Construction of random elements in Hilbert space which are almost surely orthogonal.
$begingroup$
Let $(mathcal{H},langle cdot, cdot rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(Omega,mathbb{F},P)to (mathcal{H},langle cdot, cdot rangle)$ with $text{supp}(X)not = {0}$, such that
$$
langle X(omega),Y(omega) rangle =0
$$
for almost all $omegainOmega$, i.e. $X$ and $Y$ are almost surely orthogonal.
Question:
I have shown that this can not be done for separable Hilbert spaces $mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?
probability general-topology functional-analysis probability-theory hilbert-spaces
asked Nov 6 '16 at 18:52
MartinMartin
911717
$endgroup$
add a comment |
$begingroup$
Let $(mathcal{H},langle cdot, cdot rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(Omega,mathbb{F},P)to (mathcal{H},langle cdot, cdot rangle)$ with $text{supp}(X)not = {0}$, such that
$$
langle X(omega),Y(omega) rangle =0
$$
for almost all $omegainOmega$, i.e. $X$ and $Y$ are almost surely orthogonal.
Question:
I have shown that this can not be done for separable Hilbert spaces $mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?
probability general-topology functional-analysis probability-theory hilbert-spaces
asked Nov 6 '16 at 18:52
MartinMartin
911717
$endgroup$
$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
$begingroup$
@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
$endgroup$
– Martin
Nov 8 '16 at 19:23
$begingroup$
What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
$begingroup$
Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
$endgroup$
– Martin
Nov 8 '16 at 21:18
add a comment |
$begingroup$
Let $(mathcal{H},langle cdot, cdot rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(Omega,mathbb{F},P)to (mathcal{H},langle cdot, cdot rangle)$ with $text{supp}(X)not = {0}$, such that
$$
langle X(omega),Y(omega) rangle =0
$$
for almost all $omegainOmega$, i.e. $X$ and $Y$ are almost surely orthogonal.
Question:
I have shown that this can not be done for separable Hilbert spaces $mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?
probability general-topology functional-analysis probability-theory hilbert-spaces
asked Nov 6 '16 at 18:52
MartinMartin
911717
$endgroup$
Let $(mathcal{H},langle cdot, cdot rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(Omega,mathbb{F},P)to (mathcal{H},langle cdot, cdot rangle)$ with $text{supp}(X)not = {0}$, such that
$$
langle X(omega),Y(omega) rangle =0
$$
for almost all $omegainOmega$, i.e. $X$ and $Y$ are almost surely orthogonal.
Question:
I have shown that this can not be done for separable Hilbert spaces $mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?
probability general-topology functional-analysis probability-theory hilbert-spaces
probability general-topology functional-analysis probability-theory hilbert-spaces
asked Nov 6 '16 at 18:52
MartinMartin
911717
asked Nov 6 '16 at 18:52
MartinMartin
911717
asked Nov 6 '16 at 18:52
MartinMartin
911717
asked Nov 6 '16 at 18:52
MartinMartin
911717
asked Nov 6 '16 at 18:52
MartinMartin
911717
911717
$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
$begingroup$
@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
$endgroup$
– Martin
Nov 8 '16 at 19:23
$begingroup$
What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
$begingroup$
Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
$endgroup$
– Martin
Nov 8 '16 at 21:18
add a comment |
$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
$begingroup$
@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
$endgroup$
– Martin
Nov 8 '16 at 19:23
$begingroup$
What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
$begingroup$
Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
$endgroup$
– Martin
Nov 8 '16 at 21:18
$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
$begingroup$
@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
$endgroup$
– Martin
Nov 8 '16 at 19:23
$begingroup$
@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
$endgroup$
– Martin
Nov 8 '16 at 19:23
$begingroup$
What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
$begingroup$
What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
$begingroup$
Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
$endgroup$
– Martin
Nov 8 '16 at 21:18
$begingroup$
Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
$endgroup$
– Martin
Nov 8 '16 at 21:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a ${0,1}$ valued measure on $kappa$ in which every subset is measurable.
Let $mathcal H$ be the Hilbert space whose basis is of cardinality $kappa$, and let $(e_{alpha})_{alpha in kappa}$ be a basis. Let $X,Y:kappa to mathcal H$ be the functions $X(alpha) = e_{alpha}$ and $Y(alpha) = e_{alpha'}$, where $alpha'$ denotes the successor ordinal of $alpha$.
Clearly $langle X,Yrangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
add a comment |
$begingroup$
Let $mathcal{A} = mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $mathcal H$ be $L^2(mathcal{A})$. $mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a in mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x in mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a ${0,1}$ valued measure on $kappa$ in which every subset is measurable.
Let $mathcal H$ be the Hilbert space whose basis is of cardinality $kappa$, and let $(e_{alpha})_{alpha in kappa}$ be a basis. Let $X,Y:kappa to mathcal H$ be the functions $X(alpha) = e_{alpha}$ and $Y(alpha) = e_{alpha'}$, where $alpha'$ denotes the successor ordinal of $alpha$.
Clearly $langle X,Yrangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
add a comment |
$begingroup$
Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a ${0,1}$ valued measure on $kappa$ in which every subset is measurable.
Let $mathcal H$ be the Hilbert space whose basis is of cardinality $kappa$, and let $(e_{alpha})_{alpha in kappa}$ be a basis. Let $X,Y:kappa to mathcal H$ be the functions $X(alpha) = e_{alpha}$ and $Y(alpha) = e_{alpha'}$, where $alpha'$ denotes the successor ordinal of $alpha$.
Clearly $langle X,Yrangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
add a comment |
$begingroup$
Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a ${0,1}$ valued measure on $kappa$ in which every subset is measurable.
Let $mathcal H$ be the Hilbert space whose basis is of cardinality $kappa$, and let $(e_{alpha})_{alpha in kappa}$ be a basis. Let $X,Y:kappa to mathcal H$ be the functions $X(alpha) = e_{alpha}$ and $Y(alpha) = e_{alpha'}$, where $alpha'$ denotes the successor ordinal of $alpha$.
Clearly $langle X,Yrangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
Counterexample. This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal. Thus there is a ${0,1}$ valued measure on $kappa$ in which every subset is measurable.
Let $mathcal H$ be the Hilbert space whose basis is of cardinality $kappa$, and let $(e_{alpha})_{alpha in kappa}$ be a basis. Let $X,Y:kappa to mathcal H$ be the functions $X(alpha) = e_{alpha}$ and $Y(alpha) = e_{alpha'}$, where $alpha'$ denotes the successor ordinal of $alpha$.
Clearly $langle X,Yrangle = 0$ everywhere. It remains to show that $X$ and $Y$ are independent. But this follows because any two subsets of $kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved. But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
edited Nov 11 '16 at 12:56
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Nov 10 '16 at 17:24
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
17.8k12247
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
add a comment |
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
1
1
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
$begingroup$
As an aside, if $X$ and $X'$ are independent copies of $X$ defined on the first and second coordinates of the product space $kappa times kappa$, then $langle X,Yrangle$ is non-measurable! This is essentially the point D.J.H. Garling was trying to make in his notes - separability is no mere technicality.
$endgroup$
– Stephen Montgomery-Smith
Nov 10 '16 at 20:01
add a comment |
$begingroup$
Let $mathcal{A} = mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $mathcal H$ be $L^2(mathcal{A})$. $mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a in mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x in mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
$endgroup$
add a comment |
$begingroup$
Let $mathcal{A} = mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $mathcal H$ be $L^2(mathcal{A})$. $mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a in mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x in mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
$endgroup$
add a comment |
$begingroup$
Let $mathcal{A} = mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $mathcal H$ be $L^2(mathcal{A})$. $mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a in mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x in mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
$endgroup$
Let $mathcal{A} = mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $mathcal H$ be $L^2(mathcal{A})$. $mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a in mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x in mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
edited Dec 10 '18 at 14:26
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
answered Dec 10 '18 at 14:19
NeuromathNeuromath
14710
14710
add a comment |
add a comment |
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$begingroup$
I'm not sure what you mean by a random function taking values in a Hilbert space. Usually people mean the Bochner integrable functions on a probability space. And their range is, by definition, inside a separable subspace almost surely.
$endgroup$
– Stephen Montgomery-Smith
Nov 8 '16 at 17:54
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@StephenMontgomery-Smith - $X$ and $Y$ are random Borel elements in $mathcal{H}$ (not random functions, but random variables if you will, such that each $omega$ maps to $X(omega)inmathcal{H}$). We don't assume that $X$ and $Y$ are bochner integrable. This would simply lead the the arguments for separable Hilbert spaces, in which case I have already shown non-existence. I would like to make an argument that almost surely orthogonal random i.i.d. elements are not in general degenerate, but they are if $mathcal{H}$ is separable (or the random elements are Bochner-measurable) they are.
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– Martin
Nov 8 '16 at 19:23
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What precisely do you mean by "random element?" I assume the map $omega mapsto X(omega)$ is measurable, but with respect to which topology on $mathcal H$? The norm topology or the weak topology? (I think the Borel sets generated can be different for non-separable spaces.)
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– Stephen Montgomery-Smith
Nov 8 '16 at 21:08
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Measurable with respect to the sigma algebra generated by the norm topology. Sorry if this was not explicitly written in the question.
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– Martin
Nov 8 '16 at 21:18