Are Covariance Operators based on square integrable stochastic Processes semi-positive definite?
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Given a $textbfsquare integrable stochastic process$ $X$ with $Eleft(Xleft(tright)right)=0$ $forall t $ the $textbfCovariance Operator$ is defined by
beginalign
C_X: L^2 rightarrow L^2
endalign
with
beginalign
C_Xleft(fright)left(tright)&= int Covleft(Xleft(sright),Xleft(tright)right)fleft(sright) ds \
&= int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds.
endalign
The operator $C_X$ is a $textbfHilbert-Schmidt Operator$ on the Hilbert-Space $L^2$.
I have read several times, that
beginalign
langle C_Xleft(fright),f rangle_L^2 = int int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) fleft(tright) ds dt overset!geq 0 ,,,,, forall fin L^2. tag1
endalign
Thus $C_X$ is a $textbfnon-negative operator$ and has $textbfnon-negative eigenvalues$.
I don't understand why (1) is greater than 0. Can anyone give a $textbfproof$ or a $textbfcounter example$?
stochastic-processes covariance positive-definite positive-semidefinite integral-operators
asked Aug 19 at 11:34
RobH
85
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up vote
1
down vote
favorite
Given a $textbfsquare integrable stochastic process$ $X$ with $Eleft(Xleft(tright)right)=0$ $forall t $ the $textbfCovariance Operator$ is defined by
beginalign
C_X: L^2 rightarrow L^2
endalign
with
beginalign
C_Xleft(fright)left(tright)&= int Covleft(Xleft(sright),Xleft(tright)right)fleft(sright) ds \
&= int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds.
endalign
The operator $C_X$ is a $textbfHilbert-Schmidt Operator$ on the Hilbert-Space $L^2$.
I have read several times, that
beginalign
langle C_Xleft(fright),f rangle_L^2 = int int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) fleft(tright) ds dt overset!geq 0 ,,,,, forall fin L^2. tag1
endalign
Thus $C_X$ is a $textbfnon-negative operator$ and has $textbfnon-negative eigenvalues$.
I don't understand why (1) is greater than 0. Can anyone give a $textbfproof$ or a $textbfcounter example$?
stochastic-processes covariance positive-definite positive-semidefinite integral-operators
asked Aug 19 at 11:34
RobH
85
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a $textbfsquare integrable stochastic process$ $X$ with $Eleft(Xleft(tright)right)=0$ $forall t $ the $textbfCovariance Operator$ is defined by
beginalign
C_X: L^2 rightarrow L^2
endalign
with
beginalign
C_Xleft(fright)left(tright)&= int Covleft(Xleft(sright),Xleft(tright)right)fleft(sright) ds \
&= int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds.
endalign
The operator $C_X$ is a $textbfHilbert-Schmidt Operator$ on the Hilbert-Space $L^2$.
I have read several times, that
beginalign
langle C_Xleft(fright),f rangle_L^2 = int int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) fleft(tright) ds dt overset!geq 0 ,,,,, forall fin L^2. tag1
endalign
Thus $C_X$ is a $textbfnon-negative operator$ and has $textbfnon-negative eigenvalues$.
I don't understand why (1) is greater than 0. Can anyone give a $textbfproof$ or a $textbfcounter example$?
stochastic-processes covariance positive-definite positive-semidefinite integral-operators
asked Aug 19 at 11:34
RobH
85
Given a $textbfsquare integrable stochastic process$ $X$ with $Eleft(Xleft(tright)right)=0$ $forall t $ the $textbfCovariance Operator$ is defined by
beginalign
C_X: L^2 rightarrow L^2
endalign
with
beginalign
C_Xleft(fright)left(tright)&= int Covleft(Xleft(sright),Xleft(tright)right)fleft(sright) ds \
&= int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds.
endalign
The operator $C_X$ is a $textbfHilbert-Schmidt Operator$ on the Hilbert-Space $L^2$.
I have read several times, that
beginalign
langle C_Xleft(fright),f rangle_L^2 = int int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) fleft(tright) ds dt overset!geq 0 ,,,,, forall fin L^2. tag1
endalign
Thus $C_X$ is a $textbfnon-negative operator$ and has $textbfnon-negative eigenvalues$.
I don't understand why (1) is greater than 0. Can anyone give a $textbfproof$ or a $textbfcounter example$?
stochastic-processes covariance positive-definite positive-semidefinite integral-operators
asked Aug 19 at 11:34
RobH
85
edited Aug 21 at 9:11
asked Aug 19 at 11:34
RobH
85
asked Aug 19 at 11:34
RobH
85
asked Aug 19 at 11:34
RobH
85
85
add a comment |Â
add a comment |Â
1 Answer
1
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1
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accepted
Observe that
$$
C_X(f)(t)f(t)=f(t)int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds
$$
and integrating, we get
$$
langle C_Xleft(fright),f rangle_L^2 =int C_X(f)(t)f(t) dt=int f(t)left(int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) dsright)dt
$$
hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as
$$
langle C_Xleft(fright),f rangle_L^2 =iint Eleft(fleft(sright)Xleft(sright)f(t)Xleft(tright)right) dsdt
$$
and assuming that we can switch the expectation and the integral, we get
$$
langle C_Xleft(fright),f rangle_L^2 = Eleft(iint fleft(sright)Xleft(sright)f(t)Xleft(tright)dsdtright)
$$
and the double integral in the right hand side is $left(int f(s)X(s)dsright)^2$.
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Observe that
$$
C_X(f)(t)f(t)=f(t)int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds
$$
and integrating, we get
$$
langle C_Xleft(fright),f rangle_L^2 =int C_X(f)(t)f(t) dt=int f(t)left(int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) dsright)dt
$$
hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as
$$
langle C_Xleft(fright),f rangle_L^2 =iint Eleft(fleft(sright)Xleft(sright)f(t)Xleft(tright)right) dsdt
$$
and assuming that we can switch the expectation and the integral, we get
$$
langle C_Xleft(fright),f rangle_L^2 = Eleft(iint fleft(sright)Xleft(sright)f(t)Xleft(tright)dsdtright)
$$
and the double integral in the right hand side is $left(int f(s)X(s)dsright)^2$.
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
add a comment |Â
up vote
1
down vote
accepted
Observe that
$$
C_X(f)(t)f(t)=f(t)int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds
$$
and integrating, we get
$$
langle C_Xleft(fright),f rangle_L^2 =int C_X(f)(t)f(t) dt=int f(t)left(int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) dsright)dt
$$
hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as
$$
langle C_Xleft(fright),f rangle_L^2 =iint Eleft(fleft(sright)Xleft(sright)f(t)Xleft(tright)right) dsdt
$$
and assuming that we can switch the expectation and the integral, we get
$$
langle C_Xleft(fright),f rangle_L^2 = Eleft(iint fleft(sright)Xleft(sright)f(t)Xleft(tright)dsdtright)
$$
and the double integral in the right hand side is $left(int f(s)X(s)dsright)^2$.
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Observe that
$$
C_X(f)(t)f(t)=f(t)int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds
$$
and integrating, we get
$$
langle C_Xleft(fright),f rangle_L^2 =int C_X(f)(t)f(t) dt=int f(t)left(int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) dsright)dt
$$
hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as
$$
langle C_Xleft(fright),f rangle_L^2 =iint Eleft(fleft(sright)Xleft(sright)f(t)Xleft(tright)right) dsdt
$$
and assuming that we can switch the expectation and the integral, we get
$$
langle C_Xleft(fright),f rangle_L^2 = Eleft(iint fleft(sright)Xleft(sright)f(t)Xleft(tright)dsdtright)
$$
and the double integral in the right hand side is $left(int f(s)X(s)dsright)^2$.
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
Observe that
$$
C_X(f)(t)f(t)=f(t)int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) ds
$$
and integrating, we get
$$
langle C_Xleft(fright),f rangle_L^2 =int C_X(f)(t)f(t) dt=int f(t)left(int Eleft(Xleft(sright)Xleft(tright)right)fleft(sright) dsright)dt
$$
hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as
$$
langle C_Xleft(fright),f rangle_L^2 =iint Eleft(fleft(sright)Xleft(sright)f(t)Xleft(tright)right) dsdt
$$
and assuming that we can switch the expectation and the integral, we get
$$
langle C_Xleft(fright),f rangle_L^2 = Eleft(iint fleft(sright)Xleft(sright)f(t)Xleft(tright)dsdtright)
$$
and the double integral in the right hand side is $left(int f(s)X(s)dsright)^2$.
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
edited Aug 21 at 9:49
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
answered Aug 20 at 12:59
Davide Giraudo
121k15147250
121k15147250
add a comment |Â
add a comment |Â
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