Probability Exercise: If it must happen once it happens infinitely often.

Standard

Exercise: Let \{A_n\} be a sequence of independent events with \mathbb{P}(A_n) < 1 for all n. Show that \mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(A_n \text{ i.o. }) = 1.

Solution:

\mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(\bigcap_n A_n^{c}) = 0 which by independence means that \prod_n\mathbb{P}(A_n^c) = 0. Now since \mathbb{P}(A_n^c) > 0 \; \forall n, we know that for any M, \prod_{n \geq M} \mathbb{P}(A_n^c) = 0, which means that \mathbb{P}(\bigcap_{n \geq M} A_n^c) = 0. Now suppose \textit{w} \in \{A_n\} f.o. Then for sufficiently large M, \textit{w} \in \bigcap_{n \geq M} A_n^{c}. Thus the set \{A_n\} f.o., can be written as W = \bigcup_{M} \bigcap_{n \geq M} A_n^{c}, which is a countable union of sets of measure 0 by the above reasoning. Thus \mathbb{P}(W) = 0 \implies \mathbb{P}(A_n \text{ i.o. }) = 1.

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