A question on sufficient conditions for independence of sigma algebras
Question:
Take a probability space ($Omega$,$mathcal{F}$,P) and suppose that $mathcal{I}_{i}$, $i=1,2,3$ are $pi$-systems. Assume that
begin{equation}
P(I_{1}cap I_{2}cap I_{3}) = P(I_{1})P(I_{2})P(I_{3}), forall I_i in mathcal{I_i}, i=1,2,3 (1)
end{equation}
Find a sufficient condition on the $pi$-systems $mathcal{I_i}$ such that the sigma algebras $sigma{(mathcal{I_1})}$, $sigma{(mathcal{I_2})}$, $sigma{(mathcal{I_3})}$ are independent.
My Attempt:
To be honest I thought that eq. (1) was already enough to ensure independence of the $pi$-systems and thus independence of the sigma algebras generated from them.
I know it is probably an easy question but I don't understand what I'm missing here...
measure-theory
asked Dec 2 '18 at 12:54
Giulio
273
add a comment |
Question:
Take a probability space ($Omega$,$mathcal{F}$,P) and suppose that $mathcal{I}_{i}$, $i=1,2,3$ are $pi$-systems. Assume that
begin{equation}
P(I_{1}cap I_{2}cap I_{3}) = P(I_{1})P(I_{2})P(I_{3}), forall I_i in mathcal{I_i}, i=1,2,3 (1)
end{equation}
Find a sufficient condition on the $pi$-systems $mathcal{I_i}$ such that the sigma algebras $sigma{(mathcal{I_1})}$, $sigma{(mathcal{I_2})}$, $sigma{(mathcal{I_3})}$ are independent.
My Attempt:
To be honest I thought that eq. (1) was already enough to ensure independence of the $pi$-systems and thus independence of the sigma algebras generated from them.
I know it is probably an easy question but I don't understand what I'm missing here...
measure-theory
asked Dec 2 '18 at 12:54
Giulio
273
add a comment |
Question:
Take a probability space ($Omega$,$mathcal{F}$,P) and suppose that $mathcal{I}_{i}$, $i=1,2,3$ are $pi$-systems. Assume that
begin{equation}
P(I_{1}cap I_{2}cap I_{3}) = P(I_{1})P(I_{2})P(I_{3}), forall I_i in mathcal{I_i}, i=1,2,3 (1)
end{equation}
Find a sufficient condition on the $pi$-systems $mathcal{I_i}$ such that the sigma algebras $sigma{(mathcal{I_1})}$, $sigma{(mathcal{I_2})}$, $sigma{(mathcal{I_3})}$ are independent.
My Attempt:
To be honest I thought that eq. (1) was already enough to ensure independence of the $pi$-systems and thus independence of the sigma algebras generated from them.
I know it is probably an easy question but I don't understand what I'm missing here...
measure-theory
asked Dec 2 '18 at 12:54
Giulio
273
Question:
Take a probability space ($Omega$,$mathcal{F}$,P) and suppose that $mathcal{I}_{i}$, $i=1,2,3$ are $pi$-systems. Assume that
begin{equation}
P(I_{1}cap I_{2}cap I_{3}) = P(I_{1})P(I_{2})P(I_{3}), forall I_i in mathcal{I_i}, i=1,2,3 (1)
end{equation}
Find a sufficient condition on the $pi$-systems $mathcal{I_i}$ such that the sigma algebras $sigma{(mathcal{I_1})}$, $sigma{(mathcal{I_2})}$, $sigma{(mathcal{I_3})}$ are independent.
My Attempt:
To be honest I thought that eq. (1) was already enough to ensure independence of the $pi$-systems and thus independence of the sigma algebras generated from them.
I know it is probably an easy question but I don't understand what I'm missing here...
measure-theory
measure-theory
asked Dec 2 '18 at 12:54
Giulio
273
asked Dec 2 '18 at 12:54
Giulio
273
asked Dec 2 '18 at 12:54
Giulio
273
asked Dec 2 '18 at 12:54
Giulio
273
asked Dec 2 '18 at 12:54
Giulio
273
273
add a comment |
add a comment |
1 Answer
1
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oldest
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You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = Omega$
,$I_2 = Omega$, or $I_3 = Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1cap I_2) neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $pi$-$lambda$ argument here. To be specific, let us define
$$
D_1 := {I inmathcal{F};|; P(Icap I_2cap I_3) = P(I)P(I_2)P(I_3),;forall I_iin mathcal{I}_i, i=2,3}.
$$Then we can easily see that $Omega in D_1$, $Iin D_1$ implies $I^c in D_1$, and if $A_nin D_1$ ,$ninmathbb{N}$ is a disjoint family, then $cup_n A_n in D_1$ also. That is, $D_1$ is a $lambda$-system containing $mathcal{I}_1$. By Dynkin's theorem, we have $sigma(mathcal{I}_1)subset D_1$. Next, set
$$
D_2 := {I inmathcal{F};|; P(I_1cap I_2cap I_3) = P(I_1)P(I)P(I_3),;forall I_1 in sigma(mathcal{I}_1), forall I_3in mathcal{I}_3}.
$$ By almost the same argument, we can show $sigma(mathcal{I}_2)subset D_2$. It can also be shown that $sigma(mathcal{I}_3) subset D_3$ in a very similar way.
answered Dec 2 '18 at 13:32
Song
5,225318
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
add a comment |
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1 Answer
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1 Answer
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You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = Omega$
,$I_2 = Omega$, or $I_3 = Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1cap I_2) neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $pi$-$lambda$ argument here. To be specific, let us define
$$
D_1 := {I inmathcal{F};|; P(Icap I_2cap I_3) = P(I)P(I_2)P(I_3),;forall I_iin mathcal{I}_i, i=2,3}.
$$Then we can easily see that $Omega in D_1$, $Iin D_1$ implies $I^c in D_1$, and if $A_nin D_1$ ,$ninmathbb{N}$ is a disjoint family, then $cup_n A_n in D_1$ also. That is, $D_1$ is a $lambda$-system containing $mathcal{I}_1$. By Dynkin's theorem, we have $sigma(mathcal{I}_1)subset D_1$. Next, set
$$
D_2 := {I inmathcal{F};|; P(I_1cap I_2cap I_3) = P(I_1)P(I)P(I_3),;forall I_1 in sigma(mathcal{I}_1), forall I_3in mathcal{I}_3}.
$$ By almost the same argument, we can show $sigma(mathcal{I}_2)subset D_2$. It can also be shown that $sigma(mathcal{I}_3) subset D_3$ in a very similar way.
answered Dec 2 '18 at 13:32
Song
5,225318
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
add a comment |
You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = Omega$
,$I_2 = Omega$, or $I_3 = Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1cap I_2) neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $pi$-$lambda$ argument here. To be specific, let us define
$$
D_1 := {I inmathcal{F};|; P(Icap I_2cap I_3) = P(I)P(I_2)P(I_3),;forall I_iin mathcal{I}_i, i=2,3}.
$$Then we can easily see that $Omega in D_1$, $Iin D_1$ implies $I^c in D_1$, and if $A_nin D_1$ ,$ninmathbb{N}$ is a disjoint family, then $cup_n A_n in D_1$ also. That is, $D_1$ is a $lambda$-system containing $mathcal{I}_1$. By Dynkin's theorem, we have $sigma(mathcal{I}_1)subset D_1$. Next, set
$$
D_2 := {I inmathcal{F};|; P(I_1cap I_2cap I_3) = P(I_1)P(I)P(I_3),;forall I_1 in sigma(mathcal{I}_1), forall I_3in mathcal{I}_3}.
$$ By almost the same argument, we can show $sigma(mathcal{I}_2)subset D_2$. It can also be shown that $sigma(mathcal{I}_3) subset D_3$ in a very similar way.
answered Dec 2 '18 at 13:32
Song
5,225318
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
add a comment |
You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = Omega$
,$I_2 = Omega$, or $I_3 = Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1cap I_2) neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $pi$-$lambda$ argument here. To be specific, let us define
$$
D_1 := {I inmathcal{F};|; P(Icap I_2cap I_3) = P(I)P(I_2)P(I_3),;forall I_iin mathcal{I}_i, i=2,3}.
$$Then we can easily see that $Omega in D_1$, $Iin D_1$ implies $I^c in D_1$, and if $A_nin D_1$ ,$ninmathbb{N}$ is a disjoint family, then $cup_n A_n in D_1$ also. That is, $D_1$ is a $lambda$-system containing $mathcal{I}_1$. By Dynkin's theorem, we have $sigma(mathcal{I}_1)subset D_1$. Next, set
$$
D_2 := {I inmathcal{F};|; P(I_1cap I_2cap I_3) = P(I_1)P(I)P(I_3),;forall I_1 in sigma(mathcal{I}_1), forall I_3in mathcal{I}_3}.
$$ By almost the same argument, we can show $sigma(mathcal{I}_2)subset D_2$. It can also be shown that $sigma(mathcal{I}_3) subset D_3$ in a very similar way.
answered Dec 2 '18 at 13:32
Song
5,225318
You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = Omega$
,$I_2 = Omega$, or $I_3 = Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1cap I_2) neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $pi$-$lambda$ argument here. To be specific, let us define
$$
D_1 := {I inmathcal{F};|; P(Icap I_2cap I_3) = P(I)P(I_2)P(I_3),;forall I_iin mathcal{I}_i, i=2,3}.
$$Then we can easily see that $Omega in D_1$, $Iin D_1$ implies $I^c in D_1$, and if $A_nin D_1$ ,$ninmathbb{N}$ is a disjoint family, then $cup_n A_n in D_1$ also. That is, $D_1$ is a $lambda$-system containing $mathcal{I}_1$. By Dynkin's theorem, we have $sigma(mathcal{I}_1)subset D_1$. Next, set
$$
D_2 := {I inmathcal{F};|; P(I_1cap I_2cap I_3) = P(I_1)P(I)P(I_3),;forall I_1 in sigma(mathcal{I}_1), forall I_3in mathcal{I}_3}.
$$ By almost the same argument, we can show $sigma(mathcal{I}_2)subset D_2$. It can also be shown that $sigma(mathcal{I}_3) subset D_3$ in a very similar way.
answered Dec 2 '18 at 13:32
Song
5,225318
edited Dec 2 '18 at 13:38
answered Dec 2 '18 at 13:32
Song
5,225318
answered Dec 2 '18 at 13:32
Song
5,225318
answered Dec 2 '18 at 13:32
Song
5,225318
5,225318
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
add a comment |
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
Clear. You're right! (1) has to hold true also for the case $I_i=Omega$ otherwise we woudn't be able to infer that $Omega in D_i$. One question though... Since I'm required to find a sufficient condition on the $pi$-systems rather than on P, would it be better to require, instead, that $Omega in mathcal{I_i}$ $forall i$?
– Giulio
Dec 2 '18 at 13:57
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
@Giulio If $(1)$ holds for $I_i = Omega$, then adding $Omega$ to each $mathcal{I}_i$ does no harm. So you're right.
– Song
Dec 2 '18 at 23:48
add a comment |
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