of complements of each ${E_i}$ is also fully independent, we have:

\begin{align} & P(\alpha^C) = \prod_{i = 1}^n P(E_i^C) = \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ & P(\alpha) = 1 - P(\alpha^C) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ & P(\bigcup_{i = 1}^n E_i) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \end{align}

However, even though I know (by Googling it) that my assumption is true, I'm unable to prove it so I could validly use it as a lemma. Is there another way to prove this equality?

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    $\begingroup$ If you want a proof by induction, check out math.stackexchange.com/questions/393950/… $\endgroup$
    – iamwhoiam
    Commented Jun 30, 2017 at 2:33
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    $\begingroup$ @iamwhoiam Thanks! I could see that working too, but maybe there's a cleaner, shorter way (since this was a question from a previous year's midterm). EDIT: Actually, I just saw that his proof relies on the same (unproven) assumption that if a set of events is independent, then the set of complements of said events is also independent. $\endgroup$ Commented Jun 30, 2017 at 2:38
  • $\begingroup$ Use the inclusion-exclusion formula. $\endgroup$
    – zhoraster
    Commented Jun 30, 2017 at 2:47
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    $\begingroup$ And if you're interested in proving your lemma, you might want to look at math.stackexchange.com/questions/1192151/… $\endgroup$
    – iamwhoiam
    Commented Jun 30, 2017 at 3:05

1 Answer 1

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Here's a concise proof: If $E_1,\ldots, E_n$ are fully independent, then the indicator variables $1_{E_i}$ are fully independent, i.e. $E(f_j(1_{E_j})\ldots f_k(1_{E_k}))$ factors for any functions $f_i.$ Thus $$ P(E_j^c\cap\ldots\cap E_k^c) = E((1-1_{E_j})\ldots(1-1_{E_k})) = E(1-1_{E_j})\ldots E(1-1_{E_k}) =P(E_j^c)\ldots P(E_k^c)$$

But perhaps this is cheating since we can then wonder how we know that this is how independence should work for random variables. And then we ask ourselves 'what does it actually mean for two RVs to be independent?' and 'why do the events being independent mean their indicator variables are independent?' Then we realize we are essentially making the same leap of faith as we do when we assume the complements are independent.

But we can rescue ourselves by multiplying out the expression $(1-1_{E_j})\ldots(1-1_{E_k})$ before we take the expectation. And then by linearity the problem reduces to taking expectations of products of indicators. But these are just the probabilities of certain intersections of our independent events that we know factor. Then we can refactor the expression and get the result. A little reflection reveals that this is exactly the same calculation process as using inclusion-exclusion.

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