Mixing
Calculation of variance of complicated random variable ( white noise discretization )
$begingroup$
so I have been doing some state estimation and in one part of my work it is necessary to discretize a continous time differential equation with white noise. I understood the discretization process for differntial equation ( deterministic part ) but I do not understand the last equation in deriving the covariance of discretized white noise.
Assuming that A is some matrix and that $x(t)$ is some differentiable vector-valued function mapping into $mathbb{R}$ function. Further assuming that $w(t,omega)$ is some continous time wide-sense stationary stohastic process such that $w:mathbb{R}timesOmega to mathbb{R}^4$ has mean value zero and autocorrelation of all random variables indexed by this process is delta function ( basically said $w(t,omega)$ is considered white noise ).Further random variables indexed by this process have covaraince matrix Q.Let now :
$$dot{x}(t) = Ax(t) + w(t)$$
Now using the procedure as described in Derivation of discretization on wikipedia I obtain the following equation:
$$x[k+1]=e^{AT_s}x[k] + int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
$$u[k]=int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
where $T_s$ is some discretization constant and $$x[k]=x(kT_s)$$
Now covariances of random variables indexed by $u[k]$ should be :
$$mathbb{E}[u[k]*u^{T}[k]]=mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]$$
And I am unable to calculate any further. Wherever I look they seem to use the following identity which I would like to be able to justify but cant:
$$mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]=mathbb{E}[int_{kT_s}^{(k+1)T_s}int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dldt]$$
Could someone justify this for me,or are engineers playing sneaky mathematics again?
EDIT: I understand that in current presentation this is not rigorous and that the differential equation above mathematically has no meaning.Sadly I am not knowledgable in area of Ito Intgration and Stohastic differential equations,so answers avoiding reference to such would be more useful. Material references are welcome
probability stochastic-processes covariance expected-value sde
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
$endgroup$
add a comment |
$begingroup$
so I have been doing some state estimation and in one part of my work it is necessary to discretize a continous time differential equation with white noise. I understood the discretization process for differntial equation ( deterministic part ) but I do not understand the last equation in deriving the covariance of discretized white noise.
Assuming that A is some matrix and that $x(t)$ is some differentiable vector-valued function mapping into $mathbb{R}$ function. Further assuming that $w(t,omega)$ is some continous time wide-sense stationary stohastic process such that $w:mathbb{R}timesOmega to mathbb{R}^4$ has mean value zero and autocorrelation of all random variables indexed by this process is delta function ( basically said $w(t,omega)$ is considered white noise ).Further random variables indexed by this process have covaraince matrix Q.Let now :
$$dot{x}(t) = Ax(t) + w(t)$$
Now using the procedure as described in Derivation of discretization on wikipedia I obtain the following equation:
$$x[k+1]=e^{AT_s}x[k] + int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
$$u[k]=int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
where $T_s$ is some discretization constant and $$x[k]=x(kT_s)$$
Now covariances of random variables indexed by $u[k]$ should be :
$$mathbb{E}[u[k]*u^{T}[k]]=mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]$$
And I am unable to calculate any further. Wherever I look they seem to use the following identity which I would like to be able to justify but cant:
$$mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]=mathbb{E}[int_{kT_s}^{(k+1)T_s}int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dldt]$$
Could someone justify this for me,or are engineers playing sneaky mathematics again?
EDIT: I understand that in current presentation this is not rigorous and that the differential equation above mathematically has no meaning.Sadly I am not knowledgable in area of Ito Intgration and Stohastic differential equations,so answers avoiding reference to such would be more useful. Material references are welcome
probability stochastic-processes covariance expected-value sde
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
$endgroup$
1
$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16
add a comment |
$begingroup$
so I have been doing some state estimation and in one part of my work it is necessary to discretize a continous time differential equation with white noise. I understood the discretization process for differntial equation ( deterministic part ) but I do not understand the last equation in deriving the covariance of discretized white noise.
Assuming that A is some matrix and that $x(t)$ is some differentiable vector-valued function mapping into $mathbb{R}$ function. Further assuming that $w(t,omega)$ is some continous time wide-sense stationary stohastic process such that $w:mathbb{R}timesOmega to mathbb{R}^4$ has mean value zero and autocorrelation of all random variables indexed by this process is delta function ( basically said $w(t,omega)$ is considered white noise ).Further random variables indexed by this process have covaraince matrix Q.Let now :
$$dot{x}(t) = Ax(t) + w(t)$$
Now using the procedure as described in Derivation of discretization on wikipedia I obtain the following equation:
$$x[k+1]=e^{AT_s}x[k] + int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
$$u[k]=int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
where $T_s$ is some discretization constant and $$x[k]=x(kT_s)$$
Now covariances of random variables indexed by $u[k]$ should be :
$$mathbb{E}[u[k]*u^{T}[k]]=mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]$$
And I am unable to calculate any further. Wherever I look they seem to use the following identity which I would like to be able to justify but cant:
$$mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]=mathbb{E}[int_{kT_s}^{(k+1)T_s}int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dldt]$$
Could someone justify this for me,or are engineers playing sneaky mathematics again?
EDIT: I understand that in current presentation this is not rigorous and that the differential equation above mathematically has no meaning.Sadly I am not knowledgable in area of Ito Intgration and Stohastic differential equations,so answers avoiding reference to such would be more useful. Material references are welcome
probability stochastic-processes covariance expected-value sde
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
$endgroup$
so I have been doing some state estimation and in one part of my work it is necessary to discretize a continous time differential equation with white noise. I understood the discretization process for differntial equation ( deterministic part ) but I do not understand the last equation in deriving the covariance of discretized white noise.
Assuming that A is some matrix and that $x(t)$ is some differentiable vector-valued function mapping into $mathbb{R}$ function. Further assuming that $w(t,omega)$ is some continous time wide-sense stationary stohastic process such that $w:mathbb{R}timesOmega to mathbb{R}^4$ has mean value zero and autocorrelation of all random variables indexed by this process is delta function ( basically said $w(t,omega)$ is considered white noise ).Further random variables indexed by this process have covaraince matrix Q.Let now :
$$dot{x}(t) = Ax(t) + w(t)$$
Now using the procedure as described in Derivation of discretization on wikipedia I obtain the following equation:
$$x[k+1]=e^{AT_s}x[k] + int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
$$u[k]=int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl$$
where $T_s$ is some discretization constant and $$x[k]=x(kT_s)$$
Now covariances of random variables indexed by $u[k]$ should be :
$$mathbb{E}[u[k]*u^{T}[k]]=mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]$$
And I am unable to calculate any further. Wherever I look they seem to use the following identity which I would like to be able to justify but cant:
$$mathbb{E}[int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)dl * int_{kT_s}^{(k+1)T_s}w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dt]=mathbb{E}[int_{kT_s}^{(k+1)T_s}int_{kT_s}^{(k+1)T_s}e^{A[(k+1)T_s -l]}w(l)w^{T}(t)e^{A^{T}[(k+1)T_s -t]}dldt]$$
Could someone justify this for me,or are engineers playing sneaky mathematics again?
EDIT: I understand that in current presentation this is not rigorous and that the differential equation above mathematically has no meaning.Sadly I am not knowledgable in area of Ito Intgration and Stohastic differential equations,so answers avoiding reference to such would be more useful. Material references are welcome
probability stochastic-processes covariance expected-value sde
probability stochastic-processes covariance expected-value sde
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
edited Jan 14 at 10:14
TheCoolDrop
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
asked Jan 14 at 9:33
TheCoolDropTheCoolDrop
16210
16210
1
$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16
add a comment |
1
$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16
1
1
$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
At this point you have the product of two definite integrals that are constants to each other, so you can calculate
begin{align}
int_a^b u(s),dscdot int_a^bv(t),dt
&=int_a^b u(s)cdot left[int_a^bv(t),dtright],ds
end{align}
Now $u(s)$ is a constant relative to the integral of $v$, so that one can insert this constant into the inner integral
begin{align}
...&=int_a^bleft[int_a^b u(s)cdot v(t),dtright],ds
end{align}
The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. The probability argument is usually ommited, putting it back in gives
begin{align}
{Bbb E}left[int_a^bint_a^b u(s)cdot v(t),dt,dsright]
&=int_Ωleft[int_a^bint_a^b u(omega,s)cdot v(ω,t),dt,dsright]dP(ω)
end{align}
There you need some Fubini-type result, using compactness of the time intervals and continuity in time of the integrands to obtain boundedness for each $ωinΩ$, and the finiteness and positivity of the probability measure to get the absolute boundedness of the joint integration over $[a,b]^2times Ω$. Then
begin{align}
...
&=int_a^bint_a^b int_Ωleft[u(omega,s)cdot v(ω,t)right]dP(ω),dt,ds
=int_a^bint_a^b {Bbb E}left[u(s)cdot v(t)right],dt,ds.
end{align}
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
At this point you have the product of two definite integrals that are constants to each other, so you can calculate
begin{align}
int_a^b u(s),dscdot int_a^bv(t),dt
&=int_a^b u(s)cdot left[int_a^bv(t),dtright],ds
end{align}
Now $u(s)$ is a constant relative to the integral of $v$, so that one can insert this constant into the inner integral
begin{align}
...&=int_a^bleft[int_a^b u(s)cdot v(t),dtright],ds
end{align}
The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. The probability argument is usually ommited, putting it back in gives
begin{align}
{Bbb E}left[int_a^bint_a^b u(s)cdot v(t),dt,dsright]
&=int_Ωleft[int_a^bint_a^b u(omega,s)cdot v(ω,t),dt,dsright]dP(ω)
end{align}
There you need some Fubini-type result, using compactness of the time intervals and continuity in time of the integrands to obtain boundedness for each $ωinΩ$, and the finiteness and positivity of the probability measure to get the absolute boundedness of the joint integration over $[a,b]^2times Ω$. Then
begin{align}
...
&=int_a^bint_a^b int_Ωleft[u(omega,s)cdot v(ω,t)right]dP(ω),dt,ds
=int_a^bint_a^b {Bbb E}left[u(s)cdot v(t)right],dt,ds.
end{align}
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
add a comment |
$begingroup$
At this point you have the product of two definite integrals that are constants to each other, so you can calculate
begin{align}
int_a^b u(s),dscdot int_a^bv(t),dt
&=int_a^b u(s)cdot left[int_a^bv(t),dtright],ds
end{align}
Now $u(s)$ is a constant relative to the integral of $v$, so that one can insert this constant into the inner integral
begin{align}
...&=int_a^bleft[int_a^b u(s)cdot v(t),dtright],ds
end{align}
The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. The probability argument is usually ommited, putting it back in gives
begin{align}
{Bbb E}left[int_a^bint_a^b u(s)cdot v(t),dt,dsright]
&=int_Ωleft[int_a^bint_a^b u(omega,s)cdot v(ω,t),dt,dsright]dP(ω)
end{align}
There you need some Fubini-type result, using compactness of the time intervals and continuity in time of the integrands to obtain boundedness for each $ωinΩ$, and the finiteness and positivity of the probability measure to get the absolute boundedness of the joint integration over $[a,b]^2times Ω$. Then
begin{align}
...
&=int_a^bint_a^b int_Ωleft[u(omega,s)cdot v(ω,t)right]dP(ω),dt,ds
=int_a^bint_a^b {Bbb E}left[u(s)cdot v(t)right],dt,ds.
end{align}
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
add a comment |
$begingroup$
At this point you have the product of two definite integrals that are constants to each other, so you can calculate
begin{align}
int_a^b u(s),dscdot int_a^bv(t),dt
&=int_a^b u(s)cdot left[int_a^bv(t),dtright],ds
end{align}
Now $u(s)$ is a constant relative to the integral of $v$, so that one can insert this constant into the inner integral
begin{align}
...&=int_a^bleft[int_a^b u(s)cdot v(t),dtright],ds
end{align}
The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. The probability argument is usually ommited, putting it back in gives
begin{align}
{Bbb E}left[int_a^bint_a^b u(s)cdot v(t),dt,dsright]
&=int_Ωleft[int_a^bint_a^b u(omega,s)cdot v(ω,t),dt,dsright]dP(ω)
end{align}
There you need some Fubini-type result, using compactness of the time intervals and continuity in time of the integrands to obtain boundedness for each $ωinΩ$, and the finiteness and positivity of the probability measure to get the absolute boundedness of the joint integration over $[a,b]^2times Ω$. Then
begin{align}
...
&=int_a^bint_a^b int_Ωleft[u(omega,s)cdot v(ω,t)right]dP(ω),dt,ds
=int_a^bint_a^b {Bbb E}left[u(s)cdot v(t)right],dt,ds.
end{align}
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
$endgroup$
At this point you have the product of two definite integrals that are constants to each other, so you can calculate
begin{align}
int_a^b u(s),dscdot int_a^bv(t),dt
&=int_a^b u(s)cdot left[int_a^bv(t),dtright],ds
end{align}
Now $u(s)$ is a constant relative to the integral of $v$, so that one can insert this constant into the inner integral
begin{align}
...&=int_a^bleft[int_a^b u(s)cdot v(t),dtright],ds
end{align}
The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. The probability argument is usually ommited, putting it back in gives
begin{align}
{Bbb E}left[int_a^bint_a^b u(s)cdot v(t),dt,dsright]
&=int_Ωleft[int_a^bint_a^b u(omega,s)cdot v(ω,t),dt,dsright]dP(ω)
end{align}
There you need some Fubini-type result, using compactness of the time intervals and continuity in time of the integrands to obtain boundedness for each $ωinΩ$, and the finiteness and positivity of the probability measure to get the absolute boundedness of the joint integration over $[a,b]^2times Ω$. Then
begin{align}
...
&=int_a^bint_a^b int_Ωleft[u(omega,s)cdot v(ω,t)right]dP(ω),dt,ds
=int_a^bint_a^b {Bbb E}left[u(s)cdot v(t)right],dt,ds.
end{align}
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
answered Jan 14 at 10:37
LutzLLutzL
58.7k42055
58.7k42055
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
add a comment |
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Sadly I am not on that level of measure theory. Besides solving my problem I hope in year or two I will understand completly what you meant, even though I am using the last line of your result intuitively, argumenting that this somehow comes from linearity of expected value operator.
$endgroup$
– TheCoolDrop
Jan 14 at 10:48
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
$begingroup$
Yes, linearity is the basis. The problem is that you change the order of limits, Fubini is an extension of Riemann's re-ordering theorem that allows for instance the formation of the Cauchy product of series.
$endgroup$
– LutzL
Jan 14 at 11:03
add a comment |
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$begingroup$
If you have some integral $int_a^b u(s),ds$ and some other function $v$, then $v(t)int_a^b u(s),ds=int_a^b v(t)u(s),ds$, as the $t$ is unrelated to the integrand and thus $v(t)$ acts as a constant. Nothing more happens at this step. The interesting part is in the next step where you exchange the integral over the probability space of the expectation value with the two time integrations. There you need some Fubini-type result, using compactness of the time intervals and the finiteness and positivity of the probability measure.
$endgroup$
– LutzL
Jan 14 at 10:10
$begingroup$
Sometimes I wonder is it the eyes or the brain that does not see. Thank you a milion. It would be great if you could develop your comment into full answer so that I can accept it.
$endgroup$
– TheCoolDrop
Jan 14 at 10:16