Mixing
Fourth axiom in Kolmogorov's probability axiom system?
$begingroup$
I'm reading Jaynes's "Probability Theory: The Logic of Science". In appendix A he contrasts the theory he develops in the first few chapters to the more conventional measure theory based system, A.K.A. Kolmogorov's. He lists four axioms used in this system. $F$ is a $sigma$-algebra (He calls it a "$sigma$-field")
(1) Normalization: $P(Omega) = 1$
(2) Non-negativity: $P(f_{i}) ≥ 0$ for all $f_{i}$ in $F$
(3) Additivity: if ${f_{1}ldots f_{n}}$ are disjoint elements of $F$ then
$P( f ) = sum_{i}P(f_{i})$, where $f =cup_{j}f_{j}$
(4) Continuity at zero: if a sequence $f_{1} supseteq f_{2} supseteq f_{3} supseteqldots $ tends to the empty set, then $P( f_{j} ) → 0$.
I remembered use of only the first three axioms, was the fourth once used and abandoned or is there something else going on?
probability-theory axioms
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
$endgroup$
add a comment |
$begingroup$
I'm reading Jaynes's "Probability Theory: The Logic of Science". In appendix A he contrasts the theory he develops in the first few chapters to the more conventional measure theory based system, A.K.A. Kolmogorov's. He lists four axioms used in this system. $F$ is a $sigma$-algebra (He calls it a "$sigma$-field")
(1) Normalization: $P(Omega) = 1$
(2) Non-negativity: $P(f_{i}) ≥ 0$ for all $f_{i}$ in $F$
(3) Additivity: if ${f_{1}ldots f_{n}}$ are disjoint elements of $F$ then
$P( f ) = sum_{i}P(f_{i})$, where $f =cup_{j}f_{j}$
(4) Continuity at zero: if a sequence $f_{1} supseteq f_{2} supseteq f_{3} supseteqldots $ tends to the empty set, then $P( f_{j} ) → 0$.
I remembered use of only the first three axioms, was the fourth once used and abandoned or is there something else going on?
probability-theory axioms
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
$endgroup$
3
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
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@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
1
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39
add a comment |
$begingroup$
I'm reading Jaynes's "Probability Theory: The Logic of Science". In appendix A he contrasts the theory he develops in the first few chapters to the more conventional measure theory based system, A.K.A. Kolmogorov's. He lists four axioms used in this system. $F$ is a $sigma$-algebra (He calls it a "$sigma$-field")
(1) Normalization: $P(Omega) = 1$
(2) Non-negativity: $P(f_{i}) ≥ 0$ for all $f_{i}$ in $F$
(3) Additivity: if ${f_{1}ldots f_{n}}$ are disjoint elements of $F$ then
$P( f ) = sum_{i}P(f_{i})$, where $f =cup_{j}f_{j}$
(4) Continuity at zero: if a sequence $f_{1} supseteq f_{2} supseteq f_{3} supseteqldots $ tends to the empty set, then $P( f_{j} ) → 0$.
I remembered use of only the first three axioms, was the fourth once used and abandoned or is there something else going on?
probability-theory axioms
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
$endgroup$
I'm reading Jaynes's "Probability Theory: The Logic of Science". In appendix A he contrasts the theory he develops in the first few chapters to the more conventional measure theory based system, A.K.A. Kolmogorov's. He lists four axioms used in this system. $F$ is a $sigma$-algebra (He calls it a "$sigma$-field")
(1) Normalization: $P(Omega) = 1$
(2) Non-negativity: $P(f_{i}) ≥ 0$ for all $f_{i}$ in $F$
(3) Additivity: if ${f_{1}ldots f_{n}}$ are disjoint elements of $F$ then
$P( f ) = sum_{i}P(f_{i})$, where $f =cup_{j}f_{j}$
(4) Continuity at zero: if a sequence $f_{1} supseteq f_{2} supseteq f_{3} supseteqldots $ tends to the empty set, then $P( f_{j} ) → 0$.
I remembered use of only the first three axioms, was the fourth once used and abandoned or is there something else going on?
probability-theory axioms
probability-theory axioms
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
edited Mar 10 '18 at 12:07
H.Rappeport
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
asked Mar 10 '18 at 11:27
H.RappeportH.Rappeport
6801510
6801510
3
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
$begingroup$
@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
1
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39
add a comment |
3
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
$begingroup$
@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
1
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39
3
3
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
$begingroup$
@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
$begingroup$
@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
1
1
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39
add a comment |
1 Answer
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$begingroup$
There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:
- Let $A_1$, $A_2$, $cdots$, be events defined over $Omega$. If $A_i cap A_j = emptyset$ for each $i neq j$, then
$$
Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i)
$$
In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $Omega$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $f_i$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.
See also this answer.
Edit: In response to your comment, see also this answer to see some examples of measures that are finitely additive but not countably additive.
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
$endgroup$
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
add a comment |
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$begingroup$
There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:
- Let $A_1$, $A_2$, $cdots$, be events defined over $Omega$. If $A_i cap A_j = emptyset$ for each $i neq j$, then
$$
Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i)
$$
In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $Omega$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $f_i$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.
See also this answer.
Edit: In response to your comment, see also this answer to see some examples of measures that are finitely additive but not countably additive.
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
$endgroup$
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
add a comment |
$begingroup$
There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:
- Let $A_1$, $A_2$, $cdots$, be events defined over $Omega$. If $A_i cap A_j = emptyset$ for each $i neq j$, then
$$
Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i)
$$
In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $Omega$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $f_i$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.
See also this answer.
Edit: In response to your comment, see also this answer to see some examples of measures that are finitely additive but not countably additive.
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
$endgroup$
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
add a comment |
$begingroup$
There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:
- Let $A_1$, $A_2$, $cdots$, be events defined over $Omega$. If $A_i cap A_j = emptyset$ for each $i neq j$, then
$$
Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i)
$$
In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $Omega$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $f_i$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.
See also this answer.
Edit: In response to your comment, see also this answer to see some examples of measures that are finitely additive but not countably additive.
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
$endgroup$
There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:
- Let $A_1$, $A_2$, $cdots$, be events defined over $Omega$. If $A_i cap A_j = emptyset$ for each $i neq j$, then
$$
Pleft(bigcup_{i=1}^{infty} A_iright) = sum_{i=1}^{infty} P(A_i)
$$
In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $Omega$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $f_i$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.
See also this answer.
Edit: In response to your comment, see also this answer to see some examples of measures that are finitely additive but not countably additive.
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
edited Jan 24 at 0:42
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
answered Jan 20 at 14:48
Apoorv KhurasiaApoorv Khurasia
1807
1807
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
add a comment |
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero)
$endgroup$
– H.Rappeport
Jan 22 at 17:41
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
$begingroup$
@H.Rappeport edited
$endgroup$
– Apoorv Khurasia
Jan 24 at 0:43
add a comment |
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3
$begingroup$
(3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3').
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:33
$begingroup$
@YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does?
$endgroup$
– H.Rappeport
Mar 10 '18 at 11:36
1
$begingroup$
It is an straight forward exercise. Just look at the sets $f_n=cup_{k=1}^{infty}A_ksetminuscup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(cup_{k=1}^{infty}A_k)=sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=emptyset$. To get (4) put $A_n=f_{n-1}setminus f_n$.
$endgroup$
– YAlexandrov
Mar 10 '18 at 11:39