description/proof of that for probability space, independent indexed set of events, and indexed set of complements of events, for finite subset of index set and element of 1st indexed set or 2nd indexed set for each index, measure of intersection of elements is product of measures of elements
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 knows the proposition that for any probability space and any independent indexed set of events, the indexed set of the complements of the events is independent.
- The reader admits the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set.
- The reader admits the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union 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.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any probability space, any independent indexed set of events, and the indexed set of the complements of the events, for any finite subset of the index set and any element of the 1st indexed set or the 2nd indexed set for each index of the subset, the measure of the intersection of the elements is the product of the measures of the elements.
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:
\((M, A, \mu)\): \(\in \{\text{ the probability spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S\): \(= \{a_j \in A\}_{j \in J}\), \(\in \{\text{ the independent indexed sets of events }\}\)
\(\widetilde{S}\): \(= \{M \setminus a_j\}_{j \in J}\), \(\in \{\text{ the indexed sets of events }\}\)
\(J^`\): \(\in \{\text{ the finite subsets of } J\}\)
\(\{b_j \in \{a_j, M \setminus a_j\}\}_{j \in J^`}\):
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Statements:
\(\mu (\cap_{j \in J^`} b_j) = \prod_{j \in J^`} \mu (b_j)\)
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2: Note
This proposition does not claim that "\((S \biguplus \widetilde{S})_{j \in \{1, 2\} \times J}\) is independent", which is not true in general: \(S \biguplus \widetilde{S}\) is indexed such that \((1, j)\) indexes \(a_j\) and \((2, j)\) indexes \(M \setminus a_j\).
That is because \(\mu (a_j \cap (M \setminus a_j)) = \mu (a_j) \mu (M \setminus a_j)\) does not hold in general: \(a_j \cap (M \setminus a_j) = \emptyset\), so, \(\mu (a_j \cap (M \setminus a_j)) = 0\), which does not equal \(\mu (a_j) \mu (M \setminus a_j)\) in general: it is crucial that \(b_l\) and \(b_m\) for each \(l \neq m\) are not of the same \(j \in J\).
3: Proof
Whole Strategy: Step 1: see that \(\widetilde{S}\) is independent; Step 2: deal with the case that none of \(b_j\) s is from \(\widetilde{S}\) and the case that all of \(b_j\) s are from \(\widetilde{S}\); Step 3: prove it for when only \(1\) of \(b_j\) s is from \(\widetilde{S}\); Step 4: prove it for the other cases; Step 5: conclude the proposition.
Step 1:
\(\widetilde{S}\) is independent, by the proposition that for any probability space and any independent indexed set of events, the indexed set of the complements of the events is independent.
Step 2:
When none of \(b_j\) s is from \(\widetilde{S}\), \(\mu (\cap_{j \in J^`} b_j) = \prod_{j \in J^`} \mu (b_j)\) holds, because it is directly from \(S\)'s being independent.
When all of \(b_j\) s are from \(\widetilde{S}\), \(\mu (\cap_{j \in J^`} b_j) = \prod_{j \in J^`} \mu (b_j)\) holds, because it is directly from \(\widetilde{S}\)'s being independent.
Step 3:
Let us suppose that only \(1\) of \(b_j\) s is from \(\widetilde{S}\).
Let \(J^` = \{j_1, ..., j_m\}\).
Let \(b_{j_m} = M \setminus a_{j_m}\), without loss of generality: the position of \(M \setminus a_j\) does not matter, because the operations of intersection and product are commutative.
\(\mu (\cap_{j \in J^`} b_j) = \mu (a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap (M \setminus a_{j_m})) = \mu ((a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap M) \setminus (a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap a_{j_m}))\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set.
\(= \mu ((a_{j_1} \cap ... \cap a_{j_{m - 1}}) \setminus (a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap a_{j_m})) = \mu (a_{j_1} \cap ... \cap a_{j_{m - 1}}) - \mu (a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap a_{j_m})\), because \(a_{j_1} \cap ... \cap a_{j_{m - 1}} \cap a_{j_m} \subseteq a_{j_1} \cap ... \cap a_{j_{m - 1}}\).
\(= \mu (a_{j_1}) ... \mu (a_{j_{m - 1}}) - \mu (a_{j_1}) ... \mu (a_{j_{m - 1}}) \mu (a_{j_m})\), because \(S\) is independent.
\(= \mu (a_{j_1}) ... \mu (a_{j_{m - 1}}) (1 - \mu (a_{j_m})) = \mu (a_{j_1}) ... \mu (a_{j_{m - 1}}) \mu (M \setminus a_{j_m})\).
\(= \prod_{j \in J^`} \mu (b_j)\).
So, \(\mu (\cap_{j \in J^`} b_j) = \prod_{j \in J^`} \mu (b_j)\) holds when only \(1\) of \(b_j\) s is from \(\widetilde{S}\).
Step 4:
Let \(J^` = \{j_1, ..., j_m\}\).
Let us suppose that \(b_{j_1} = a_{j_1}, ..., b_{j_n} = a_{j_n}, b_{j_{n + 1}} = M \setminus a_{j_{n + 1}}, ..., b_{j_m} = M \setminus a_{j_m}\), without loss of generality: the positions of \(M \setminus a_j\) s do not matter, because the operations of intersection and product are commutative.
\(\mu (\cap_{j \in J^`} b_j) = \mu (a_{j_1} \cap ... \cap a_{j_n} \cap (M \setminus a_{j_{n + 1}}) \cap ... \cap (M \setminus a_{j_m})) = \mu (a_{j_1} \cap ... \cap a_{j_n} \cap (M \setminus (a_{j_{n + 1}} \cup ... \cup a_{j_m})))\), by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets.
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 and Step 3, \(= \mu (a_{j_1}) ... \mu (a_{j_n}) \mu (M \setminus (a_{j_{n + 1}} \cup ... \cup a_{j_m})) = \mu (a_{j_1}) ... \mu (a_{j_n}) (1 - \mu (a_{j_{n + 1}} \cup ... \cup a_{j_m})) = \mu (a_{j_1}) ... \mu (a_{j_n}) (1 - \mu (a_{j_{n + 1}})) ... (1 - \mu (a_{j_m}))\), by 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.
\(= \mu (a_{j_1}) ... \mu (a_{j_n}) \mu (M \setminus a_{j_{n + 1}}) ... \mu (M \setminus a_{j_m})\).
\(= \prod_{j \in J^`} \mu (b_j)\).
Step 5:
So, in any case, \(\mu (\cap_{j \in J^`} b_j) = \prod_{j \in J^`} \mu (b_j)\) holds.
