A problem about conditional probablity

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Consider probability space
$left( Omegamathcal,F,mathbbP right)$. Given an event
$Amathcalin F$ and a $sigma$-algebra $mathcalG subseteq F$, we
define the conditional probability given $sigma$-algebra
$mathcalG$, denoted by $mathbbPleft( A|mathcalG right)$, as a
non-negative
$left( Omegamathcal,G right) rightarrow left( mathbbRmathcal,Bleft( mathbbR right) right)$
random variable s.t.
$forall G in mathcalG,mathbbPleft( A cap G right) = int_G^ mathcalG right)dmathbbP$.
Existence and uniqueness is proved below.

For any $A in mathcalF$ and $mathcalG subseteq F$, the
conditional probability exists and is a.s. unique. Notice
$mathbbP_Aleft( cdot right):=mathbbPleft( Abigcap_^left( cdot right) right)$
is a measure for $left( Omegamathcal,G right)$ by checking those
axioms, and we can also verify $mathbbP_Amathbbll P$. Also,
$mathbbP$ is also a valid measure for
$left( Omegamathcal,G right)$, then by Randon-Nikodym
Theorem we have a non-negative $mathcalG$-measurable function
$mathbbPleft( A middle| mathcalG right) = fracdmathbbP_AdmathbbP$
exists and is a.s. unique.

We can further show given any $omega in Omega$, then
$mathbbP_mathcalG^left( omega right)left( cdot right):=mathbbPleft( cdot middle| mathcalG right)left( omega right)$
is a.s. a valid measure on $left( Omegamathcal,F right)$.

First, $mathbbP_mathcalG^left( omega right) geq 0$
by definition. Secondly, $mathbbP_mathcalG^left( omega right)left( varnothing right) = 0$ because by definition $0 = mathbbPleft( varnothing cap G right) = int_G^mathbbPleft( varnothing middle$ a.s.
Thirdly, for any
$A_1,A_2mathcalin F,A_1bigcap A_2 = varnothing$, then
$A_1bigcap G$ is disjoint form $A_2bigcap G$ and

$$int_G^mathbbPleft( A_1bigcup A_2 middlemathbb= Pleft( left( A_1bigcup A_2 right) cap G right)mathbb= Pleft( left( A_1bigcap G right)bigcupleft( A_2bigcap G right) right)mathbb= Pleft( A_1bigcap G right)mathbb+ Pleft( A_2bigcap G right) = int_G^ mathcalG right)dmathbbP + int_G^ mathcalG right)dmathbbP = int_G^ mathcalG right) right)dmathbbP$$

Then by uniqueness we have a.s.
$mathbbPleft( A_1bigcup A_2 middle| mathcalG right) = mathbbPleft( A_1 middle| mathcalG right)mathbb+ Pleft( A_2 middle| mathcalG right)$.

I need help with showing the following,

Given any random variable $Y:left( Omegamathcal,F right) rightarrow left( S,mathcalE right)$, let $sigma(Y)$ be its generated $sigma$-algebra, then $mathbbP_sigmaleft( Y right)^left( omega_1 right) = mathbbP_sigmaleft( Y right)^left( omega_2 right)$ a.s.
if $Yleft( omega_1 right) = Yleft( omega_2 right)$

probability-theory measure-theory

share|cite|improve this question

edited Jul 25 at 6:32

asked Jul 25 at 6:10

Ralph B.

1,012517

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
  – Did
  Jul 25 at 6:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
  – Kavi Rama Murthy
  Jul 25 at 6:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
  – Did
  Jul 25 at 6:19
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

Consider probability space
$left( Omegamathcal,F,mathbbP right)$. Given an event
$Amathcalin F$ and a $sigma$-algebra $mathcalG subseteq F$, we
define the conditional probability given $sigma$-algebra
$mathcalG$, denoted by $mathbbPleft( A|mathcalG right)$, as a
non-negative
$left( Omegamathcal,G right) rightarrow left( mathbbRmathcal,Bleft( mathbbR right) right)$
random variable s.t.
$forall G in mathcalG,mathbbPleft( A cap G right) = int_G^ mathcalG right)dmathbbP$.
Existence and uniqueness is proved below.

For any $A in mathcalF$ and $mathcalG subseteq F$, the
conditional probability exists and is a.s. unique. Notice
$mathbbP_Aleft( cdot right):=mathbbPleft( Abigcap_^left( cdot right) right)$
is a measure for $left( Omegamathcal,G right)$ by checking those
axioms, and we can also verify $mathbbP_Amathbbll P$. Also,
$mathbbP$ is also a valid measure for
$left( Omegamathcal,G right)$, then by Randon-Nikodym
Theorem we have a non-negative $mathcalG$-measurable function
$mathbbPleft( A middle| mathcalG right) = fracdmathbbP_AdmathbbP$
exists and is a.s. unique.

We can further show given any $omega in Omega$, then
$mathbbP_mathcalG^left( omega right)left( cdot right):=mathbbPleft( cdot middle| mathcalG right)left( omega right)$
is a.s. a valid measure on $left( Omegamathcal,F right)$.

First, $mathbbP_mathcalG^left( omega right) geq 0$
by definition. Secondly, $mathbbP_mathcalG^left( omega right)left( varnothing right) = 0$ because by definition $0 = mathbbPleft( varnothing cap G right) = int_G^mathbbPleft( varnothing middle$ a.s.
Thirdly, for any
$A_1,A_2mathcalin F,A_1bigcap A_2 = varnothing$, then
$A_1bigcap G$ is disjoint form $A_2bigcap G$ and

$$int_G^mathbbPleft( A_1bigcup A_2 middlemathbb= Pleft( left( A_1bigcup A_2 right) cap G right)mathbb= Pleft( left( A_1bigcap G right)bigcupleft( A_2bigcap G right) right)mathbb= Pleft( A_1bigcap G right)mathbb+ Pleft( A_2bigcap G right) = int_G^ mathcalG right)dmathbbP + int_G^ mathcalG right)dmathbbP = int_G^ mathcalG right) right)dmathbbP$$

Then by uniqueness we have a.s.
$mathbbPleft( A_1bigcup A_2 middle| mathcalG right) = mathbbPleft( A_1 middle| mathcalG right)mathbb+ Pleft( A_2 middle| mathcalG right)$.

I need help with showing the following,

Given any random variable $Y:left( Omegamathcal,F right) rightarrow left( S,mathcalE right)$, let $sigma(Y)$ be its generated $sigma$-algebra, then $mathbbP_sigmaleft( Y right)^left( omega_1 right) = mathbbP_sigmaleft( Y right)^left( omega_2 right)$ a.s.
if $Yleft( omega_1 right) = Yleft( omega_2 right)$

probability-theory measure-theory

share|cite|improve this question

edited Jul 25 at 6:32

asked Jul 25 at 6:10

Ralph B.

1,012517

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
  – Did
  Jul 25 at 6:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
  – Kavi Rama Murthy
  Jul 25 at 6:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
  – Did
  Jul 25 at 6:19
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

Consider probability space
$left( Omegamathcal,F,mathbbP right)$. Given an event
$Amathcalin F$ and a $sigma$-algebra $mathcalG subseteq F$, we
define the conditional probability given $sigma$-algebra
$mathcalG$, denoted by $mathbbPleft( A|mathcalG right)$, as a
non-negative
$left( Omegamathcal,G right) rightarrow left( mathbbRmathcal,Bleft( mathbbR right) right)$
random variable s.t.
$forall G in mathcalG,mathbbPleft( A cap G right) = int_G^ mathcalG right)dmathbbP$.
Existence and uniqueness is proved below.

For any $A in mathcalF$ and $mathcalG subseteq F$, the
conditional probability exists and is a.s. unique. Notice
$mathbbP_Aleft( cdot right):=mathbbPleft( Abigcap_^left( cdot right) right)$
is a measure for $left( Omegamathcal,G right)$ by checking those
axioms, and we can also verify $mathbbP_Amathbbll P$. Also,
$mathbbP$ is also a valid measure for
$left( Omegamathcal,G right)$, then by Randon-Nikodym
Theorem we have a non-negative $mathcalG$-measurable function
$mathbbPleft( A middle| mathcalG right) = fracdmathbbP_AdmathbbP$
exists and is a.s. unique.

We can further show given any $omega in Omega$, then
$mathbbP_mathcalG^left( omega right)left( cdot right):=mathbbPleft( cdot middle| mathcalG right)left( omega right)$
is a.s. a valid measure on $left( Omegamathcal,F right)$.

First, $mathbbP_mathcalG^left( omega right) geq 0$
by definition. Secondly, $mathbbP_mathcalG^left( omega right)left( varnothing right) = 0$ because by definition $0 = mathbbPleft( varnothing cap G right) = int_G^mathbbPleft( varnothing middle$ a.s.
Thirdly, for any
$A_1,A_2mathcalin F,A_1bigcap A_2 = varnothing$, then
$A_1bigcap G$ is disjoint form $A_2bigcap G$ and

$$int_G^mathbbPleft( A_1bigcup A_2 middlemathbb= Pleft( left( A_1bigcup A_2 right) cap G right)mathbb= Pleft( left( A_1bigcap G right)bigcupleft( A_2bigcap G right) right)mathbb= Pleft( A_1bigcap G right)mathbb+ Pleft( A_2bigcap G right) = int_G^ mathcalG right)dmathbbP + int_G^ mathcalG right)dmathbbP = int_G^ mathcalG right) right)dmathbbP$$

Then by uniqueness we have a.s.
$mathbbPleft( A_1bigcup A_2 middle| mathcalG right) = mathbbPleft( A_1 middle| mathcalG right)mathbb+ Pleft( A_2 middle| mathcalG right)$.

I need help with showing the following,

Given any random variable $Y:left( Omegamathcal,F right) rightarrow left( S,mathcalE right)$, let $sigma(Y)$ be its generated $sigma$-algebra, then $mathbbP_sigmaleft( Y right)^left( omega_1 right) = mathbbP_sigmaleft( Y right)^left( omega_2 right)$ a.s.
if $Yleft( omega_1 right) = Yleft( omega_2 right)$

probability-theory measure-theory

share|cite|improve this question

edited Jul 25 at 6:32

asked Jul 25 at 6:10

Ralph B.

1,012517

Consider probability space
$left( Omegamathcal,F,mathbbP right)$. Given an event
$Amathcalin F$ and a $sigma$-algebra $mathcalG subseteq F$, we
define the conditional probability given $sigma$-algebra
$mathcalG$, denoted by $mathbbPleft( A|mathcalG right)$, as a
non-negative
$left( Omegamathcal,G right) rightarrow left( mathbbRmathcal,Bleft( mathbbR right) right)$
random variable s.t.
$forall G in mathcalG,mathbbPleft( A cap G right) = int_G^ mathcalG right)dmathbbP$.
Existence and uniqueness is proved below.

For any $A in mathcalF$ and $mathcalG subseteq F$, the
conditional probability exists and is a.s. unique. Notice
$mathbbP_Aleft( cdot right):=mathbbPleft( Abigcap_^left( cdot right) right)$
is a measure for $left( Omegamathcal,G right)$ by checking those
axioms, and we can also verify $mathbbP_Amathbbll P$. Also,
$mathbbP$ is also a valid measure for
$left( Omegamathcal,G right)$, then by Randon-Nikodym
Theorem we have a non-negative $mathcalG$-measurable function
$mathbbPleft( A middle| mathcalG right) = fracdmathbbP_AdmathbbP$
exists and is a.s. unique.

We can further show given any $omega in Omega$, then
$mathbbP_mathcalG^left( omega right)left( cdot right):=mathbbPleft( cdot middle| mathcalG right)left( omega right)$
is a.s. a valid measure on $left( Omegamathcal,F right)$.

First, $mathbbP_mathcalG^left( omega right) geq 0$
by definition. Secondly, $mathbbP_mathcalG^left( omega right)left( varnothing right) = 0$ because by definition $0 = mathbbPleft( varnothing cap G right) = int_G^mathbbPleft( varnothing middle$ a.s.
Thirdly, for any
$A_1,A_2mathcalin F,A_1bigcap A_2 = varnothing$, then
$A_1bigcap G$ is disjoint form $A_2bigcap G$ and

$$int_G^mathbbPleft( A_1bigcup A_2 middlemathbb= Pleft( left( A_1bigcup A_2 right) cap G right)mathbb= Pleft( left( A_1bigcap G right)bigcupleft( A_2bigcap G right) right)mathbb= Pleft( A_1bigcap G right)mathbb+ Pleft( A_2bigcap G right) = int_G^ mathcalG right)dmathbbP + int_G^ mathcalG right)dmathbbP = int_G^ mathcalG right) right)dmathbbP$$

Then by uniqueness we have a.s.
$mathbbPleft( A_1bigcup A_2 middle| mathcalG right) = mathbbPleft( A_1 middle| mathcalG right)mathbb+ Pleft( A_2 middle| mathcalG right)$.

I need help with showing the following,

Given any random variable $Y:left( Omegamathcal,F right) rightarrow left( S,mathcalE right)$, let $sigma(Y)$ be its generated $sigma$-algebra, then $mathbbP_sigmaleft( Y right)^left( omega_1 right) = mathbbP_sigmaleft( Y right)^left( omega_2 right)$ a.s.
if $Yleft( omega_1 right) = Yleft( omega_2 right)$

probability-theory measure-theory

share|cite|improve this question

edited Jul 25 at 6:32

asked Jul 25 at 6:10

Ralph B.

1,012517

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 6:32

edited Jul 25 at 6:32

edited Jul 25 at 6:32

asked Jul 25 at 6:10

Ralph B.

1,012517

asked Jul 25 at 6:10

Ralph B.

1,012517

asked Jul 25 at 6:10

Ralph B.

1,012517

1,012517

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
  – Did
  Jul 25 at 6:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
  – Kavi Rama Murthy
  Jul 25 at 6:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
  – Did
  Jul 25 at 6:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
  – Did
  Jul 25 at 6:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
  – Kavi Rama Murthy
  Jul 25 at 6:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
  – Did
  Jul 25 at 6:19
  

  
  
  
  

1

1

The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
– Did
Jul 25 at 6:17

The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(varnothingmidmathcal G)(omega)=0$$ is not guaranteed for every $omega$ in $Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(varnothingmidmathcal G)$, one has $$P(P(varnothingmidmathcal G)ne0)=0$$ Likewise, the statement you are asking about does not hold.
– Did
Jul 25 at 6:17

1

1

I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
– Kavi Rama Murthy
Jul 25 at 6:19

I think there are many wrong sattements here. $P^(omega )_mathcal G$ is defined only almost everywhere so we cannot say that for fixed $omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'.
– Kavi Rama Murthy
Jul 25 at 6:19

Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
– Did
Jul 25 at 6:19

Unrelated: you might want to replace every bigcap and bigcup in your post by cap and cup.
– Did
Jul 25 at 6:19

add a comment | 

1 Answer
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down vote



accepted

What you can show is that for any $AinmathcalF$,

$$
mathbbP(Amid Y)(omega_1)=mathbbP(Amid Y)(omega_2)
$$
when $Y(omega_1)=Y(omega_2)$.

---

Since $mathbbP(Amid Y)=mathsfE[1_Amid Y]$ and the latter is $sigma(Y)$-measuralbe, there exists a Borel function $phi$ such that

$$
mathbbP(Amid Y)(omega)=phi(Y(omega))
$$
for all $omegainOmega$.

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

add a comment | 

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

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oldest

votes

up vote
1
down vote



accepted

What you can show is that for any $AinmathcalF$,

$$
mathbbP(Amid Y)(omega_1)=mathbbP(Amid Y)(omega_2)
$$
when $Y(omega_1)=Y(omega_2)$.

---

Since $mathbbP(Amid Y)=mathsfE[1_Amid Y]$ and the latter is $sigma(Y)$-measuralbe, there exists a Borel function $phi$ such that

$$
mathbbP(Amid Y)(omega)=phi(Y(omega))
$$
for all $omegainOmega$.

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

add a comment | 

up vote
1
down vote



accepted

What you can show is that for any $AinmathcalF$,

$$
mathbbP(Amid Y)(omega_1)=mathbbP(Amid Y)(omega_2)
$$
when $Y(omega_1)=Y(omega_2)$.

---

Since $mathbbP(Amid Y)=mathsfE[1_Amid Y]$ and the latter is $sigma(Y)$-measuralbe, there exists a Borel function $phi$ such that

$$
mathbbP(Amid Y)(omega)=phi(Y(omega))
$$
for all $omegainOmega$.

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

What you can show is that for any $AinmathcalF$,

$$
mathbbP(Amid Y)(omega_1)=mathbbP(Amid Y)(omega_2)
$$
when $Y(omega_1)=Y(omega_2)$.

---

Since $mathbbP(Amid Y)=mathsfE[1_Amid Y]$ and the latter is $sigma(Y)$-measuralbe, there exists a Borel function $phi$ such that

$$
mathbbP(Amid Y)(omega)=phi(Y(omega))
$$
for all $omegainOmega$.

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

What you can show is that for any $AinmathcalF$,

$$
mathbbP(Amid Y)(omega_1)=mathbbP(Amid Y)(omega_2)
$$
when $Y(omega_1)=Y(omega_2)$.

---

Since $mathbbP(Amid Y)=mathsfE[1_Amid Y]$ and the latter is $sigma(Y)$-measuralbe, there exists a Borel function $phi$ such that

$$
mathbbP(Amid Y)(omega)=phi(Y(omega))
$$
for all $omegainOmega$.

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 7:18

d.k.o.

7,709526

answered Jul 25 at 7:18

d.k.o.

7,709526

answered Jul 25 at 7:18

d.k.o.

7,709526

7,709526

add a comment | 

add a comment | 

 

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