description/proof of that for probability space and independent indexed set of events, indexed set of events by taking union of some finite elements is independent
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 set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- 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.
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, the indexed set of events by taking the union of some finite elements is independent.
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:
\((M, A, \mu)\): \(\in \{\text{ the probability spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S\): \(= \{a_j \in A\}_{j \in J}\), \(\in \{\text{ the independent indexed sets of events }\}\)
\(J^`\): \(\in \{\text{ the finite subsets of } J\}\)
\(\widetilde{S}\): \(= \{\cup_{j \in J^`} a_j\} \cup \{a_j \in S \vert j \in J \setminus J^`\}\)
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Statements:
\(\widetilde{S} \in \{\text{ the independent indexed sets of events }\}\)
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2: Note
By applying this proposition iteratively, \(\{\cup_{j \in J_1} a_j, ..., \cup_{j \in J_n} a_j\} \cup \{a_j \in S \vert j \in J \setminus (J_1 \cup ... \cup J_n)\}\), where \(\{J_l \subseteq J \vert l \in \{1, ..., n\}\}\) is any set of some disjoint (being disjoint is crucial) finite subsets, is independent: as \(\{\cup_{j \in J_1} a_j\} \cup \{a_j \in S \vert j \in J \setminus (J_1)\}\) is independent, \(\{\cup_{j \in J_1} a_j, \cup_{j \in J_2} a_j\} \cup \{a_j \in S \vert j \in (J \setminus (J_1)) \setminus J_2 = J \setminus (J_1 \cup J_2)\}\) is independent, and so on.
3: 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 prove it for when \(\vert J^` \vert = n'\).
Step 1:
It holds when \(\vert J^` \vert = 1\), obviously, because \(\widetilde{S} = S\).
Let us suppose that \(\vert J^` \vert = 2\).
Let \(J^` = \{j_1, j_2\}\).
The index set of \(\widetilde{S}\) is \(\{J^`\} \cup (J \setminus J^`)\).
Any finite subset of the index set may or may not contain \(J^`\).
When the finite subset does not contain \(J^`\), the condition for being independent is satisfied, because it is a finite subset of \(S\).
Let us suppose that the finite subset contains \(J^`\).
Let the finite subset be \(\{J^`, j_3, ..., j_m\}\).
We are going to see that \(\mu ((a_{j_1} \cup a_{j_2}) \cap a_{j_3} \cap ... \cap a_{j_m}) = \mu (a_{j_1} \cup a_{j_2}) \mu (a_{j_3}) ... \mu (a_{j_m})\).
\(\mu ((a_{j_1} \cup a_{j_2}) \cap a_{j_3} \cap ... \cap a_{j_m}) = \mu ((a_{j_1} \cap a_{j_3} \cap ... \cap a_{j_m}) \cup (a_{j_2} \cap a_{j_3} \cap ... \cap a_{j_m}))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(= \mu (a_{j_1} \cap a_{j_3} \cap ... \cap a_{j_m}) + \mu (a_{j_2} \cap a_{j_3} \cap ... \cap a_{j_m}) - \mu ((a_{j_1} \cap a_{j_3} \cap ... \cap a_{j_m}) \cap (a_{j_2} \cap a_{j_3} \cap ... \cap a_{j_m}))\), 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, \(= \mu (a_{j_1} \cap a_{j_3} \cap ... \cap a_{j_m}) + \mu (a_{j_2} \cap a_{j_3} \cap ... \cap a_{j_m}) - \mu (a_{j_1} \cap a_{j_2} \cap a_{j_3} \cap ... \cap a_{j_m})\).
\(= \mu (a_{j_1}) \mu (a_{j_3}) ... \mu (a_{j_m}) + \mu (a_{j_2}) \mu (a_{j_3}) ... \mu (a_{j_m}) - \mu (a_{j_1}) \mu (a_{j_2}) \mu (a_{j_3}) ... \mu (a_{j_m})\), because \(S\) is independent.
\(= \mu (a_{j_3}) ... \mu (a_{j_m}) (\mu (a_{j_1}) + \mu (a_{j_2}) - \mu (a_{j_1}) \mu (a_{j_2})) = \mu (a_{j_3}) ... \mu (a_{j_m}) (\mu (a_{j_1}) + \mu (a_{j_2}) - \mu (a_{j_1} \cap a_{j_2}))\), because \(S\) is independent, \(= \mu (a_{j_3}) ... \mu (a_{j_m}) \mu (a_{j_1} \cup a_{j_2})\), 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, \(= \mu (a_{j_1} \cup a_{j_2}) \mu (a_{j_3}) ... \mu (a_{j_m})\).
So, \(\widetilde{S}\) is independent.
Step 2:
Let us suppose that it holds when \(\vert J^` \vert \in \{1, ..., n' - 1\}\) where \(3 \le n'\).
Let us suppose that \(\vert J^` \vert = n'\).
Let \(J^` = \{j_1, ..., j_{n'}\}\).
The index set of \(\widetilde{S}\) is \(\{J^`\} \cup (J \setminus J^`)\).
Any finite subset of the index set may or may not contain \(J^`\).
When the finite subset does not contain \(J^`\), the condition for being independent is satisfied, because it is a finite subset of \(S\).
Let us suppose that the finite subset contains \(J^`\).
Let the finite subset be \(\{J^`, j_{n' + 1}, ..., j_m\}\).
We are going to see that \(\mu ((a_{j_1} \cup ... \cup a_{j_{n'}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) = \mu (a_{j_1} \cup ... \cup a_{j_{n'}}) \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m})\).
\(\mu ((a_{j_1} \cup ... \cup a_{j_{n'}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) = \mu (((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) \cup (a_{j_{n'}} \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(= \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) + \mu (a_{j_{n'}} \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) - \mu (((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) \cap (a_{j_{n'}} \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}))\), 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, \(= \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) + \mu (a_{j_{n'}} \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m}) - \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n'}} \cap a_{j_{n' + 1}} \cap ... \cap a_{j_m})\).
\(= \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) + \mu (a_{j_{n'}}) \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \mu (a_{j_{n'}}) \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m})\), by the induction hypothesis.
\(= \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) (\mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) + \mu (a_{j_{n'}}) - \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \mu (a_{j_{n'}})) = \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) (\mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}}) + \mu (a_{j_{n'}}) - \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cap a_{j_{n'}}))\), by the induction hypothesis, \(= \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) \mu ((a_{j_1} \cup ... \cup a_{j_{n' - 1}}) \cup a_{j_{n'}})\), 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, \(= \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m}) \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}} \cup a_{j_{n'}}) = \mu (a_{j_1} \cup ... \cup a_{j_{n' - 1}} \cup a_{j_{n'}}) \mu (a_{j_{n' + 1}}) ... \mu (a_{j_m})\).
So, \(\widetilde{S}\) is independent.
Step 3:
So, by the induction principle, the proposition holds for each \(\vert J^` \vert \in \mathbb{N} \setminus \{0\}\).
