Quest for a non-inductive proof of the addition theorem of probability.
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♦
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
add a comment |
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♦
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
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♦
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
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♦
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
edited Nov 16 '18 at 13:15
Asaf Karagila♦
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
edited Nov 16 '18 at 13:15
Asaf Karagila♦
302k32426756
edited Nov 16 '18 at 13:15
Asaf Karagila♦
302k32426756
edited Nov 16 '18 at 13:15
Asaf Karagila♦
302k32426756
302k32426756
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
asked Nov 16 '18 at 12:42
Awe Kumar Jha
38613
38613
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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20
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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20
add a comment |
1 Answer
1
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oldest
votes
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
Tusky
644618
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
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
Tusky
644618
add a comment |
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
Tusky
644618
add a comment |
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
Tusky
644618
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
Tusky
644618
edited yesterday
answered Nov 16 '18 at 13:34
Tusky
644618
answered Nov 16 '18 at 13:34
Tusky
644618
answered Nov 16 '18 at 13:34
Tusky
644618
644618
add a comment |
add a comment |
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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 $P(A) = P(A cap B) + P(A cap B^c)$.
– Stockfish
Nov 16 '18 at 12:46
Of course, that's not really an answer to your question ...
– Stockfish
Nov 16 '18 at 12:53
@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.
– Awe Kumar Jha
Nov 16 '18 at 12:58
@Tusky, Ah, that was very helpful.
– Awe Kumar Jha
Nov 16 '18 at 13:20