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How the Bayes rule for density functions is formulated in probability theory?
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Given a probability space
$left( Omegamathcal,F,mathbbP right)$, and two
$mathcalF$-measurable real-valued random variables $X,Y$, then the
joint random variable $left( X,Y right)$ can be defined on a product space
$left( Omega^2,sigmaleft( mathcalF^2 right),mathbbP times P right)$
where $mathbbP times P$ is the product measure of $mathbbP$. Let
$fleft( x,y right),f_Xleft( x,y right),f_Yleft( y right)$ be
the density functions (Randon-Nikodym derivatives) of
$left( X,Y right),X,Y$ respectively, and let
$f_Yleft( x,y right)$ be the density function of $X$ conditioned on $Y$.
Anyone can help with a construction, or proof or related materials about the Bayes rule
$f_Yleft( x|y right) = fracfleft( x,y right)f_Yleft( y right)$?
We may also instead consider the other version $f_Yleft( x|y right) = fracf_Xleft( yf_Yleft( y right)$ which does involve the joint random variable. I do not understand how the this Bayes rule is formulated in measure theory. This is a widely used formula, while I cannot find any construction or proof from my probability books.
I can find related definition for "conditional density" in the following way. There could be other definitions.
We denote the integration w.r.t. the measure
$mathbbP circX^- 1$ of a RV as
$int_B^dX := int_B^dleft( mathbbP circX^- 1 right)$
for simplicity. Define the conditional probability measures
$mathbbP_y,y in Yleft( Omega right)$ as a family of probability
measures on $left( Omegamathcal,F right)$ s.t. two axioms hold: 1)
$mathbbP_yleft( A right)$ is
$left( mathbbR,mathcalBleft( mathbbR right) right)$-measurable
for any $A in mathcalF$ (given a fixed
$A in mathcalF$,$ mathbbP_yleft( A right)$ is a
$mathbbR rightarrowleftlbrack 0,1 rightrbrack$ function w.r.t.
index $y$); and 2) the general version of law of total
probability
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall A in mathcalF, B in mathcalBleft ( mathbb R right)$$
We then denote
$mathbbPleft( A|Y = y right) = mathbbP_yleft( A right),forall A in mathcalF$
as the conditional probability measure given event $Y = y$. Then for any RV $X$, the conditional probability density function
$f_Yleft( x|y right)$ is the Radon-Nikodym derivative of
distribution $mathbbP_y circ X^- 1$
I list all relations I can conceive, based on above definition,
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall Amathcalin F,B in mathcalBleft( mathbbR right)$$
$$int_B^mathbbP_xleft( A right)dYmathbb= Pleft( Abigcap X^- 1left( B right) right),forall Amathcalin F,X in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( x = mathbbP_yleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^x right) = mathbbP_xleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( y right) = mathbbPleft Y^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Xleft( x right) = mathbbPleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
real-analysis probability probability-theory measure-theory
asked Jul 23 at 0:49
Tony
2,1241625
add a comment |Â
up vote
3
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Given a probability space
$left( Omegamathcal,F,mathbbP right)$, and two
$mathcalF$-measurable real-valued random variables $X,Y$, then the
joint random variable $left( X,Y right)$ can be defined on a product space
$left( Omega^2,sigmaleft( mathcalF^2 right),mathbbP times P right)$
where $mathbbP times P$ is the product measure of $mathbbP$. Let
$fleft( x,y right),f_Xleft( x,y right),f_Yleft( y right)$ be
the density functions (Randon-Nikodym derivatives) of
$left( X,Y right),X,Y$ respectively, and let
$f_Yleft( x,y right)$ be the density function of $X$ conditioned on $Y$.
Anyone can help with a construction, or proof or related materials about the Bayes rule
$f_Yleft( x|y right) = fracfleft( x,y right)f_Yleft( y right)$?
We may also instead consider the other version $f_Yleft( x|y right) = fracf_Xleft( yf_Yleft( y right)$ which does involve the joint random variable. I do not understand how the this Bayes rule is formulated in measure theory. This is a widely used formula, while I cannot find any construction or proof from my probability books.
I can find related definition for "conditional density" in the following way. There could be other definitions.
We denote the integration w.r.t. the measure
$mathbbP circX^- 1$ of a RV as
$int_B^dX := int_B^dleft( mathbbP circX^- 1 right)$
for simplicity. Define the conditional probability measures
$mathbbP_y,y in Yleft( Omega right)$ as a family of probability
measures on $left( Omegamathcal,F right)$ s.t. two axioms hold: 1)
$mathbbP_yleft( A right)$ is
$left( mathbbR,mathcalBleft( mathbbR right) right)$-measurable
for any $A in mathcalF$ (given a fixed
$A in mathcalF$,$ mathbbP_yleft( A right)$ is a
$mathbbR rightarrowleftlbrack 0,1 rightrbrack$ function w.r.t.
index $y$); and 2) the general version of law of total
probability
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall A in mathcalF, B in mathcalBleft ( mathbb R right)$$
We then denote
$mathbbPleft( A|Y = y right) = mathbbP_yleft( A right),forall A in mathcalF$
as the conditional probability measure given event $Y = y$. Then for any RV $X$, the conditional probability density function
$f_Yleft( x|y right)$ is the Radon-Nikodym derivative of
distribution $mathbbP_y circ X^- 1$
I list all relations I can conceive, based on above definition,
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall Amathcalin F,B in mathcalBleft( mathbbR right)$$
$$int_B^mathbbP_xleft( A right)dYmathbb= Pleft( Abigcap X^- 1left( B right) right),forall Amathcalin F,X in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( x = mathbbP_yleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^x right) = mathbbP_xleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( y right) = mathbbPleft Y^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Xleft( x right) = mathbbPleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
real-analysis probability probability-theory measure-theory
asked Jul 23 at 0:49
Tony
2,1241625
Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
1
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a probability space
$left( Omegamathcal,F,mathbbP right)$, and two
$mathcalF$-measurable real-valued random variables $X,Y$, then the
joint random variable $left( X,Y right)$ can be defined on a product space
$left( Omega^2,sigmaleft( mathcalF^2 right),mathbbP times P right)$
where $mathbbP times P$ is the product measure of $mathbbP$. Let
$fleft( x,y right),f_Xleft( x,y right),f_Yleft( y right)$ be
the density functions (Randon-Nikodym derivatives) of
$left( X,Y right),X,Y$ respectively, and let
$f_Yleft( x,y right)$ be the density function of $X$ conditioned on $Y$.
Anyone can help with a construction, or proof or related materials about the Bayes rule
$f_Yleft( x|y right) = fracfleft( x,y right)f_Yleft( y right)$?
We may also instead consider the other version $f_Yleft( x|y right) = fracf_Xleft( yf_Yleft( y right)$ which does involve the joint random variable. I do not understand how the this Bayes rule is formulated in measure theory. This is a widely used formula, while I cannot find any construction or proof from my probability books.
I can find related definition for "conditional density" in the following way. There could be other definitions.
We denote the integration w.r.t. the measure
$mathbbP circX^- 1$ of a RV as
$int_B^dX := int_B^dleft( mathbbP circX^- 1 right)$
for simplicity. Define the conditional probability measures
$mathbbP_y,y in Yleft( Omega right)$ as a family of probability
measures on $left( Omegamathcal,F right)$ s.t. two axioms hold: 1)
$mathbbP_yleft( A right)$ is
$left( mathbbR,mathcalBleft( mathbbR right) right)$-measurable
for any $A in mathcalF$ (given a fixed
$A in mathcalF$,$ mathbbP_yleft( A right)$ is a
$mathbbR rightarrowleftlbrack 0,1 rightrbrack$ function w.r.t.
index $y$); and 2) the general version of law of total
probability
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall A in mathcalF, B in mathcalBleft ( mathbb R right)$$
We then denote
$mathbbPleft( A|Y = y right) = mathbbP_yleft( A right),forall A in mathcalF$
as the conditional probability measure given event $Y = y$. Then for any RV $X$, the conditional probability density function
$f_Yleft( x|y right)$ is the Radon-Nikodym derivative of
distribution $mathbbP_y circ X^- 1$
I list all relations I can conceive, based on above definition,
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall Amathcalin F,B in mathcalBleft( mathbbR right)$$
$$int_B^mathbbP_xleft( A right)dYmathbb= Pleft( Abigcap X^- 1left( B right) right),forall Amathcalin F,X in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( x = mathbbP_yleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^x right) = mathbbP_xleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( y right) = mathbbPleft Y^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Xleft( x right) = mathbbPleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
real-analysis probability probability-theory measure-theory
asked Jul 23 at 0:49
Tony
2,1241625
Given a probability space
$left( Omegamathcal,F,mathbbP right)$, and two
$mathcalF$-measurable real-valued random variables $X,Y$, then the
joint random variable $left( X,Y right)$ can be defined on a product space
$left( Omega^2,sigmaleft( mathcalF^2 right),mathbbP times P right)$
where $mathbbP times P$ is the product measure of $mathbbP$. Let
$fleft( x,y right),f_Xleft( x,y right),f_Yleft( y right)$ be
the density functions (Randon-Nikodym derivatives) of
$left( X,Y right),X,Y$ respectively, and let
$f_Yleft( x,y right)$ be the density function of $X$ conditioned on $Y$.
Anyone can help with a construction, or proof or related materials about the Bayes rule
$f_Yleft( x|y right) = fracfleft( x,y right)f_Yleft( y right)$?
We may also instead consider the other version $f_Yleft( x|y right) = fracf_Xleft( yf_Yleft( y right)$ which does involve the joint random variable. I do not understand how the this Bayes rule is formulated in measure theory. This is a widely used formula, while I cannot find any construction or proof from my probability books.
I can find related definition for "conditional density" in the following way. There could be other definitions.
We denote the integration w.r.t. the measure
$mathbbP circX^- 1$ of a RV as
$int_B^dX := int_B^dleft( mathbbP circX^- 1 right)$
for simplicity. Define the conditional probability measures
$mathbbP_y,y in Yleft( Omega right)$ as a family of probability
measures on $left( Omegamathcal,F right)$ s.t. two axioms hold: 1)
$mathbbP_yleft( A right)$ is
$left( mathbbR,mathcalBleft( mathbbR right) right)$-measurable
for any $A in mathcalF$ (given a fixed
$A in mathcalF$,$ mathbbP_yleft( A right)$ is a
$mathbbR rightarrowleftlbrack 0,1 rightrbrack$ function w.r.t.
index $y$); and 2) the general version of law of total
probability
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall A in mathcalF, B in mathcalBleft ( mathbb R right)$$
We then denote
$mathbbPleft( A|Y = y right) = mathbbP_yleft( A right),forall A in mathcalF$
as the conditional probability measure given event $Y = y$. Then for any RV $X$, the conditional probability density function
$f_Yleft( x|y right)$ is the Radon-Nikodym derivative of
distribution $mathbbP_y circ X^- 1$
I list all relations I can conceive, based on above definition,
$$int_B^mathbbP_yleft( A right)dYmathbb= Pleft( Abigcap Y^- 1left( B right) right),forall Amathcalin F,B in mathcalBleft( mathbbR right)$$
$$int_B^mathbbP_xleft( A right)dYmathbb= Pleft( Abigcap X^- 1left( B right) right),forall Amathcalin F,X in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( x = mathbbP_yleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^x right) = mathbbP_xleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Yleft( y right) = mathbbPleft Y^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
$$int_B^f_Xleft( x right) = mathbbPleft X^- 1left( B right) right,forall B in mathcalBleft( mathbbR right)$$
real-analysis probability probability-theory measure-theory
asked Jul 23 at 0:49
Tony
2,1241625
edited Jul 23 at 4:28
asked Jul 23 at 0:49
Tony
2,1241625
asked Jul 23 at 0:49
Tony
2,1241625
asked Jul 23 at 0:49
Tony
2,1241625
2,1241625
Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
1
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30
add a comment |Â
Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
1
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30
Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
1
1
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Here an outline about how you can get to the result you're looking for.
- Given a sub-$sigma$-algebra $mathcalG$ of $mathcalF$ and given $FinmathcalF$, define $$mathbbP(F|mathcalG) : left(Omega,mathcalGright) rightarrow left(mathbbR,mathcalB_mathbbRright)$$
as the (essentially) unique $left(Omega,mathcalGright) - left(mathbbR,mathcalB_mathbbRright)$ -measurable function such that $$forall Gin mathcalG, mathbbP(Fcap G)=int_G mathbbP(F|mathcalG)operatornamedmathbbP,$$
via Radon-Nikodym theorem. - If $mathcalG$ is a sub-$sigma$-algebra of $mathcalF$ and $X : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $$mathbbP_mathcalG(A):=mathbbP(Xin A | mathcalG).$$
- If $Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $FinmathcalF$, the map $mathbbP(F|sigma(Y))$ is $left(Omega,sigma(Y)right)-left(mathbbR,mathcalB_mathbbRright)$ - measurable, and so there exists $varphi :(mathbbR,mathcalB_mathbbR)rightarrow (mathbbR,mathcalB_mathbbR)$ such that $varphi circ Y = mathbbP(F|sigma(Y))$. Notice that if $psi$ is another map that does the same work, then $$varphi=psi$$ $mathbbP_Y$-a.e.. So, define $mathbbP(F|Y):=varphi$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $mathbbP_Y(A):=mathbbP(Xin A|Y)$. Then we have $$mathbbP_Y(A)circ Y=mathbbP_X(A).$$ If $yinmathbbR$, let's denote $mathbbP_Y(A)(y)$ with the less clumsy notation $mathbbP_Y=y(A)$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $A,BinmathcalB_mathbbR$, then: $$mathbbP(Xin Acap Yin B)= int_Y^-1(B)mathbbPleft(Xin A|sigma(Y)right) operatornamedmathbbP = int_Y^-1(B)mathbbP_sigma(Y)(A) operatornamedmathbbP \ = int_Y^-1(B)mathbbP_Y(A)circ Y operatornamedmathbbP = int_BmathbbP_Y(A) operatornamedmathbbP_Y = int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y).$$
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $mathbbP_(X,Y)$ has a density w.r.t. the Lebesgue measure on $mathbbR^2$, say $f_(X,Y)$, then also $Y$ has one with respect to Lebesgue measure on $mathbbR$, say $f_Y$ and:
$$forall AinmathcalB_mathbbR, textfor mathbbP_Y-a.e. yinmathbbR, mathbbP_Y=y(A)=int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx.$$
In order to prove that, fix $AinmathcalB_mathbbR$, and notice that $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right)operatornamedmathbbP_Y(y) = int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right) f_Y(y) operatornamedy \ = int_B left( int_A f_(X,Y)(x,y)operatorname dx right) operatornamedy = int_Atimes B f_(X,Y)(x,y)operatornamedxoperatornamedy = mathbbP_(X,Y)(Atimes B) = mathbbP(Xin Acap Yin B)= int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y)$$
and so $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A) right)operatornamedmathbbP_Y(y)=0,$$
and then $$int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A)=0$$
for $mathbbP_Y$-a.e. $yinmathbbR$.
answered Jul 23 at 5:23
Bob
1,517522
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
add a comment |Â
up vote
2
down vote
Here is a very elementary practical problems about election polling that illustrates how to get a
posterior probability interval (credible interval) from a prior and data.
Prior. Expert's prior on proportion $theta$ in favor of Candidate A is $mathsfBeta(alpha_0=330, beta_0=270),$ which has $P(0.51 < theta < 0.59) approx 0.95.$ Mean, median, mode 0.55. (Expert thinks candidate will win, but not by much.)
Data. $x = 620$ of $n = 1000$ prospective voters polled favor the candidate.
Bayes' Theorem gives Posterior.
$$ p(theta | x) propto p(theta) times p(x|theta) propto theta^alpha_0-1(1-theta)^beta_0 - 1 times
theta^x(1-theta)^n-x \
= theta^alpha_0 + x -1(1-theta)^beta_0 + n - x -1
= theta^alpha_n - 1(1-theta)^beta_n -1.$$
Notice that constants of integration are omitted, hence the use of
$propto$ ('proportional to') instead of $=.$
Because prior and likelihood are 'conjugate' (mathematically compatible) we can notice that the posterior has the kernel of $mathsfBeta(alpha_n=alpha_0 + x, beta_n = beta_0 + n - x),$ so we can identify the exact posterior distribution $mathsfBetaalpha_n = 950, beta_n = 650$ without having to evaluate the integral in the denominator of the right-hand side of Bayes' Theorem.
Posterior probability interval. One way to get a 95% credible interval is to take quantiles 0.025 and 0.975 of $mathsfBeta(950, 650)$ to obtain
$(0.570, 0.618),$ using R statistical software.
qbeta(c(.025, .975), 950, 650)
[1] 0.5695848 0.6176932
answered Jul 23 at 2:38
BruceET
33.2k61440
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here an outline about how you can get to the result you're looking for.
- Given a sub-$sigma$-algebra $mathcalG$ of $mathcalF$ and given $FinmathcalF$, define $$mathbbP(F|mathcalG) : left(Omega,mathcalGright) rightarrow left(mathbbR,mathcalB_mathbbRright)$$
as the (essentially) unique $left(Omega,mathcalGright) - left(mathbbR,mathcalB_mathbbRright)$ -measurable function such that $$forall Gin mathcalG, mathbbP(Fcap G)=int_G mathbbP(F|mathcalG)operatornamedmathbbP,$$
via Radon-Nikodym theorem. - If $mathcalG$ is a sub-$sigma$-algebra of $mathcalF$ and $X : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $$mathbbP_mathcalG(A):=mathbbP(Xin A | mathcalG).$$
- If $Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $FinmathcalF$, the map $mathbbP(F|sigma(Y))$ is $left(Omega,sigma(Y)right)-left(mathbbR,mathcalB_mathbbRright)$ - measurable, and so there exists $varphi :(mathbbR,mathcalB_mathbbR)rightarrow (mathbbR,mathcalB_mathbbR)$ such that $varphi circ Y = mathbbP(F|sigma(Y))$. Notice that if $psi$ is another map that does the same work, then $$varphi=psi$$ $mathbbP_Y$-a.e.. So, define $mathbbP(F|Y):=varphi$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $mathbbP_Y(A):=mathbbP(Xin A|Y)$. Then we have $$mathbbP_Y(A)circ Y=mathbbP_X(A).$$ If $yinmathbbR$, let's denote $mathbbP_Y(A)(y)$ with the less clumsy notation $mathbbP_Y=y(A)$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $A,BinmathcalB_mathbbR$, then: $$mathbbP(Xin Acap Yin B)= int_Y^-1(B)mathbbPleft(Xin A|sigma(Y)right) operatornamedmathbbP = int_Y^-1(B)mathbbP_sigma(Y)(A) operatornamedmathbbP \ = int_Y^-1(B)mathbbP_Y(A)circ Y operatornamedmathbbP = int_BmathbbP_Y(A) operatornamedmathbbP_Y = int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y).$$
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $mathbbP_(X,Y)$ has a density w.r.t. the Lebesgue measure on $mathbbR^2$, say $f_(X,Y)$, then also $Y$ has one with respect to Lebesgue measure on $mathbbR$, say $f_Y$ and:
$$forall AinmathcalB_mathbbR, textfor mathbbP_Y-a.e. yinmathbbR, mathbbP_Y=y(A)=int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx.$$
In order to prove that, fix $AinmathcalB_mathbbR$, and notice that $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right)operatornamedmathbbP_Y(y) = int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right) f_Y(y) operatornamedy \ = int_B left( int_A f_(X,Y)(x,y)operatorname dx right) operatornamedy = int_Atimes B f_(X,Y)(x,y)operatornamedxoperatornamedy = mathbbP_(X,Y)(Atimes B) = mathbbP(Xin Acap Yin B)= int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y)$$
and so $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A) right)operatornamedmathbbP_Y(y)=0,$$
and then $$int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A)=0$$
for $mathbbP_Y$-a.e. $yinmathbbR$.
answered Jul 23 at 5:23
Bob
1,517522
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
add a comment |Â
up vote
2
down vote
accepted
Here an outline about how you can get to the result you're looking for.
- Given a sub-$sigma$-algebra $mathcalG$ of $mathcalF$ and given $FinmathcalF$, define $$mathbbP(F|mathcalG) : left(Omega,mathcalGright) rightarrow left(mathbbR,mathcalB_mathbbRright)$$
as the (essentially) unique $left(Omega,mathcalGright) - left(mathbbR,mathcalB_mathbbRright)$ -measurable function such that $$forall Gin mathcalG, mathbbP(Fcap G)=int_G mathbbP(F|mathcalG)operatornamedmathbbP,$$
via Radon-Nikodym theorem. - If $mathcalG$ is a sub-$sigma$-algebra of $mathcalF$ and $X : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $$mathbbP_mathcalG(A):=mathbbP(Xin A | mathcalG).$$
- If $Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $FinmathcalF$, the map $mathbbP(F|sigma(Y))$ is $left(Omega,sigma(Y)right)-left(mathbbR,mathcalB_mathbbRright)$ - measurable, and so there exists $varphi :(mathbbR,mathcalB_mathbbR)rightarrow (mathbbR,mathcalB_mathbbR)$ such that $varphi circ Y = mathbbP(F|sigma(Y))$. Notice that if $psi$ is another map that does the same work, then $$varphi=psi$$ $mathbbP_Y$-a.e.. So, define $mathbbP(F|Y):=varphi$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $mathbbP_Y(A):=mathbbP(Xin A|Y)$. Then we have $$mathbbP_Y(A)circ Y=mathbbP_X(A).$$ If $yinmathbbR$, let's denote $mathbbP_Y(A)(y)$ with the less clumsy notation $mathbbP_Y=y(A)$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $A,BinmathcalB_mathbbR$, then: $$mathbbP(Xin Acap Yin B)= int_Y^-1(B)mathbbPleft(Xin A|sigma(Y)right) operatornamedmathbbP = int_Y^-1(B)mathbbP_sigma(Y)(A) operatornamedmathbbP \ = int_Y^-1(B)mathbbP_Y(A)circ Y operatornamedmathbbP = int_BmathbbP_Y(A) operatornamedmathbbP_Y = int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y).$$
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $mathbbP_(X,Y)$ has a density w.r.t. the Lebesgue measure on $mathbbR^2$, say $f_(X,Y)$, then also $Y$ has one with respect to Lebesgue measure on $mathbbR$, say $f_Y$ and:
$$forall AinmathcalB_mathbbR, textfor mathbbP_Y-a.e. yinmathbbR, mathbbP_Y=y(A)=int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx.$$
In order to prove that, fix $AinmathcalB_mathbbR$, and notice that $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right)operatornamedmathbbP_Y(y) = int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right) f_Y(y) operatornamedy \ = int_B left( int_A f_(X,Y)(x,y)operatorname dx right) operatornamedy = int_Atimes B f_(X,Y)(x,y)operatornamedxoperatornamedy = mathbbP_(X,Y)(Atimes B) = mathbbP(Xin Acap Yin B)= int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y)$$
and so $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A) right)operatornamedmathbbP_Y(y)=0,$$
and then $$int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A)=0$$
for $mathbbP_Y$-a.e. $yinmathbbR$.
answered Jul 23 at 5:23
Bob
1,517522
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here an outline about how you can get to the result you're looking for.
- Given a sub-$sigma$-algebra $mathcalG$ of $mathcalF$ and given $FinmathcalF$, define $$mathbbP(F|mathcalG) : left(Omega,mathcalGright) rightarrow left(mathbbR,mathcalB_mathbbRright)$$
as the (essentially) unique $left(Omega,mathcalGright) - left(mathbbR,mathcalB_mathbbRright)$ -measurable function such that $$forall Gin mathcalG, mathbbP(Fcap G)=int_G mathbbP(F|mathcalG)operatornamedmathbbP,$$
via Radon-Nikodym theorem. - If $mathcalG$ is a sub-$sigma$-algebra of $mathcalF$ and $X : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $$mathbbP_mathcalG(A):=mathbbP(Xin A | mathcalG).$$
- If $Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $FinmathcalF$, the map $mathbbP(F|sigma(Y))$ is $left(Omega,sigma(Y)right)-left(mathbbR,mathcalB_mathbbRright)$ - measurable, and so there exists $varphi :(mathbbR,mathcalB_mathbbR)rightarrow (mathbbR,mathcalB_mathbbR)$ such that $varphi circ Y = mathbbP(F|sigma(Y))$. Notice that if $psi$ is another map that does the same work, then $$varphi=psi$$ $mathbbP_Y$-a.e.. So, define $mathbbP(F|Y):=varphi$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $mathbbP_Y(A):=mathbbP(Xin A|Y)$. Then we have $$mathbbP_Y(A)circ Y=mathbbP_X(A).$$ If $yinmathbbR$, let's denote $mathbbP_Y(A)(y)$ with the less clumsy notation $mathbbP_Y=y(A)$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $A,BinmathcalB_mathbbR$, then: $$mathbbP(Xin Acap Yin B)= int_Y^-1(B)mathbbPleft(Xin A|sigma(Y)right) operatornamedmathbbP = int_Y^-1(B)mathbbP_sigma(Y)(A) operatornamedmathbbP \ = int_Y^-1(B)mathbbP_Y(A)circ Y operatornamedmathbbP = int_BmathbbP_Y(A) operatornamedmathbbP_Y = int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y).$$
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $mathbbP_(X,Y)$ has a density w.r.t. the Lebesgue measure on $mathbbR^2$, say $f_(X,Y)$, then also $Y$ has one with respect to Lebesgue measure on $mathbbR$, say $f_Y$ and:
$$forall AinmathcalB_mathbbR, textfor mathbbP_Y-a.e. yinmathbbR, mathbbP_Y=y(A)=int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx.$$
In order to prove that, fix $AinmathcalB_mathbbR$, and notice that $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right)operatornamedmathbbP_Y(y) = int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right) f_Y(y) operatornamedy \ = int_B left( int_A f_(X,Y)(x,y)operatorname dx right) operatornamedy = int_Atimes B f_(X,Y)(x,y)operatornamedxoperatornamedy = mathbbP_(X,Y)(Atimes B) = mathbbP(Xin Acap Yin B)= int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y)$$
and so $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A) right)operatornamedmathbbP_Y(y)=0,$$
and then $$int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A)=0$$
for $mathbbP_Y$-a.e. $yinmathbbR$.
answered Jul 23 at 5:23
Bob
1,517522
Here an outline about how you can get to the result you're looking for.
- Given a sub-$sigma$-algebra $mathcalG$ of $mathcalF$ and given $FinmathcalF$, define $$mathbbP(F|mathcalG) : left(Omega,mathcalGright) rightarrow left(mathbbR,mathcalB_mathbbRright)$$
as the (essentially) unique $left(Omega,mathcalGright) - left(mathbbR,mathcalB_mathbbRright)$ -measurable function such that $$forall Gin mathcalG, mathbbP(Fcap G)=int_G mathbbP(F|mathcalG)operatornamedmathbbP,$$
via Radon-Nikodym theorem. - If $mathcalG$ is a sub-$sigma$-algebra of $mathcalF$ and $X : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $$mathbbP_mathcalG(A):=mathbbP(Xin A | mathcalG).$$
- If $Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $FinmathcalF$, the map $mathbbP(F|sigma(Y))$ is $left(Omega,sigma(Y)right)-left(mathbbR,mathcalB_mathbbRright)$ - measurable, and so there exists $varphi :(mathbbR,mathcalB_mathbbR)rightarrow (mathbbR,mathcalB_mathbbR)$ such that $varphi circ Y = mathbbP(F|sigma(Y))$. Notice that if $psi$ is another map that does the same work, then $$varphi=psi$$ $mathbbP_Y$-a.e.. So, define $mathbbP(F|Y):=varphi$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $AinmathcalB_mathbbR$, define $mathbbP_Y(A):=mathbbP(Xin A|Y)$. Then we have $$mathbbP_Y(A)circ Y=mathbbP_X(A).$$ If $yinmathbbR$, let's denote $mathbbP_Y(A)(y)$ with the less clumsy notation $mathbbP_Y=y(A)$.
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $A,BinmathcalB_mathbbR$, then: $$mathbbP(Xin Acap Yin B)= int_Y^-1(B)mathbbPleft(Xin A|sigma(Y)right) operatornamedmathbbP = int_Y^-1(B)mathbbP_sigma(Y)(A) operatornamedmathbbP \ = int_Y^-1(B)mathbbP_Y(A)circ Y operatornamedmathbbP = int_BmathbbP_Y(A) operatornamedmathbbP_Y = int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y).$$
- If $X,Y : left(Omega,mathcalFright) rightarrow left(mathbbR,mathcalB_mathbbRright)$ and $mathbbP_(X,Y)$ has a density w.r.t. the Lebesgue measure on $mathbbR^2$, say $f_(X,Y)$, then also $Y$ has one with respect to Lebesgue measure on $mathbbR$, say $f_Y$ and:
$$forall AinmathcalB_mathbbR, textfor mathbbP_Y-a.e. yinmathbbR, mathbbP_Y=y(A)=int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx.$$
In order to prove that, fix $AinmathcalB_mathbbR$, and notice that $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right)operatornamedmathbbP_Y(y) = int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx right) f_Y(y) operatornamedy \ = int_B left( int_A f_(X,Y)(x,y)operatorname dx right) operatornamedy = int_Atimes B f_(X,Y)(x,y)operatornamedxoperatornamedy = mathbbP_(X,Y)(Atimes B) = mathbbP(Xin Acap Yin B)= int_BmathbbP_Y=y(A) operatornamedmathbbP_Y(y)$$
and so $$forall BinmathcalB_mathbbR, int_B left( int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A) right)operatornamedmathbbP_Y(y)=0,$$
and then $$int_A fracf_(X,Y)(x,y)f_Y(y)operatorname dx-mathbbP_Y=y(A)=0$$
for $mathbbP_Y$-a.e. $yinmathbbR$.
answered Jul 23 at 5:23
Bob
1,517522
edited Jul 23 at 5:29
answered Jul 23 at 5:23
Bob
1,517522
answered Jul 23 at 5:23
Bob
1,517522
answered Jul 23 at 5:23
Bob
1,517522
1,517522
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
add a comment |Â
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
Thanks a lot! Could I ask why the existence and a.s. uniqueness of the third step holds?
– Tony
Jul 25 at 4:26
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the existence, you can find a proof in the book Probability with martingales by David Williams (the only lemma in the appendix to chapter 3). It is a beautiful result, because basically states that it is true what intuitively you are expecting to happens: if a random variable is known if you know all the information in $Y$, then this random variable is a function of $Y$. It is a factorization theorem, however, I think about such a result as "Radon-Nikodym theorem for information valued measures".
– Bob
Jul 25 at 4:56
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
About the uniqueness, suppose $varphi, psi$ satisfy that relation. Then $varphicirc Y$ and $psicirc Y$ are both a version of $mathbbP(F|sigma(Y))$, so they differ at most on a set of $mathbbP$ measure zero. Suppose to get a contradiction that $varphi$ and $psi$ differ on a set of non-null $mathbbP_Y$ measure, say $A$. Then $0neqmathbbP_Y(A)= mathbbP(Y^-1(A))$ and $forallomegain Y^-1(A), varphicirc Y (omega) = varphi( Y (omega))neqpsi( Y (omega)) = psicirc Y (omega)$, and so they differ on a set of positive $mathbbP$ measure, absurd.
– Bob
Jul 25 at 4:57
add a comment |Â
up vote
2
down vote
Here is a very elementary practical problems about election polling that illustrates how to get a
posterior probability interval (credible interval) from a prior and data.
Prior. Expert's prior on proportion $theta$ in favor of Candidate A is $mathsfBeta(alpha_0=330, beta_0=270),$ which has $P(0.51 < theta < 0.59) approx 0.95.$ Mean, median, mode 0.55. (Expert thinks candidate will win, but not by much.)
Data. $x = 620$ of $n = 1000$ prospective voters polled favor the candidate.
Bayes' Theorem gives Posterior.
$$ p(theta | x) propto p(theta) times p(x|theta) propto theta^alpha_0-1(1-theta)^beta_0 - 1 times
theta^x(1-theta)^n-x \
= theta^alpha_0 + x -1(1-theta)^beta_0 + n - x -1
= theta^alpha_n - 1(1-theta)^beta_n -1.$$
Notice that constants of integration are omitted, hence the use of
$propto$ ('proportional to') instead of $=.$
Because prior and likelihood are 'conjugate' (mathematically compatible) we can notice that the posterior has the kernel of $mathsfBeta(alpha_n=alpha_0 + x, beta_n = beta_0 + n - x),$ so we can identify the exact posterior distribution $mathsfBetaalpha_n = 950, beta_n = 650$ without having to evaluate the integral in the denominator of the right-hand side of Bayes' Theorem.
Posterior probability interval. One way to get a 95% credible interval is to take quantiles 0.025 and 0.975 of $mathsfBeta(950, 650)$ to obtain
$(0.570, 0.618),$ using R statistical software.
qbeta(c(.025, .975), 950, 650)
[1] 0.5695848 0.6176932
answered Jul 23 at 2:38
BruceET
33.2k61440
add a comment |Â
up vote
2
down vote
Here is a very elementary practical problems about election polling that illustrates how to get a
posterior probability interval (credible interval) from a prior and data.
Prior. Expert's prior on proportion $theta$ in favor of Candidate A is $mathsfBeta(alpha_0=330, beta_0=270),$ which has $P(0.51 < theta < 0.59) approx 0.95.$ Mean, median, mode 0.55. (Expert thinks candidate will win, but not by much.)
Data. $x = 620$ of $n = 1000$ prospective voters polled favor the candidate.
Bayes' Theorem gives Posterior.
$$ p(theta | x) propto p(theta) times p(x|theta) propto theta^alpha_0-1(1-theta)^beta_0 - 1 times
theta^x(1-theta)^n-x \
= theta^alpha_0 + x -1(1-theta)^beta_0 + n - x -1
= theta^alpha_n - 1(1-theta)^beta_n -1.$$
Notice that constants of integration are omitted, hence the use of
$propto$ ('proportional to') instead of $=.$
Because prior and likelihood are 'conjugate' (mathematically compatible) we can notice that the posterior has the kernel of $mathsfBeta(alpha_n=alpha_0 + x, beta_n = beta_0 + n - x),$ so we can identify the exact posterior distribution $mathsfBetaalpha_n = 950, beta_n = 650$ without having to evaluate the integral in the denominator of the right-hand side of Bayes' Theorem.
Posterior probability interval. One way to get a 95% credible interval is to take quantiles 0.025 and 0.975 of $mathsfBeta(950, 650)$ to obtain
$(0.570, 0.618),$ using R statistical software.
qbeta(c(.025, .975), 950, 650)
[1] 0.5695848 0.6176932
answered Jul 23 at 2:38
BruceET
33.2k61440
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is a very elementary practical problems about election polling that illustrates how to get a
posterior probability interval (credible interval) from a prior and data.
Prior. Expert's prior on proportion $theta$ in favor of Candidate A is $mathsfBeta(alpha_0=330, beta_0=270),$ which has $P(0.51 < theta < 0.59) approx 0.95.$ Mean, median, mode 0.55. (Expert thinks candidate will win, but not by much.)
Data. $x = 620$ of $n = 1000$ prospective voters polled favor the candidate.
Bayes' Theorem gives Posterior.
$$ p(theta | x) propto p(theta) times p(x|theta) propto theta^alpha_0-1(1-theta)^beta_0 - 1 times
theta^x(1-theta)^n-x \
= theta^alpha_0 + x -1(1-theta)^beta_0 + n - x -1
= theta^alpha_n - 1(1-theta)^beta_n -1.$$
Notice that constants of integration are omitted, hence the use of
$propto$ ('proportional to') instead of $=.$
Because prior and likelihood are 'conjugate' (mathematically compatible) we can notice that the posterior has the kernel of $mathsfBeta(alpha_n=alpha_0 + x, beta_n = beta_0 + n - x),$ so we can identify the exact posterior distribution $mathsfBetaalpha_n = 950, beta_n = 650$ without having to evaluate the integral in the denominator of the right-hand side of Bayes' Theorem.
Posterior probability interval. One way to get a 95% credible interval is to take quantiles 0.025 and 0.975 of $mathsfBeta(950, 650)$ to obtain
$(0.570, 0.618),$ using R statistical software.
qbeta(c(.025, .975), 950, 650)
[1] 0.5695848 0.6176932
answered Jul 23 at 2:38
BruceET
33.2k61440
Here is a very elementary practical problems about election polling that illustrates how to get a
posterior probability interval (credible interval) from a prior and data.
Prior. Expert's prior on proportion $theta$ in favor of Candidate A is $mathsfBeta(alpha_0=330, beta_0=270),$ which has $P(0.51 < theta < 0.59) approx 0.95.$ Mean, median, mode 0.55. (Expert thinks candidate will win, but not by much.)
Data. $x = 620$ of $n = 1000$ prospective voters polled favor the candidate.
Bayes' Theorem gives Posterior.
$$ p(theta | x) propto p(theta) times p(x|theta) propto theta^alpha_0-1(1-theta)^beta_0 - 1 times
theta^x(1-theta)^n-x \
= theta^alpha_0 + x -1(1-theta)^beta_0 + n - x -1
= theta^alpha_n - 1(1-theta)^beta_n -1.$$
Notice that constants of integration are omitted, hence the use of
$propto$ ('proportional to') instead of $=.$
Because prior and likelihood are 'conjugate' (mathematically compatible) we can notice that the posterior has the kernel of $mathsfBeta(alpha_n=alpha_0 + x, beta_n = beta_0 + n - x),$ so we can identify the exact posterior distribution $mathsfBetaalpha_n = 950, beta_n = 650$ without having to evaluate the integral in the denominator of the right-hand side of Bayes' Theorem.
Posterior probability interval. One way to get a 95% credible interval is to take quantiles 0.025 and 0.975 of $mathsfBeta(950, 650)$ to obtain
$(0.570, 0.618),$ using R statistical software.
qbeta(c(.025, .975), 950, 650)
[1] 0.5695848 0.6176932
answered Jul 23 at 2:38
BruceET
33.2k61440
edited Jul 23 at 2:44
answered Jul 23 at 2:38
BruceET
33.2k61440
answered Jul 23 at 2:38
BruceET
33.2k61440
answered Jul 23 at 2:38
BruceET
33.2k61440
33.2k61440
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Out of curiosity, where is this definition of conditional probability density function quoted from?
– littleO
Jul 23 at 2:46
1
@littleO It is from lecture notes. I am not aware of its original source. This answer seems to use the same definition. math.stackexchange.com/questions/496608/…
– Tony
Jul 23 at 4:29
I am very glad to know if you have any other definition for the conditional density function? Maybe some others are more friendly with the Bayes rule.
– Tony
Jul 23 at 4:30