Mixing
What does it mean to define events in probability space?
$begingroup$
Assume for any set $Ssubseteq [n]$, there is a parameter $q_Sge 0$. And these parameters satisfy $sum_{Ssubseteq [n]}q_S=1$.
A claim states that we can define a collection of events $A_1,dots,A_n$ such that
$$mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$$
for every $Ssubseteq [n]$.
My question is,
(a) what is the definition of "define a collection of events"? How do we know it is well-defined?
(b) Why there exists such a collection of events of this form satisfying $mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$? Is there any quantitative relation $q_S$ should satisfy?
probability combinatorics probability-theory
asked Jan 30 at 18:33
ConnorConnor
933514
$endgroup$
add a comment |
$begingroup$
Assume for any set $Ssubseteq [n]$, there is a parameter $q_Sge 0$. And these parameters satisfy $sum_{Ssubseteq [n]}q_S=1$.
A claim states that we can define a collection of events $A_1,dots,A_n$ such that
$$mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$$
for every $Ssubseteq [n]$.
My question is,
(a) what is the definition of "define a collection of events"? How do we know it is well-defined?
(b) Why there exists such a collection of events of this form satisfying $mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$? Is there any quantitative relation $q_S$ should satisfy?
probability combinatorics probability-theory
asked Jan 30 at 18:33
ConnorConnor
933514
$endgroup$
$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38
add a comment |
$begingroup$
Assume for any set $Ssubseteq [n]$, there is a parameter $q_Sge 0$. And these parameters satisfy $sum_{Ssubseteq [n]}q_S=1$.
A claim states that we can define a collection of events $A_1,dots,A_n$ such that
$$mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$$
for every $Ssubseteq [n]$.
My question is,
(a) what is the definition of "define a collection of events"? How do we know it is well-defined?
(b) Why there exists such a collection of events of this form satisfying $mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$? Is there any quantitative relation $q_S$ should satisfy?
probability combinatorics probability-theory
asked Jan 30 at 18:33
ConnorConnor
933514
$endgroup$
Assume for any set $Ssubseteq [n]$, there is a parameter $q_Sge 0$. And these parameters satisfy $sum_{Ssubseteq [n]}q_S=1$.
A claim states that we can define a collection of events $A_1,dots,A_n$ such that
$$mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$$
for every $Ssubseteq [n]$.
My question is,
(a) what is the definition of "define a collection of events"? How do we know it is well-defined?
(b) Why there exists such a collection of events of this form satisfying $mathbb{P}left(left(bigcap_{iin S}A_iright)cap left(bigcap_{jin [n]setminus S} A_j^cright)right)=q_S$? Is there any quantitative relation $q_S$ should satisfy?
probability combinatorics probability-theory
probability combinatorics probability-theory
asked Jan 30 at 18:33
ConnorConnor
933514
asked Jan 30 at 18:33
ConnorConnor
933514
edited Jan 30 at 20:51
Connor
asked Jan 30 at 18:33
ConnorConnor
933514
asked Jan 30 at 18:33
ConnorConnor
933514
asked Jan 30 at 18:33
ConnorConnor
933514
933514
$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38
add a comment |
$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38
$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38
add a comment |
1 Answer
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oldest
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$begingroup$
A simple (and minimal) way to construct such a probability space is to choose $Omega={0,1}^n$, $mathcal F=2^Omega$, and $P({omega_S})=q_S$ for every $Ssubseteq[n]$, where each $omega_S$ is defined as $omega_S=(omega^S_k)_{1leqslant kleqslant n}$ with $omega^S_k=mathbf 1_{kin S}$.
Thus, $Omega={omega_Smid Ssubseteq[n]}$ and the only relevant condition on some family of real numbers $(q_S)$ for this space $(Omega,mathcal F,P)$ to define a bona fide probability space, is that every $q_S$ should be nonnegative and that $sumlimits_{Ssubseteq[n]}q_S=1$.
Then, the events $A_k$ are simply $A_k={omega_Smid Sni k}$ and, for every $Ssubseteq[n]$, $$left(bigcap_{kin S}A_kright)cap left(bigcap_{kin [n]setminus S} A_k^cright)={omega_S}$$
answered Jan 30 at 21:02
DidDid
249k23227466
$endgroup$
add a comment |
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$begingroup$
A simple (and minimal) way to construct such a probability space is to choose $Omega={0,1}^n$, $mathcal F=2^Omega$, and $P({omega_S})=q_S$ for every $Ssubseteq[n]$, where each $omega_S$ is defined as $omega_S=(omega^S_k)_{1leqslant kleqslant n}$ with $omega^S_k=mathbf 1_{kin S}$.
Thus, $Omega={omega_Smid Ssubseteq[n]}$ and the only relevant condition on some family of real numbers $(q_S)$ for this space $(Omega,mathcal F,P)$ to define a bona fide probability space, is that every $q_S$ should be nonnegative and that $sumlimits_{Ssubseteq[n]}q_S=1$.
Then, the events $A_k$ are simply $A_k={omega_Smid Sni k}$ and, for every $Ssubseteq[n]$, $$left(bigcap_{kin S}A_kright)cap left(bigcap_{kin [n]setminus S} A_k^cright)={omega_S}$$
answered Jan 30 at 21:02
DidDid
249k23227466
$endgroup$
add a comment |
$begingroup$
A simple (and minimal) way to construct such a probability space is to choose $Omega={0,1}^n$, $mathcal F=2^Omega$, and $P({omega_S})=q_S$ for every $Ssubseteq[n]$, where each $omega_S$ is defined as $omega_S=(omega^S_k)_{1leqslant kleqslant n}$ with $omega^S_k=mathbf 1_{kin S}$.
Thus, $Omega={omega_Smid Ssubseteq[n]}$ and the only relevant condition on some family of real numbers $(q_S)$ for this space $(Omega,mathcal F,P)$ to define a bona fide probability space, is that every $q_S$ should be nonnegative and that $sumlimits_{Ssubseteq[n]}q_S=1$.
Then, the events $A_k$ are simply $A_k={omega_Smid Sni k}$ and, for every $Ssubseteq[n]$, $$left(bigcap_{kin S}A_kright)cap left(bigcap_{kin [n]setminus S} A_k^cright)={omega_S}$$
answered Jan 30 at 21:02
DidDid
249k23227466
$endgroup$
add a comment |
$begingroup$
A simple (and minimal) way to construct such a probability space is to choose $Omega={0,1}^n$, $mathcal F=2^Omega$, and $P({omega_S})=q_S$ for every $Ssubseteq[n]$, where each $omega_S$ is defined as $omega_S=(omega^S_k)_{1leqslant kleqslant n}$ with $omega^S_k=mathbf 1_{kin S}$.
Thus, $Omega={omega_Smid Ssubseteq[n]}$ and the only relevant condition on some family of real numbers $(q_S)$ for this space $(Omega,mathcal F,P)$ to define a bona fide probability space, is that every $q_S$ should be nonnegative and that $sumlimits_{Ssubseteq[n]}q_S=1$.
Then, the events $A_k$ are simply $A_k={omega_Smid Sni k}$ and, for every $Ssubseteq[n]$, $$left(bigcap_{kin S}A_kright)cap left(bigcap_{kin [n]setminus S} A_k^cright)={omega_S}$$
answered Jan 30 at 21:02
DidDid
249k23227466
$endgroup$
A simple (and minimal) way to construct such a probability space is to choose $Omega={0,1}^n$, $mathcal F=2^Omega$, and $P({omega_S})=q_S$ for every $Ssubseteq[n]$, where each $omega_S$ is defined as $omega_S=(omega^S_k)_{1leqslant kleqslant n}$ with $omega^S_k=mathbf 1_{kin S}$.
Thus, $Omega={omega_Smid Ssubseteq[n]}$ and the only relevant condition on some family of real numbers $(q_S)$ for this space $(Omega,mathcal F,P)$ to define a bona fide probability space, is that every $q_S$ should be nonnegative and that $sumlimits_{Ssubseteq[n]}q_S=1$.
Then, the events $A_k$ are simply $A_k={omega_Smid Sni k}$ and, for every $Ssubseteq[n]$, $$left(bigcap_{kin S}A_kright)cap left(bigcap_{kin [n]setminus S} A_k^cright)={omega_S}$$
answered Jan 30 at 21:02
DidDid
249k23227466
edited Jan 30 at 21:07
answered Jan 30 at 21:02
DidDid
249k23227466
answered Jan 30 at 21:02
DidDid
249k23227466
answered Jan 30 at 21:02
DidDid
249k23227466
249k23227466
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$begingroup$
$q_S$ is a probability measure on $2^{[n]}$, the set of subsets of $[n]$. An event in this space is a subset of $2^{[n]}$. "Define a collection of events" just asks you to show that there exist events $A_1,dots,A_n$ which satisfy that equation.
$endgroup$
– Mike Earnest
Jan 30 at 18:51
$begingroup$
But should we know $q_S$ satisfies if $S_1cap S_2=emptyset$, then $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$, to make it a probability measure?
$endgroup$
– Connor
Jan 30 at 19:06
$begingroup$
I think they want you to define the probabilities of the events, as well the events themselves.
$endgroup$
– saulspatz
Jan 30 at 19:35
$begingroup$
Given any function $p:Xto mathbb R^+$, where $X$ is finite and $sum_{xin X}p(x)=1$, then there is an induced probability measure $mathbb P$ defined on $X$, where for any $Asubset X$, $mathbb P(A)=sum_{ain A}p(a)$. This new function $mathbb P$ satisfies $mathbb P(Acup B)=mathbb P(A)+mathbb P(B)$ when $A,B$ are disjoint. In this case, $X=2^{[n]}$ and $p=q$. There is no requirement on $q$ to satisfy $q_{S_1cup S_2}=q_{S_1}+q_{S_2}$. Only the induced probability measure $mathbb P(A) = sum_{Sin A}q_S$ must satisfy this, and it does, as it always will.
$endgroup$
– Mike Earnest
Jan 30 at 19:38