Mixing
Computing variance of sum
0
$begingroup$
I want to find an expression for the variance of $hat{b}=frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}$ where $u_t$ is the deterministic input signal of the process $y_t=bu_t+e_t$ and $e_t$ is white Gaussian noise with variance $sigma_e^2$. Also $lambda$ is a constant value.
Im stuck since I dont know how to handle the summations when dealing with variance
statistics signal-processing
asked Jan 16 at 18:27
99ghegh99ghegh
216
$endgroup$
add a comment |
0
$begingroup$
I want to find an expression for the variance of $hat{b}=frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}$ where $u_t$ is the deterministic input signal of the process $y_t=bu_t+e_t$ and $e_t$ is white Gaussian noise with variance $sigma_e^2$. Also $lambda$ is a constant value.
Im stuck since I dont know how to handle the summations when dealing with variance
statistics signal-processing
asked Jan 16 at 18:27
99ghegh99ghegh
216
$endgroup$
add a comment |
0
0
0
$begingroup$
I want to find an expression for the variance of $hat{b}=frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}$ where $u_t$ is the deterministic input signal of the process $y_t=bu_t+e_t$ and $e_t$ is white Gaussian noise with variance $sigma_e^2$. Also $lambda$ is a constant value.
Im stuck since I dont know how to handle the summations when dealing with variance
statistics signal-processing
asked Jan 16 at 18:27
99ghegh99ghegh
216
$endgroup$
I want to find an expression for the variance of $hat{b}=frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2}$ where $u_t$ is the deterministic input signal of the process $y_t=bu_t+e_t$ and $e_t$ is white Gaussian noise with variance $sigma_e^2$. Also $lambda$ is a constant value.
Im stuck since I dont know how to handle the summations when dealing with variance
statistics signal-processing
statistics signal-processing
asked Jan 16 at 18:27
99ghegh99ghegh
216
asked Jan 16 at 18:27
99ghegh99ghegh
216
asked Jan 16 at 18:27
99ghegh99ghegh
216
asked Jan 16 at 18:27
99ghegh99ghegh
216
asked Jan 16 at 18:27
99ghegh99ghegh
216
216
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
One thing you need to clarify first is the following. Are $y_t=bu_t+e_t$ with $t=1,2,...,N$ uncorrelated? If it's the case you should know the general equation for sum of random variable 1. You only need to work with the upper term.
Cheers
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
$endgroup$
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
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
});
}
});
draft saved
draft discarded
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%2f3076097%2fcomputing-variance-of-sum%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
0
$begingroup$
One thing you need to clarify first is the following. Are $y_t=bu_t+e_t$ with $t=1,2,...,N$ uncorrelated? If it's the case you should know the general equation for sum of random variable 1. You only need to work with the upper term.
Cheers
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
$endgroup$
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
add a comment |
0
$begingroup$
One thing you need to clarify first is the following. Are $y_t=bu_t+e_t$ with $t=1,2,...,N$ uncorrelated? If it's the case you should know the general equation for sum of random variable 1. You only need to work with the upper term.
Cheers
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
$endgroup$
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
add a comment |
0
0
0
$begingroup$
One thing you need to clarify first is the following. Are $y_t=bu_t+e_t$ with $t=1,2,...,N$ uncorrelated? If it's the case you should know the general equation for sum of random variable 1. You only need to work with the upper term.
Cheers
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
$endgroup$
One thing you need to clarify first is the following. Are $y_t=bu_t+e_t$ with $t=1,2,...,N$ uncorrelated? If it's the case you should know the general equation for sum of random variable 1. You only need to work with the upper term.
Cheers
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
answered Jan 16 at 18:45
Alvaro Joaquín GaonaAlvaro Joaquín Gaona
1
1
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
add a comment |
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
It's not mentioned that they are uncorrelated, so I would assume no but not really sure
$endgroup$
– 99ghegh
Jan 16 at 18:55
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Could you help me out? Why is it enough to only work with the upper term, I dont understand?
$endgroup$
– 99ghegh
Jan 16 at 19:36
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
Yes, of course. You have to work only with the upper term in terms of variance because the lower one is deterministic. It's a number (depends on t but still a number at the end). The upper term is the only one that has a random variable, $y_t$. Is it clear? Regarding variables correlation, you can procede without assuming uncorrelation until you need the covariance that I showed you in the picture, then make a comment if you assume that they are uncorrelated.
$endgroup$
– Alvaro Joaquín Gaona
Jan 17 at 23:10
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
So you are saying that $Vleft[ frac{sum_{t=1}^Nlambda^{N-t} y_t u_t}{sum_{t=1}^{N}lambda^{N-t}u_t^2} right]=frac{Vleft[sum_{t=1}^Nlambda^{N-t} y_t u_tright]}{sum_{t=1}^{N}lambda^{N-t}u_t^2} $ ?
$endgroup$
– 99ghegh
Jan 17 at 23:16
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
$begingroup$
Almost. Since lower term is a number that only depends on time, you can name it $b(t)$, so remember that constants get out squared from variance operator. $V(frac{X}{b(t)})=frac{1}{b^2(t)}V(X)$.
$endgroup$
– Alvaro Joaquín Gaona
Jan 18 at 23:35
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3076097%2fcomputing-variance-of-sum%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
