Demonstrate the following propierties (State-space representation)
$begingroup$
Having this properties:
$E[x(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
$E[z(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
Demonstrate it with the following assumptions:
$x(k+1)=Phi (k+1,k)x(k)+Gamma (k+1,k)w(k)~$, $kgeq 0$; $x(0)=x_0~$
(is State equation)
$z(k)=H(k)x(k)+v(k), kgeq0$ (observation equation)- Initial state $x_0$, vector n-dimensional gaussian with cov. matrix
$E[x_0x^T_o]=P_0$
The process ${w(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[w(k)w^T(k)]=Q(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[v(k)v^T(k)]=R(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
The initial state and additive noises are mutually independent.
I think it has to be done with Doob theorem but in my case is not working. Thanks in advance,
ordinary-differential-equations stochastic-processes
asked Dec 20 '18 at 13:47
finafina
33
$endgroup$
add a comment |
$begingroup$
Having this properties:
$E[x(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
$E[z(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
Demonstrate it with the following assumptions:
$x(k+1)=Phi (k+1,k)x(k)+Gamma (k+1,k)w(k)~$, $kgeq 0$; $x(0)=x_0~$
(is State equation)
$z(k)=H(k)x(k)+v(k), kgeq0$ (observation equation)- Initial state $x_0$, vector n-dimensional gaussian with cov. matrix
$E[x_0x^T_o]=P_0$
The process ${w(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[w(k)w^T(k)]=Q(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[v(k)v^T(k)]=R(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
The initial state and additive noises are mutually independent.
I think it has to be done with Doob theorem but in my case is not working. Thanks in advance,
ordinary-differential-equations stochastic-processes
asked Dec 20 '18 at 13:47
finafina
33
$endgroup$
add a comment |
$begingroup$
Having this properties:
$E[x(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
$E[z(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
Demonstrate it with the following assumptions:
$x(k+1)=Phi (k+1,k)x(k)+Gamma (k+1,k)w(k)~$, $kgeq 0$; $x(0)=x_0~$
(is State equation)
$z(k)=H(k)x(k)+v(k), kgeq0$ (observation equation)- Initial state $x_0$, vector n-dimensional gaussian with cov. matrix
$E[x_0x^T_o]=P_0$
The process ${w(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[w(k)w^T(k)]=Q(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[v(k)v^T(k)]=R(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
The initial state and additive noises are mutually independent.
I think it has to be done with Doob theorem but in my case is not working. Thanks in advance,
ordinary-differential-equations stochastic-processes
asked Dec 20 '18 at 13:47
finafina
33
$endgroup$
Having this properties:
$E[x(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
$E[z(j)w^T(k)]=0$; $j,kgeq 0, jleq k$.
Demonstrate it with the following assumptions:
$x(k+1)=Phi (k+1,k)x(k)+Gamma (k+1,k)w(k)~$, $kgeq 0$; $x(0)=x_0~$
(is State equation)
$z(k)=H(k)x(k)+v(k), kgeq0$ (observation equation)- Initial state $x_0$, vector n-dimensional gaussian with cov. matrix
$E[x_0x^T_o]=P_0$
The process ${w(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[w(k)w^T(k)]=Q(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
with cov. matrix $E[v(k)v^T(k)]=R(k), kgeq0$
- The process ${v(k);kgeq 0}$ is a white noise succession, centred,
The initial state and additive noises are mutually independent.
I think it has to be done with Doob theorem but in my case is not working. Thanks in advance,
ordinary-differential-equations stochastic-processes
ordinary-differential-equations stochastic-processes
asked Dec 20 '18 at 13:47
finafina
33
asked Dec 20 '18 at 13:47
finafina
33
edited Dec 20 '18 at 17:55
fina
asked Dec 20 '18 at 13:47
finafina
33
asked Dec 20 '18 at 13:47
finafina
33
asked Dec 20 '18 at 13:47
finafina
33
33
add a comment |
add a comment |
1 Answer
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$begingroup$
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $kge j$.
Therefore, the claim.
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
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active
oldest
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votes
$begingroup$
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $kge j$.
Therefore, the claim.
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
add a comment |
$begingroup$
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $kge j$.
Therefore, the claim.
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
add a comment |
$begingroup$
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $kge j$.
Therefore, the claim.
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
$endgroup$
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $kge j$.
Therefore, the claim.
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
answered Dec 20 '18 at 15:23
LutzLLutzL
58.7k42055
58.7k42055
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
add a comment |
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
I add some more information, just to know if it is right. I need some help about this, I'm quite new in stochastic systems.
$endgroup$
– fina
Dec 20 '18 at 15:45
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
What is $v_k$? The same applies, also $z_j$ is composed of data from times $<j$ that are independent of $w_k$, $k>j$.
$endgroup$
– LutzL
Dec 20 '18 at 16:07
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Thanks @LutzL I have corrected this in the question area
$endgroup$
– fina
Dec 20 '18 at 17:55
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
Then $z_j$ is obviously correlated to $v_j$, but still not to $w_k$, $kge j$.
$endgroup$
– LutzL
Dec 20 '18 at 18:02
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
$begingroup$
You're right, thank you
$endgroup$
– fina
Dec 20 '18 at 22:13
add a comment |
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