Deriving covariance of Gaussian of linear combination of latent variable + noise
$begingroup$
I'm trying to understand this part of Bishop's PRML (Equations 12.31 - 12.38).
If we have a pdf $p(mathbf{z}) = mathcal{N}(mathbf{z}|mathbf{0}, mathbf{I})$ and a random variable $mathbf{x} = mathbf{Wz} + boldsymbolmu + boldsymbolepsilon$, then the covariance of $mathbf{x}$ can be calculated as
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T] = mathbf{WW}^T + sigma^2mathbf{I}
end{align}
where Bishop writes he has used the fact that $mathbf{z}$ and $boldsymbolepsilon$ are independent random variables and hence are uncorrelated.
If we expand the first row of the covariance calculations, we get:
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] + mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T]
end{align}
This must mean that the two middle terms of this last expression evaluate to zero because $mathbf{z}$ and $boldsymbolepsilon$ are uncorrelated. But I fail to see why this is the case. How do they become zero?
probability-theory
asked Dec 29 '18 at 16:11
SandiSandi
262112
$endgroup$
add a comment |
$begingroup$
I'm trying to understand this part of Bishop's PRML (Equations 12.31 - 12.38).
If we have a pdf $p(mathbf{z}) = mathcal{N}(mathbf{z}|mathbf{0}, mathbf{I})$ and a random variable $mathbf{x} = mathbf{Wz} + boldsymbolmu + boldsymbolepsilon$, then the covariance of $mathbf{x}$ can be calculated as
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T] = mathbf{WW}^T + sigma^2mathbf{I}
end{align}
where Bishop writes he has used the fact that $mathbf{z}$ and $boldsymbolepsilon$ are independent random variables and hence are uncorrelated.
If we expand the first row of the covariance calculations, we get:
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] + mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T]
end{align}
This must mean that the two middle terms of this last expression evaluate to zero because $mathbf{z}$ and $boldsymbolepsilon$ are uncorrelated. But I fail to see why this is the case. How do they become zero?
probability-theory
asked Dec 29 '18 at 16:11
SandiSandi
262112
$endgroup$
add a comment |
$begingroup$
I'm trying to understand this part of Bishop's PRML (Equations 12.31 - 12.38).
If we have a pdf $p(mathbf{z}) = mathcal{N}(mathbf{z}|mathbf{0}, mathbf{I})$ and a random variable $mathbf{x} = mathbf{Wz} + boldsymbolmu + boldsymbolepsilon$, then the covariance of $mathbf{x}$ can be calculated as
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T] = mathbf{WW}^T + sigma^2mathbf{I}
end{align}
where Bishop writes he has used the fact that $mathbf{z}$ and $boldsymbolepsilon$ are independent random variables and hence are uncorrelated.
If we expand the first row of the covariance calculations, we get:
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] + mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T]
end{align}
This must mean that the two middle terms of this last expression evaluate to zero because $mathbf{z}$ and $boldsymbolepsilon$ are uncorrelated. But I fail to see why this is the case. How do they become zero?
probability-theory
asked Dec 29 '18 at 16:11
SandiSandi
262112
$endgroup$
I'm trying to understand this part of Bishop's PRML (Equations 12.31 - 12.38).
If we have a pdf $p(mathbf{z}) = mathcal{N}(mathbf{z}|mathbf{0}, mathbf{I})$ and a random variable $mathbf{x} = mathbf{Wz} + boldsymbolmu + boldsymbolepsilon$, then the covariance of $mathbf{x}$ can be calculated as
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T] = mathbf{WW}^T + sigma^2mathbf{I}
end{align}
where Bishop writes he has used the fact that $mathbf{z}$ and $boldsymbolepsilon$ are independent random variables and hence are uncorrelated.
If we expand the first row of the covariance calculations, we get:
begin{align}
cov[mathbf{x}] &= mathbb{E}[(mathbf{Wz} + boldsymbolepsilon)(mathbf{Wz} + boldsymbolepsilon)^T]\
&= mathbb{E}[mathbf{Wzz}^Tmathbf{W}^T] + mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] + mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] + mathbb{E}[boldsymbolepsilonboldsymbolepsilon^T]
end{align}
This must mean that the two middle terms of this last expression evaluate to zero because $mathbf{z}$ and $boldsymbolepsilon$ are uncorrelated. But I fail to see why this is the case. How do they become zero?
probability-theory
probability-theory
asked Dec 29 '18 at 16:11
SandiSandi
262112
asked Dec 29 '18 at 16:11
SandiSandi
262112
asked Dec 29 '18 at 16:11
SandiSandi
262112
asked Dec 29 '18 at 16:11
SandiSandi
262112
asked Dec 29 '18 at 16:11
SandiSandi
262112
262112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe I've found the answer myself. Since $mathbf{W}$ is a parameter and not a random variable, and $mathbf{z}$ and $boldsymbolepsilon$ have zero-mean, we have:
begin{align}
mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}]mathbb{E}[boldsymbolepsilon^T] = mathbf{W}mathbf{0}mathbf{0}^T
end{align}
Which is a matrix of zeros. We can factorize the expected value of $mathbf{z}boldsymbolepsilon^T$ as above because they are uncorrelated.
Similarly,
begin{align}
mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] = mathbb{E}[boldsymbolepsilonmathbf{z}^T]mathbf{W}^T = mathbb{E}[boldsymbolepsilon]mathbb{E}[mathbf{z}^T]mathbf{W}^T = mathbf{0}mathbf{0}^Tmathbf{W}^T
end{align}
Which also is a matrix of zeros.
answered Dec 29 '18 at 16:29
SandiSandi
262112
$endgroup$
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%2f3055985%2fderiving-covariance-of-gaussian-of-linear-combination-of-latent-variable-noise%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
$begingroup$
I believe I've found the answer myself. Since $mathbf{W}$ is a parameter and not a random variable, and $mathbf{z}$ and $boldsymbolepsilon$ have zero-mean, we have:
begin{align}
mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}]mathbb{E}[boldsymbolepsilon^T] = mathbf{W}mathbf{0}mathbf{0}^T
end{align}
Which is a matrix of zeros. We can factorize the expected value of $mathbf{z}boldsymbolepsilon^T$ as above because they are uncorrelated.
Similarly,
begin{align}
mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] = mathbb{E}[boldsymbolepsilonmathbf{z}^T]mathbf{W}^T = mathbb{E}[boldsymbolepsilon]mathbb{E}[mathbf{z}^T]mathbf{W}^T = mathbf{0}mathbf{0}^Tmathbf{W}^T
end{align}
Which also is a matrix of zeros.
answered Dec 29 '18 at 16:29
SandiSandi
262112
$endgroup$
add a comment |
$begingroup$
I believe I've found the answer myself. Since $mathbf{W}$ is a parameter and not a random variable, and $mathbf{z}$ and $boldsymbolepsilon$ have zero-mean, we have:
begin{align}
mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}]mathbb{E}[boldsymbolepsilon^T] = mathbf{W}mathbf{0}mathbf{0}^T
end{align}
Which is a matrix of zeros. We can factorize the expected value of $mathbf{z}boldsymbolepsilon^T$ as above because they are uncorrelated.
Similarly,
begin{align}
mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] = mathbb{E}[boldsymbolepsilonmathbf{z}^T]mathbf{W}^T = mathbb{E}[boldsymbolepsilon]mathbb{E}[mathbf{z}^T]mathbf{W}^T = mathbf{0}mathbf{0}^Tmathbf{W}^T
end{align}
Which also is a matrix of zeros.
answered Dec 29 '18 at 16:29
SandiSandi
262112
$endgroup$
add a comment |
$begingroup$
I believe I've found the answer myself. Since $mathbf{W}$ is a parameter and not a random variable, and $mathbf{z}$ and $boldsymbolepsilon$ have zero-mean, we have:
begin{align}
mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}]mathbb{E}[boldsymbolepsilon^T] = mathbf{W}mathbf{0}mathbf{0}^T
end{align}
Which is a matrix of zeros. We can factorize the expected value of $mathbf{z}boldsymbolepsilon^T$ as above because they are uncorrelated.
Similarly,
begin{align}
mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] = mathbb{E}[boldsymbolepsilonmathbf{z}^T]mathbf{W}^T = mathbb{E}[boldsymbolepsilon]mathbb{E}[mathbf{z}^T]mathbf{W}^T = mathbf{0}mathbf{0}^Tmathbf{W}^T
end{align}
Which also is a matrix of zeros.
answered Dec 29 '18 at 16:29
SandiSandi
262112
$endgroup$
I believe I've found the answer myself. Since $mathbf{W}$ is a parameter and not a random variable, and $mathbf{z}$ and $boldsymbolepsilon$ have zero-mean, we have:
begin{align}
mathbb{E}[mathbf{Wz}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}boldsymbolepsilon^T] = mathbf{W}mathbb{E}[mathbf{z}]mathbb{E}[boldsymbolepsilon^T] = mathbf{W}mathbf{0}mathbf{0}^T
end{align}
Which is a matrix of zeros. We can factorize the expected value of $mathbf{z}boldsymbolepsilon^T$ as above because they are uncorrelated.
Similarly,
begin{align}
mathbb{E}[boldsymbolepsilonmathbf{z}^Tmathbf{W}^T] = mathbb{E}[boldsymbolepsilonmathbf{z}^T]mathbf{W}^T = mathbb{E}[boldsymbolepsilon]mathbb{E}[mathbf{z}^T]mathbf{W}^T = mathbf{0}mathbf{0}^Tmathbf{W}^T
end{align}
Which also is a matrix of zeros.
answered Dec 29 '18 at 16:29
SandiSandi
262112
answered Dec 29 '18 at 16:29
SandiSandi
262112
answered Dec 29 '18 at 16:29
SandiSandi
262112
answered Dec 29 '18 at 16:29
SandiSandi
262112
262112
add a comment |
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%2f3055985%2fderiving-covariance-of-gaussian-of-linear-combination-of-latent-variable-noise%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
