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I want to prove that the events $A_{1}, \dots, A_{n}$ are independent iff

$$P(B_{1}\cap \dots \cap B_{n})=P(B_{1}) \cap P(B_{2}) \dots P(B_{n})$$

for $B_{i}=A_{i}$ or $B_{i}=A_{i}^{c}$

Well, I feel that for the case where $B_{i}=A_{i}$ is something obvious by definition, what would it take to put it in the proof, while for the other case, let $A_{1}, \dots, A_{n}$ independent events, note that:

\begin{align*} P(A_{1}^{c} \cap \cdots \cap A_{n}^{c}) &= 1-P(A_{1} \cup \cdots \cup A_{n}) \\ &= 1-[P(A_{1})+ \cdots-P(A_{1} \cap \cdots \cap A_{n} )] \\ &= 1-P(A_{1})- \cdots+P(A_{1} \cap \cdots \cap A_{n} ) \\ &=(1-P(A_{1}))(1-P(A_{2})) \cdots (1-P(A_{n})) \end{align*}

And from the last equality one would have, is this a valid reasoning?

The reverse side, I don't know how to give an argument for the first case, while for the second we could assume that $A_{1}^{c}, \dots, A_{n}^{c}$ are independent events, and

\begin{align*} P(A_{1} \cap \cdots \cap A_{n}) &= P(A_{1}) + \cdots +P(A_{n}) - P(A_{1} \cup \cdots \cup A_{n}) \\ &=P(A_{1}) + \cdots +P(A_{n}) -1+ P(A_{1}^{c}) \cdots P(A_{n}^{c}) \\ &= P(A_{1}) \cdots P(A_{1}) \end{align*}

Is this valid again?

It would be of great help to receive a suggestion for the case that was not possible for me, thank you.

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  • $\begingroup$ I can't follow your reasoning because your notation is opaque: what exactly is buried in all those "$\cdots$"? Regardless, all proofs proceed by logical deduction from a set of assumptions to results, invoking definitions and theorems along the way. For many people there's nothing to prove because your statement is the definition. Please, then, state what definition of independence you are working with. $\endgroup$
    – whuber
    Nov 28, 2023 at 4:20
  • $\begingroup$ math.stackexchange.com/questions/4499206/… $\endgroup$
    – Zhanxiong
    Nov 28, 2023 at 4:23
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    $\begingroup$ Why don't you use induction instead? $\endgroup$
    – Tran Khanh
    Nov 28, 2023 at 5:19

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