How to derive the form of the posterior for regression?
$begingroup$
I have seen the general form of posterior for a regression $y = f(x)$ defined as
$$
P(theta|y,x) = frac{ P(y | x, theta) P(theta) }{ P(y|x) }
$$
I would like to know how to arrive at this form, starting from Bayes rule and the laws of probability.
The approach I tried is to write Bayes theorem notated as $p(theta|v) = frac{ p(v|theta) p(theta) }{ p(v) }$ and then try subsituting something for $v$.
Approach A. With $v rightarrow y|x$, this gives
$$
" P(theta|y|x) = frac{ P(y|x | theta) P(theta) }{ P(y|x) } "
$$
and then if there is a rule that $a|b|c rightarrow a|b,c$ it gives the result.
However I have not seen such a rule. Does it exist?
Approach B. Start with $v rightarrow y$ giving
$$
P(theta|y) = frac{ P(y | theta) P(theta) }{ P(y) }
$$
and then assume there is a rule that you can condition every factor on some other variable $x$ (see rule below),
giving
$$
P(theta|y,x) = frac{ P(y | theta,x) P(theta|x) }{ P(y|x) }
$$
and lastly assume that $P(theta|x) = P(theta)$.
But here I have not seen a rule that allows
$$
text{if} quad p(A) = P(B)P(C)cdots quadtext{then}quad p(A|X) = P(B|X)P(C|X)cdots
$$
Does such a rule exist?
What is the right approach?
probability bayesian
$endgroup$
add a comment |
$begingroup$
I have seen the general form of posterior for a regression $y = f(x)$ defined as
$$
P(theta|y,x) = frac{ P(y | x, theta) P(theta) }{ P(y|x) }
$$
I would like to know how to arrive at this form, starting from Bayes rule and the laws of probability.
The approach I tried is to write Bayes theorem notated as $p(theta|v) = frac{ p(v|theta) p(theta) }{ p(v) }$ and then try subsituting something for $v$.
Approach A. With $v rightarrow y|x$, this gives
$$
" P(theta|y|x) = frac{ P(y|x | theta) P(theta) }{ P(y|x) } "
$$
and then if there is a rule that $a|b|c rightarrow a|b,c$ it gives the result.
However I have not seen such a rule. Does it exist?
Approach B. Start with $v rightarrow y$ giving
$$
P(theta|y) = frac{ P(y | theta) P(theta) }{ P(y) }
$$
and then assume there is a rule that you can condition every factor on some other variable $x$ (see rule below),
giving
$$
P(theta|y,x) = frac{ P(y | theta,x) P(theta|x) }{ P(y|x) }
$$
and lastly assume that $P(theta|x) = P(theta)$.
But here I have not seen a rule that allows
$$
text{if} quad p(A) = P(B)P(C)cdots quadtext{then}quad p(A|X) = P(B|X)P(C|X)cdots
$$
Does such a rule exist?
What is the right approach?
probability bayesian
$endgroup$
3
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53
add a comment |
$begingroup$
I have seen the general form of posterior for a regression $y = f(x)$ defined as
$$
P(theta|y,x) = frac{ P(y | x, theta) P(theta) }{ P(y|x) }
$$
I would like to know how to arrive at this form, starting from Bayes rule and the laws of probability.
The approach I tried is to write Bayes theorem notated as $p(theta|v) = frac{ p(v|theta) p(theta) }{ p(v) }$ and then try subsituting something for $v$.
Approach A. With $v rightarrow y|x$, this gives
$$
" P(theta|y|x) = frac{ P(y|x | theta) P(theta) }{ P(y|x) } "
$$
and then if there is a rule that $a|b|c rightarrow a|b,c$ it gives the result.
However I have not seen such a rule. Does it exist?
Approach B. Start with $v rightarrow y$ giving
$$
P(theta|y) = frac{ P(y | theta) P(theta) }{ P(y) }
$$
and then assume there is a rule that you can condition every factor on some other variable $x$ (see rule below),
giving
$$
P(theta|y,x) = frac{ P(y | theta,x) P(theta|x) }{ P(y|x) }
$$
and lastly assume that $P(theta|x) = P(theta)$.
But here I have not seen a rule that allows
$$
text{if} quad p(A) = P(B)P(C)cdots quadtext{then}quad p(A|X) = P(B|X)P(C|X)cdots
$$
Does such a rule exist?
What is the right approach?
probability bayesian
$endgroup$
I have seen the general form of posterior for a regression $y = f(x)$ defined as
$$
P(theta|y,x) = frac{ P(y | x, theta) P(theta) }{ P(y|x) }
$$
I would like to know how to arrive at this form, starting from Bayes rule and the laws of probability.
The approach I tried is to write Bayes theorem notated as $p(theta|v) = frac{ p(v|theta) p(theta) }{ p(v) }$ and then try subsituting something for $v$.
Approach A. With $v rightarrow y|x$, this gives
$$
" P(theta|y|x) = frac{ P(y|x | theta) P(theta) }{ P(y|x) } "
$$
and then if there is a rule that $a|b|c rightarrow a|b,c$ it gives the result.
However I have not seen such a rule. Does it exist?
Approach B. Start with $v rightarrow y$ giving
$$
P(theta|y) = frac{ P(y | theta) P(theta) }{ P(y) }
$$
and then assume there is a rule that you can condition every factor on some other variable $x$ (see rule below),
giving
$$
P(theta|y,x) = frac{ P(y | theta,x) P(theta|x) }{ P(y|x) }
$$
and lastly assume that $P(theta|x) = P(theta)$.
But here I have not seen a rule that allows
$$
text{if} quad p(A) = P(B)P(C)cdots quadtext{then}quad p(A|X) = P(B|X)P(C|X)cdots
$$
Does such a rule exist?
What is the right approach?
probability bayesian
probability bayesian
asked Jan 2 at 5:30
basicideabasicidea
383
383
3
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53
add a comment |
3
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53
3
3
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
begin{align*}
P(theta mid y,x) &= frac{P(theta, y,x)}{P(y,x)} tag{defn. condtl. prob} \
&= frac{P(y mid theta, x)P(theta mid x)P(x)}{P(y mid x)P(x)} tag{defn. condtl. prob}\
&= frac{P(y mid theta, x)P(theta mid x)}{P(y mid x)} tag{cancellation}\
&= frac{ P(y | x, theta) P(theta) }{ P(y|x) } tag{indep.}
end{align*}
$endgroup$
add a comment |
$begingroup$
Your initial idea is correct, but you should have used $x,y$ as $v$. Then $p(v)=p(x,y)=p(y|x)p(x)$, and:
$$
p(thetavert v)=p(thetavert x,y)
=frac{p(vverttheta)p(theta)}{p(v)}
=frac{p(yvert x,theta)p(xverttheta)p(theta)}{p(yvert x)p(x)}=frac{p(yvert x,theta)p(theta)}{p(yvert x)}
$$
Where we used the fact that $p(xverttheta)=p(x)$
$endgroup$
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: "65"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
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%2fstats.stackexchange.com%2fquestions%2f385239%2fhow-to-derive-the-form-of-the-posterior-for-regression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
begin{align*}
P(theta mid y,x) &= frac{P(theta, y,x)}{P(y,x)} tag{defn. condtl. prob} \
&= frac{P(y mid theta, x)P(theta mid x)P(x)}{P(y mid x)P(x)} tag{defn. condtl. prob}\
&= frac{P(y mid theta, x)P(theta mid x)}{P(y mid x)} tag{cancellation}\
&= frac{ P(y | x, theta) P(theta) }{ P(y|x) } tag{indep.}
end{align*}
$endgroup$
add a comment |
$begingroup$
begin{align*}
P(theta mid y,x) &= frac{P(theta, y,x)}{P(y,x)} tag{defn. condtl. prob} \
&= frac{P(y mid theta, x)P(theta mid x)P(x)}{P(y mid x)P(x)} tag{defn. condtl. prob}\
&= frac{P(y mid theta, x)P(theta mid x)}{P(y mid x)} tag{cancellation}\
&= frac{ P(y | x, theta) P(theta) }{ P(y|x) } tag{indep.}
end{align*}
$endgroup$
add a comment |
$begingroup$
begin{align*}
P(theta mid y,x) &= frac{P(theta, y,x)}{P(y,x)} tag{defn. condtl. prob} \
&= frac{P(y mid theta, x)P(theta mid x)P(x)}{P(y mid x)P(x)} tag{defn. condtl. prob}\
&= frac{P(y mid theta, x)P(theta mid x)}{P(y mid x)} tag{cancellation}\
&= frac{ P(y | x, theta) P(theta) }{ P(y|x) } tag{indep.}
end{align*}
$endgroup$
begin{align*}
P(theta mid y,x) &= frac{P(theta, y,x)}{P(y,x)} tag{defn. condtl. prob} \
&= frac{P(y mid theta, x)P(theta mid x)P(x)}{P(y mid x)P(x)} tag{defn. condtl. prob}\
&= frac{P(y mid theta, x)P(theta mid x)}{P(y mid x)} tag{cancellation}\
&= frac{ P(y | x, theta) P(theta) }{ P(y|x) } tag{indep.}
end{align*}
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
11.8k11945
add a comment |
add a comment |
$begingroup$
Your initial idea is correct, but you should have used $x,y$ as $v$. Then $p(v)=p(x,y)=p(y|x)p(x)$, and:
$$
p(thetavert v)=p(thetavert x,y)
=frac{p(vverttheta)p(theta)}{p(v)}
=frac{p(yvert x,theta)p(xverttheta)p(theta)}{p(yvert x)p(x)}=frac{p(yvert x,theta)p(theta)}{p(yvert x)}
$$
Where we used the fact that $p(xverttheta)=p(x)$
$endgroup$
add a comment |
$begingroup$
Your initial idea is correct, but you should have used $x,y$ as $v$. Then $p(v)=p(x,y)=p(y|x)p(x)$, and:
$$
p(thetavert v)=p(thetavert x,y)
=frac{p(vverttheta)p(theta)}{p(v)}
=frac{p(yvert x,theta)p(xverttheta)p(theta)}{p(yvert x)p(x)}=frac{p(yvert x,theta)p(theta)}{p(yvert x)}
$$
Where we used the fact that $p(xverttheta)=p(x)$
$endgroup$
add a comment |
$begingroup$
Your initial idea is correct, but you should have used $x,y$ as $v$. Then $p(v)=p(x,y)=p(y|x)p(x)$, and:
$$
p(thetavert v)=p(thetavert x,y)
=frac{p(vverttheta)p(theta)}{p(v)}
=frac{p(yvert x,theta)p(xverttheta)p(theta)}{p(yvert x)p(x)}=frac{p(yvert x,theta)p(theta)}{p(yvert x)}
$$
Where we used the fact that $p(xverttheta)=p(x)$
$endgroup$
Your initial idea is correct, but you should have used $x,y$ as $v$. Then $p(v)=p(x,y)=p(y|x)p(x)$, and:
$$
p(thetavert v)=p(thetavert x,y)
=frac{p(vverttheta)p(theta)}{p(v)}
=frac{p(yvert x,theta)p(xverttheta)p(theta)}{p(yvert x)p(x)}=frac{p(yvert x,theta)p(theta)}{p(yvert x)}
$$
Where we used the fact that $p(xverttheta)=p(x)$
edited Jan 2 at 8:16
answered Jan 2 at 8:04
BlackBearBlackBear
1616
1616
add a comment |
add a comment |
Thanks for contributing an answer to Cross Validated!
- 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%2fstats.stackexchange.com%2fquestions%2f385239%2fhow-to-derive-the-form-of-the-posterior-for-regression%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
3
$begingroup$
To write that $nu=y|x$ does not make sense: $y$ is a random variable that has both a marginal distribution (if irrelevant here) and a conditional distribution given the random variable $x$. The notation $P(y|x|theta)$ does not make sense either.
$endgroup$
– Xi'an
Jan 2 at 7:53