description/proof of that for probability space and independent indexed set of events, for finite indexed subset of indexed set, \(1\) minus probability of union of indexed subset is product of \(1\) minus probabilities of elements of indexed subset
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of independent indexed set of events of probability space.
- The reader admits the proposition that for any measure space and any \(2\) measurable subsets, the measure of the union of the subsets is the sum of the measures of the subsets minus the measure of the intersection of the subsets.
- The reader admits the proposition that for any probability space and any independent indexed set of events, the indexed set of events by taking the union of some finite elements is independent.
Target Context
- The reader will have a description and a proof of the proposition that for any probability space and any independent indexed set of events, for any finite indexed subset of the indexed set, \(1\) minus the probability of the union of the indexed subset is the product of \(1\) minus the probabilities of the elements of the indexed subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((M, A, \mu)\): \(\in \{\text{ the probability spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S\): \(= \{a_j \in A\}_{j \in J}\), \(\in \{\text{ the independent indexed sets of events }\}\)
\(J^`\): \(\in \{\text{ the finite subsets of } J\}\)
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Statements:
\(1 - \mu (\cup_{j \in J^`} a_j) = \prod_{j \in J^`} (1 - \mu (a_j))\)
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2: Proof
Whole Strategy: prove it inductively with respect to \(\vert J^` \vert\); Step 1: prove it for when \(\vert J^` \vert = 2\); Step 2: suppose that it holds when \(\vert J^` \vert \in \{1, ..., n' - 1\}\), and see that it holds when \(\vert J^` \vert = n'\); Step 3: conclude the proposition.
Step 1:
When \(\vert J^` \vert = 1\), it holds obviously.
Let us suppose that \(\vert J^` \vert = 2\).
Let \(J^` = \{j_1, j_2\}\).
\(1 - \mu (a_{j_1} \cup a_{j_2}) = 1 - (\mu (a_{j_1}) + \mu (a_{j_2}) - \mu (a_{j_1} \cap a_{j_2}))\), by the proposition that for any measure space and any \(2\) measurable subsets, the measure of the union of the subsets is the sum of the measures of the subsets minus the measure of the intersection of the subsets.
\(= 1 - \mu (a_{j_1}) - \mu (a_{j_2}) + \mu (a_{j_1}) \mu (a_{j_2})\), because \(S\) is independent.
\(= 1 - \mu (a_{j_1}) - \mu (a_{j_2}) (1 - \mu (a_{j_1})) = (1 - \mu (a_{j_1})) (1 - \mu (a_{j_2}))\).
So, it holds when \(\vert J^` \vert = 2\).
Step 2:
Let us suppose that it holds when \(\vert J^` \vert \in \{1, ..., n' - 1\}\), where \(3 \le n'\).
Let \(\vert J^` \vert = n'\).
Let \(J^` = \{j_1, ..., j_{n'}\}\).
\(1 - \mu (a_{j_1} \cup ... \cup a_{j_{n'}}) = 1 - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}} \cup a_{j_{n'}}) = 1 - (\mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) + \mu (a_{j_{n'}}) - \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n'}}))\), by the proposition that for any measure space and any \(2\) measurable subsets, the measure of the union of the subsets is the sum of the measures of the subsets minus the measure of the intersection of the subsets.
\(\mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n'}}) = \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \mu (a_{j_{n'}})\), by the proposition that for any probability space and any independent indexed set of events, the indexed set of events by taking the union of some finite elements is independent.
So, \(= 1 - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) - \mu (a_{j_{n'}}) + \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \mu (a_{j_{n'}}) = 1 - \mu (a_{j_{n'}}) - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) (1 - \mu (a_{j_{n'}})) = (1 - \mu (a_{j_{n'}})) (1 - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}})) = (1 - \mu (a_{j_{n'}})) (1 - \mu (a_{j_1})) ... (1 - \mu (a_{j_{n' - 1}}))\), by the induction hypothesis, \(= (1 - \mu (a_{j_1})) ... (1 - \mu (a_{j_{n'}}))\).
Step 3:
So, by the induction principle, the proposition holds for each \(\vert J^` \vert \in \mathbb{N} \setminus \{0\}\).
