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
1
active
oldest
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
57.6k42054
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047555%2fdemonstrate-the-following-propierties-state-space-representation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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
57.6k42054
$endgroup$
$begingroup$
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
57.6k42054
$endgroup$
$begingroup$
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
57.6k42054
$endgroup$
$x_j$ 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
57.6k42054
answered Dec 20 '18 at 15:23
LutzLLutzL
57.6k42054
answered Dec 20 '18 at 15:23
LutzLLutzL
57.6k42054
answered Dec 20 '18 at 15:23
LutzLLutzL
57.6k42054
57.6k42054
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047555%2fdemonstrate-the-following-propierties-state-space-representation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
