Proof of an Abstract Bayes' Theorem The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Two definitions of Bayes SufficiencyExponential Families defined by Radon-Nikodym TheoremConditional expectations with different measuresProof of the monotone convergence theorem for the conditional expectationSpecific Radon-Nikodym Derivative InterpretationVariance of Conditional Expectation From First PrinciplesAlternative definition of a submartingale, problem with the Radon-Nikodym theorem.Prove that E(E(Y|X)*X)=E(Y*X)Equality of conditional expectations of random variablesConditional expectation for a random variable with infinite expectation
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Proof of an Abstract Bayes' Theorem
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Two definitions of Bayes SufficiencyExponential Families defined by Radon-Nikodym TheoremConditional expectations with different measuresProof of the monotone convergence theorem for the conditional expectationSpecific Radon-Nikodym Derivative InterpretationVariance of Conditional Expectation From First PrinciplesAlternative definition of a submartingale, problem with the Radon-Nikodym theorem.Prove that E(E(Y|X)*X)=E(Y*X)Equality of conditional expectations of random variablesConditional expectation for a random variable with infinite expectation
$begingroup$
In Björk (2009) a Bayes' theorem is given by
Assume that $X$ is a random variable on $(Omega, mathcal F, P)$ and let $Q$ be another probability measure on $(Omega, mathcal F)$ with Radon-Nikodym derivative
$$L = fracdQdP ; ; texton mathcal F.$$
Let $G$ be a sigma-algebra such that $G subset mathcal F$. Then
$$E^Q[Xmid G] = fracE^P[LXmid G]E^P[Lmid G].$$
The proof is to show that the integral over any $A in G$ of the both sides of
$$E^Q[Xmid G]E^P[Lmid G] = E^P[LXmid G]$$
is the same quantity.
Starting with the left hand side,
$$beginaligned
int_AE^Q[Xmid G]E^P[Lmid G]dP &= int_AE^Pleft[Lcdot E^Q[Xmid G]mid Gright]dP \
&=int_ALcdot E^Q[Xmid G]dP
\
&= int_AE^Q[Xmid G]dQ
\
&= int_AXdQ.
endaligned$$
The first equality follows from the fact that $E[Xmid G]$ is $G$-measurable. Why the second equality follows? My guess is that he moves $L$ inside the expectation under $Q$ and then applies the law of iterated expectations. However for this $L$ has to be $G$-measurable which is not generally the case since it is defined of $mathcal F$ which is bigger than $G$.
probability-theory self-learning conditional-expectation bayes-theorem radon-nikodym
asked Mar 31 at 13:35
tosiktosik
1504
$endgroup$
add a comment |
$begingroup$
In Björk (2009) a Bayes' theorem is given by
Assume that $X$ is a random variable on $(Omega, mathcal F, P)$ and let $Q$ be another probability measure on $(Omega, mathcal F)$ with Radon-Nikodym derivative
$$L = fracdQdP ; ; texton mathcal F.$$
Let $G$ be a sigma-algebra such that $G subset mathcal F$. Then
$$E^Q[Xmid G] = fracE^P[LXmid G]E^P[Lmid G].$$
The proof is to show that the integral over any $A in G$ of the both sides of
$$E^Q[Xmid G]E^P[Lmid G] = E^P[LXmid G]$$
is the same quantity.
Starting with the left hand side,
$$beginaligned
int_AE^Q[Xmid G]E^P[Lmid G]dP &= int_AE^Pleft[Lcdot E^Q[Xmid G]mid Gright]dP \
&=int_ALcdot E^Q[Xmid G]dP
\
&= int_AE^Q[Xmid G]dQ
\
&= int_AXdQ.
endaligned$$
The first equality follows from the fact that $E[Xmid G]$ is $G$-measurable. Why the second equality follows? My guess is that he moves $L$ inside the expectation under $Q$ and then applies the law of iterated expectations. However for this $L$ has to be $G$-measurable which is not generally the case since it is defined of $mathcal F$ which is bigger than $G$.
probability-theory self-learning conditional-expectation bayes-theorem radon-nikodym
asked Mar 31 at 13:35
tosiktosik
1504
$endgroup$
add a comment |
$begingroup$
In Björk (2009) a Bayes' theorem is given by
Assume that $X$ is a random variable on $(Omega, mathcal F, P)$ and let $Q$ be another probability measure on $(Omega, mathcal F)$ with Radon-Nikodym derivative
$$L = fracdQdP ; ; texton mathcal F.$$
Let $G$ be a sigma-algebra such that $G subset mathcal F$. Then
$$E^Q[Xmid G] = fracE^P[LXmid G]E^P[Lmid G].$$
The proof is to show that the integral over any $A in G$ of the both sides of
$$E^Q[Xmid G]E^P[Lmid G] = E^P[LXmid G]$$
is the same quantity.
Starting with the left hand side,
$$beginaligned
int_AE^Q[Xmid G]E^P[Lmid G]dP &= int_AE^Pleft[Lcdot E^Q[Xmid G]mid Gright]dP \
&=int_ALcdot E^Q[Xmid G]dP
\
&= int_AE^Q[Xmid G]dQ
\
&= int_AXdQ.
endaligned$$
The first equality follows from the fact that $E[Xmid G]$ is $G$-measurable. Why the second equality follows? My guess is that he moves $L$ inside the expectation under $Q$ and then applies the law of iterated expectations. However for this $L$ has to be $G$-measurable which is not generally the case since it is defined of $mathcal F$ which is bigger than $G$.
probability-theory self-learning conditional-expectation bayes-theorem radon-nikodym
asked Mar 31 at 13:35
tosiktosik
1504
$endgroup$
In Björk (2009) a Bayes' theorem is given by
Assume that $X$ is a random variable on $(Omega, mathcal F, P)$ and let $Q$ be another probability measure on $(Omega, mathcal F)$ with Radon-Nikodym derivative
$$L = fracdQdP ; ; texton mathcal F.$$
Let $G$ be a sigma-algebra such that $G subset mathcal F$. Then
$$E^Q[Xmid G] = fracE^P[LXmid G]E^P[Lmid G].$$
The proof is to show that the integral over any $A in G$ of the both sides of
$$E^Q[Xmid G]E^P[Lmid G] = E^P[LXmid G]$$
is the same quantity.
Starting with the left hand side,
$$beginaligned
int_AE^Q[Xmid G]E^P[Lmid G]dP &= int_AE^Pleft[Lcdot E^Q[Xmid G]mid Gright]dP \
&=int_ALcdot E^Q[Xmid G]dP
\
&= int_AE^Q[Xmid G]dQ
\
&= int_AXdQ.
endaligned$$
The first equality follows from the fact that $E[Xmid G]$ is $G$-measurable. Why the second equality follows? My guess is that he moves $L$ inside the expectation under $Q$ and then applies the law of iterated expectations. However for this $L$ has to be $G$-measurable which is not generally the case since it is defined of $mathcal F$ which is bigger than $G$.
probability-theory self-learning conditional-expectation bayes-theorem radon-nikodym
probability-theory self-learning conditional-expectation bayes-theorem radon-nikodym
asked Mar 31 at 13:35
tosiktosik
1504
asked Mar 31 at 13:35
tosiktosik
1504
asked Mar 31 at 13:35
tosiktosik
1504
asked Mar 31 at 13:35
tosiktosik
1504
asked Mar 31 at 13:35
tosiktosik
1504
1504
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.
answered Mar 31 at 15:41
C. DavideC. Davide
812
$endgroup$
add a comment |
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$begingroup$
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.
answered Mar 31 at 15:41
C. DavideC. Davide
812
$endgroup$
add a comment |
$begingroup$
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.
answered Mar 31 at 15:41
C. DavideC. Davide
812
$endgroup$
add a comment |
$begingroup$
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.
answered Mar 31 at 15:41
C. DavideC. Davide
812
$endgroup$
Try to read the second equality backwards, what it says is nothing but the definition of the conditional expectation.
answered Mar 31 at 15:41
C. DavideC. Davide
812
answered Mar 31 at 15:41
C. DavideC. Davide
812
answered Mar 31 at 15:41
C. DavideC. Davide
812
answered Mar 31 at 15:41
C. DavideC. Davide
812
812
add a comment |
add a comment |
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