Quest for a non-inductive proof of the addition theorem of probability.
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The addition theorem of probability states that:
$$P(A_1cup A_2cup … cup A_n) = sum P(A_i) - sum P(A_icap A_j) + sum P(A_icap A_jcap A_k) - … + (-1)^nP(A_1cap A_2cap … cap A_n)$$
Although this result is proved using the method of induction (as in the case of most textbooks), can it be proved directly?
The only thing which I see is its structural similarity with the general product of binomials :
$$(1+alpha)(1+beta)……(1+nu) = 1 + (alpha + beta + … + nu) + (alpha beta + ……) + ……… + (alpha beta gamma …nu)$$
However I am unable to get a link between the two. Any help is appreciated.
probability
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
$endgroup$
add a comment |
$begingroup$
The addition theorem of probability states that:
$$P(A_1cup A_2cup … cup A_n) = sum P(A_i) - sum P(A_icap A_j) + sum P(A_icap A_jcap A_k) - … + (-1)^nP(A_1cap A_2cap … cap A_n)$$
Although this result is proved using the method of induction (as in the case of most textbooks), can it be proved directly?
The only thing which I see is its structural similarity with the general product of binomials :
$$(1+alpha)(1+beta)……(1+nu) = 1 + (alpha + beta + … + nu) + (alpha beta + ……) + ……… + (alpha beta gamma …nu)$$
However I am unable to get a link between the two. Any help is appreciated.
probability
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
$endgroup$
$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
$begingroup$
The addition theorem of probability states that:
$$P(A_1cup A_2cup … cup A_n) = sum P(A_i) - sum P(A_icap A_j) + sum P(A_icap A_jcap A_k) - … + (-1)^nP(A_1cap A_2cap … cap A_n)$$
Although this result is proved using the method of induction (as in the case of most textbooks), can it be proved directly?
The only thing which I see is its structural similarity with the general product of binomials :
$$(1+alpha)(1+beta)……(1+nu) = 1 + (alpha + beta + … + nu) + (alpha beta + ……) + ……… + (alpha beta gamma …nu)$$
However I am unable to get a link between the two. Any help is appreciated.
probability
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
$endgroup$
The addition theorem of probability states that:
$$P(A_1cup A_2cup … cup A_n) = sum P(A_i) - sum P(A_icap A_j) + sum P(A_icap A_jcap A_k) - … + (-1)^nP(A_1cap A_2cap … cap A_n)$$
Although this result is proved using the method of induction (as in the case of most textbooks), can it be proved directly?
The only thing which I see is its structural similarity with the general product of binomials :
$$(1+alpha)(1+beta)……(1+nu) = 1 + (alpha + beta + … + nu) + (alpha beta + ……) + ……… + (alpha beta gamma …nu)$$
However I am unable to get a link between the two. Any help is appreciated.
probability
probability
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
edited Nov 16 '18 at 13:15
Asaf Karagila♦
307k33440773
307k33440773
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
asked Nov 16 '18 at 12:42
Awe Kumar JhaAwe Kumar Jha
594113
594113
$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20
$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your very statement:
$$ P(A1∪A2∪…∪An)=∑P(Ai)−∑P(Ai∩Aj)+∑P(Ai∩Aj∩Ak)−…+(−1)nP(A1∩A2∩…∩An) $$
is what we arrive at from what we call the Inclusion-Exclusion Principle.
Since you are looking for a non-induction based proof, there exists a proof using the Binomial Theorem here and an algebraic proof using indicator functions over here.
answered Nov 16 '18 at 13:34
TuskyTusky
664618
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Your very statement:
$$ P(A1∪A2∪…∪An)=∑P(Ai)−∑P(Ai∩Aj)+∑P(Ai∩Aj∩Ak)−…+(−1)nP(A1∩A2∩…∩An) $$
is what we arrive at from what we call the Inclusion-Exclusion Principle.
Since you are looking for a non-induction based proof, there exists a proof using the Binomial Theorem here and an algebraic proof using indicator functions over here.
answered Nov 16 '18 at 13:34
TuskyTusky
664618
$endgroup$
add a comment |
$begingroup$
Your very statement:
$$ P(A1∪A2∪…∪An)=∑P(Ai)−∑P(Ai∩Aj)+∑P(Ai∩Aj∩Ak)−…+(−1)nP(A1∩A2∩…∩An) $$
is what we arrive at from what we call the Inclusion-Exclusion Principle.
Since you are looking for a non-induction based proof, there exists a proof using the Binomial Theorem here and an algebraic proof using indicator functions over here.
answered Nov 16 '18 at 13:34
TuskyTusky
664618
$endgroup$
add a comment |
$begingroup$
Your very statement:
$$ P(A1∪A2∪…∪An)=∑P(Ai)−∑P(Ai∩Aj)+∑P(Ai∩Aj∩Ak)−…+(−1)nP(A1∩A2∩…∩An) $$
is what we arrive at from what we call the Inclusion-Exclusion Principle.
Since you are looking for a non-induction based proof, there exists a proof using the Binomial Theorem here and an algebraic proof using indicator functions over here.
answered Nov 16 '18 at 13:34
TuskyTusky
664618
$endgroup$
Your very statement:
$$ P(A1∪A2∪…∪An)=∑P(Ai)−∑P(Ai∩Aj)+∑P(Ai∩Aj∩Ak)−…+(−1)nP(A1∩A2∩…∩An) $$
is what we arrive at from what we call the Inclusion-Exclusion Principle.
Since you are looking for a non-induction based proof, there exists a proof using the Binomial Theorem here and an algebraic proof using indicator functions over here.
answered Nov 16 '18 at 13:34
TuskyTusky
664618
edited Jan 3 at 14:25
answered Nov 16 '18 at 13:34
TuskyTusky
664618
answered Nov 16 '18 at 13:34
TuskyTusky
664618
answered Nov 16 '18 at 13:34
TuskyTusky
664618
664618
add a comment |
add a comment |
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$begingroup$
You could decompose the union into disjoint sets. For instance, $A cup B = (A cap B) cup (A cap B^c) cup (A^c cap B)$. Now rearrange $P(A) = P(A cap B) + P(A cap B^c)$.
$endgroup$
– Stockfish
Nov 16 '18 at 12:46
$begingroup$
Of course, that's not really an answer to your question ...
$endgroup$
– Stockfish
Nov 16 '18 at 12:53
$begingroup$
@Stockfish, I think it may involve polynomial fields and other concepts of linear algebra, when we are looking for the decomposition of the general union operator and further rearrangement.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 12:58
$begingroup$
@Tusky, Ah, that was very helpful.
$endgroup$
– Awe Kumar Jha
Nov 16 '18 at 13:20