Differentials of Measures in the context of Radon Nikodym Derivatives.Calculating Radon Nikodym...
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Differentials of Measures in the context of Radon Nikodym Derivatives.
Calculating Radon Nikodym derivativeNotation when integrating with respect to a measureConfusions about Radon-Nikodym derivative and dominating measuresFinding an explicit Radon-Nikodym derivativeFinding Radon-Nikodym derivativeRadon-Nikodym derivative of an atomic measureFinding the Radon-Nikodym DerivativeCalculating Explicit Radon-Nikodym DerivativesReweighting Radon Nikodym DerivativeConstruction of a “density” or a Radon-Nikodym for a Semicontinuous DistributionRadon-Nikodym derivatives. Royden, Problem 33.Radon-Nikodym derivative of a convolution of signed measures
$begingroup$
I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.
Over measurable space $([0,1],mathcal{B}_{[0,1]}), mu:=nu+delta_{0}.$
$forall Ainmathcal{B}_{[0,1]}$ define $nu(A):=|A|.$
Finally, $delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.
$mu >>nu$ is clear and therefore $dnu/dmu$ exists.
$dnu/dmu overset{*}= frac{nu(dy)}{mu(dy)}=frac{|dy|}{|dy|+delta_{0}(dy)} = 1/frac{|dy|+delta_{0}(dy)}{|dy|} =1/(1+delta_{0}(dy)/|dy|)$.
It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.
UPDATE:
It is clear that $nu(A)=int_{A}dnu= int_{A}frac{dnu}{dmu}dmuLeftarrow frac{dnu}{dmu}=mathbf{1}_{{X/0}}$ to cancel out the dirac measure part in $mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?
probability-theory measure-theory radon-nikodym
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
$endgroup$
add a comment |
$begingroup$
I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.
Over measurable space $([0,1],mathcal{B}_{[0,1]}), mu:=nu+delta_{0}.$
$forall Ainmathcal{B}_{[0,1]}$ define $nu(A):=|A|.$
Finally, $delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.
$mu >>nu$ is clear and therefore $dnu/dmu$ exists.
$dnu/dmu overset{*}= frac{nu(dy)}{mu(dy)}=frac{|dy|}{|dy|+delta_{0}(dy)} = 1/frac{|dy|+delta_{0}(dy)}{|dy|} =1/(1+delta_{0}(dy)/|dy|)$.
It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.
UPDATE:
It is clear that $nu(A)=int_{A}dnu= int_{A}frac{dnu}{dmu}dmuLeftarrow frac{dnu}{dmu}=mathbf{1}_{{X/0}}$ to cancel out the dirac measure part in $mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?
probability-theory measure-theory radon-nikodym
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
$endgroup$
$begingroup$
The "differential" is just notation and is defined by the integral equation.
$endgroup$
– user251257
Mar 20 at 0:17
$begingroup$
@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
$endgroup$
– BayesIsBae
Mar 20 at 1:27
$begingroup$
As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
$endgroup$
– user251257
Mar 20 at 1:32
$begingroup$
@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
$endgroup$
– BayesIsBae
Mar 20 at 1:59
add a comment |
$begingroup$
I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.
Over measurable space $([0,1],mathcal{B}_{[0,1]}), mu:=nu+delta_{0}.$
$forall Ainmathcal{B}_{[0,1]}$ define $nu(A):=|A|.$
Finally, $delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.
$mu >>nu$ is clear and therefore $dnu/dmu$ exists.
$dnu/dmu overset{*}= frac{nu(dy)}{mu(dy)}=frac{|dy|}{|dy|+delta_{0}(dy)} = 1/frac{|dy|+delta_{0}(dy)}{|dy|} =1/(1+delta_{0}(dy)/|dy|)$.
It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.
UPDATE:
It is clear that $nu(A)=int_{A}dnu= int_{A}frac{dnu}{dmu}dmuLeftarrow frac{dnu}{dmu}=mathbf{1}_{{X/0}}$ to cancel out the dirac measure part in $mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?
probability-theory measure-theory radon-nikodym
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
$endgroup$
I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.
Over measurable space $([0,1],mathcal{B}_{[0,1]}), mu:=nu+delta_{0}.$
$forall Ainmathcal{B}_{[0,1]}$ define $nu(A):=|A|.$
Finally, $delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.
$mu >>nu$ is clear and therefore $dnu/dmu$ exists.
$dnu/dmu overset{*}= frac{nu(dy)}{mu(dy)}=frac{|dy|}{|dy|+delta_{0}(dy)} = 1/frac{|dy|+delta_{0}(dy)}{|dy|} =1/(1+delta_{0}(dy)/|dy|)$.
It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.
UPDATE:
It is clear that $nu(A)=int_{A}dnu= int_{A}frac{dnu}{dmu}dmuLeftarrow frac{dnu}{dmu}=mathbf{1}_{{X/0}}$ to cancel out the dirac measure part in $mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?
probability-theory measure-theory radon-nikodym
probability-theory measure-theory radon-nikodym
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
edited Mar 20 at 0:14
BayesIsBae
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
asked Mar 19 at 23:40
BayesIsBaeBayesIsBae
887
887
$begingroup$
The "differential" is just notation and is defined by the integral equation.
$endgroup$
– user251257
Mar 20 at 0:17
$begingroup$
@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
$endgroup$
– BayesIsBae
Mar 20 at 1:27
$begingroup$
As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
$endgroup$
– user251257
Mar 20 at 1:32
$begingroup$
@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
$endgroup$
– BayesIsBae
Mar 20 at 1:59
add a comment |
$begingroup$
The "differential" is just notation and is defined by the integral equation.
$endgroup$
– user251257
Mar 20 at 0:17
$begingroup$
@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
$endgroup$
– BayesIsBae
Mar 20 at 1:27
$begingroup$
As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
$endgroup$
– user251257
Mar 20 at 1:32
$begingroup$
@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
$endgroup$
– BayesIsBae
Mar 20 at 1:59
$begingroup$
The "differential" is just notation and is defined by the integral equation.
$endgroup$
– user251257
Mar 20 at 0:17
$begingroup$
The "differential" is just notation and is defined by the integral equation.
$endgroup$
– user251257
Mar 20 at 0:17
$begingroup$
@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
$endgroup$
– BayesIsBae
Mar 20 at 1:27
$begingroup$
@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
$endgroup$
– BayesIsBae
Mar 20 at 1:27
$begingroup$
As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
$endgroup$
– user251257
Mar 20 at 1:32
$begingroup$
As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
$endgroup$
– user251257
Mar 20 at 1:32
$begingroup$
@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
$endgroup$
– BayesIsBae
Mar 20 at 1:59
$begingroup$
@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
$endgroup$
– BayesIsBae
Mar 20 at 1:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your analysis was basically correct, and will be made rigorous below. At
$$
frac1{1+frac{delta_0(dy)}{|dy|}}
$$
look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $delta_0(dy)=1$. Therefore, the above is $frac1{1+frac10}=frac1{infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/text{stuff}$ could not be zero, but you forgot about stuff $=infty$.
To get a differential characterization of $frac{dnu}{dmu}$, it can shown that for $mu$ a.e. $x$, you have
$$
frac{dnu}{dmu}(x)=lim_{epsilonto 0}frac{nu((x-epsilon,x+epsilon)cap [0,1])}{mu((x-epsilon,x+epsilon)cap [0,1])}.
$$
This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $xneq 0$, then both the numerator and denominator are equal for all small enough $epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that
$$
frac{dnu}{dmu}=1_{(0,1]}qquad mu ;a.e.
$$
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
$endgroup$
$begingroup$
I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
$endgroup$
– BayesIsBae
Mar 20 at 23:26
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
$endgroup$
– Mike Earnest
Mar 21 at 0:08
$begingroup$
That is very interesting!
$endgroup$
– BayesIsBae
Mar 21 at 2:30
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Sorry about that - I realized only after asking in a previous comment
$endgroup$
– BayesIsBae
Apr 4 at 1:24
add a comment |
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1 Answer
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votes
$begingroup$
Your analysis was basically correct, and will be made rigorous below. At
$$
frac1{1+frac{delta_0(dy)}{|dy|}}
$$
look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $delta_0(dy)=1$. Therefore, the above is $frac1{1+frac10}=frac1{infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/text{stuff}$ could not be zero, but you forgot about stuff $=infty$.
To get a differential characterization of $frac{dnu}{dmu}$, it can shown that for $mu$ a.e. $x$, you have
$$
frac{dnu}{dmu}(x)=lim_{epsilonto 0}frac{nu((x-epsilon,x+epsilon)cap [0,1])}{mu((x-epsilon,x+epsilon)cap [0,1])}.
$$
This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $xneq 0$, then both the numerator and denominator are equal for all small enough $epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that
$$
frac{dnu}{dmu}=1_{(0,1]}qquad mu ;a.e.
$$
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
$endgroup$
$begingroup$
I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
$endgroup$
– BayesIsBae
Mar 20 at 23:26
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
$endgroup$
– Mike Earnest
Mar 21 at 0:08
$begingroup$
That is very interesting!
$endgroup$
– BayesIsBae
Mar 21 at 2:30
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Sorry about that - I realized only after asking in a previous comment
$endgroup$
– BayesIsBae
Apr 4 at 1:24
add a comment |
$begingroup$
Your analysis was basically correct, and will be made rigorous below. At
$$
frac1{1+frac{delta_0(dy)}{|dy|}}
$$
look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $delta_0(dy)=1$. Therefore, the above is $frac1{1+frac10}=frac1{infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/text{stuff}$ could not be zero, but you forgot about stuff $=infty$.
To get a differential characterization of $frac{dnu}{dmu}$, it can shown that for $mu$ a.e. $x$, you have
$$
frac{dnu}{dmu}(x)=lim_{epsilonto 0}frac{nu((x-epsilon,x+epsilon)cap [0,1])}{mu((x-epsilon,x+epsilon)cap [0,1])}.
$$
This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $xneq 0$, then both the numerator and denominator are equal for all small enough $epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that
$$
frac{dnu}{dmu}=1_{(0,1]}qquad mu ;a.e.
$$
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
$endgroup$
$begingroup$
I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
$endgroup$
– BayesIsBae
Mar 20 at 23:26
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
$endgroup$
– Mike Earnest
Mar 21 at 0:08
$begingroup$
That is very interesting!
$endgroup$
– BayesIsBae
Mar 21 at 2:30
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Sorry about that - I realized only after asking in a previous comment
$endgroup$
– BayesIsBae
Apr 4 at 1:24
add a comment |
$begingroup$
Your analysis was basically correct, and will be made rigorous below. At
$$
frac1{1+frac{delta_0(dy)}{|dy|}}
$$
look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $delta_0(dy)=1$. Therefore, the above is $frac1{1+frac10}=frac1{infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/text{stuff}$ could not be zero, but you forgot about stuff $=infty$.
To get a differential characterization of $frac{dnu}{dmu}$, it can shown that for $mu$ a.e. $x$, you have
$$
frac{dnu}{dmu}(x)=lim_{epsilonto 0}frac{nu((x-epsilon,x+epsilon)cap [0,1])}{mu((x-epsilon,x+epsilon)cap [0,1])}.
$$
This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $xneq 0$, then both the numerator and denominator are equal for all small enough $epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that
$$
frac{dnu}{dmu}=1_{(0,1]}qquad mu ;a.e.
$$
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
$endgroup$
Your analysis was basically correct, and will be made rigorous below. At
$$
frac1{1+frac{delta_0(dy)}{|dy|}}
$$
look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $delta_0(dy)=1$. Therefore, the above is $frac1{1+frac10}=frac1{infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/text{stuff}$ could not be zero, but you forgot about stuff $=infty$.
To get a differential characterization of $frac{dnu}{dmu}$, it can shown that for $mu$ a.e. $x$, you have
$$
frac{dnu}{dmu}(x)=lim_{epsilonto 0}frac{nu((x-epsilon,x+epsilon)cap [0,1])}{mu((x-epsilon,x+epsilon)cap [0,1])}.
$$
This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $xneq 0$, then both the numerator and denominator are equal for all small enough $epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that
$$
frac{dnu}{dmu}=1_{(0,1]}qquad mu ;a.e.
$$
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
answered Mar 20 at 5:09
Mike EarnestMike Earnest
27k22152
27k22152
$begingroup$
I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
$endgroup$
– BayesIsBae
Mar 20 at 23:26
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
$endgroup$
– Mike Earnest
Mar 21 at 0:08
$begingroup$
That is very interesting!
$endgroup$
– BayesIsBae
Mar 21 at 2:30
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Sorry about that - I realized only after asking in a previous comment
$endgroup$
– BayesIsBae
Apr 4 at 1:24
add a comment |
$begingroup$
I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
$endgroup$
– BayesIsBae
Mar 20 at 23:26
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
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– Mike Earnest
Mar 21 at 0:08
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That is very interesting!
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– BayesIsBae
Mar 21 at 2:30
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Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
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– Mike Earnest
Apr 4 at 1:22
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Sorry about that - I realized only after asking in a previous comment
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– BayesIsBae
Apr 4 at 1:24
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I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
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– BayesIsBae
Mar 20 at 23:26
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I see. This is surprising as I thought it would be the case that because $mu$ is not absolutely continuous w.r.t. $nu$ that perhaps $ (frac{nu(dy)}{mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper?
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– BayesIsBae
Mar 20 at 23:26
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$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
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– Mike Earnest
Mar 21 at 0:08
$begingroup$
$(frac{nu(dy)}{mu(dy)})^{-1}$ always exists when $frac{nu(dy)}{mu(dy)}$ exists, though it might be infinite. It is only $frac{mu(dy)}{nu(dy)}$ which may not exist. If $frac{mu(dy)}{nu(dy)}$ exists, then it equals $(frac{nu(dy)}{mu(dy)})^{-1}$. @BayesIsBae
$endgroup$
– Mike Earnest
Mar 21 at 0:08
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That is very interesting!
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– BayesIsBae
Mar 21 at 2:30
$begingroup$
That is very interesting!
$endgroup$
– BayesIsBae
Mar 21 at 2:30
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Well, we are working over the measurable space $([0,1],mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space.
$endgroup$
– Mike Earnest
Apr 4 at 1:22
$begingroup$
Sorry about that - I realized only after asking in a previous comment
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– BayesIsBae
Apr 4 at 1:24
$begingroup$
Sorry about that - I realized only after asking in a previous comment
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– BayesIsBae
Apr 4 at 1:24
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The "differential" is just notation and is defined by the integral equation.
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– user251257
Mar 20 at 0:17
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@user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…?
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– BayesIsBae
Mar 20 at 1:27
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As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation.
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– user251257
Mar 20 at 1:32
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@user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now.
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– BayesIsBae
Mar 20 at 1:59