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
asked Jan 2 at 5:30
basicideabasicidea
383
$endgroup$
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
asked Jan 2 at 5:30
basicideabasicidea
383
$endgroup$
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
asked Jan 2 at 5:30
basicideabasicidea
383
$endgroup$
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
asked Jan 2 at 5:30
basicideabasicidea
383
asked Jan 2 at 5:30
basicideabasicidea
383
asked Jan 2 at 5:30
basicideabasicidea
383
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*}
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
$endgroup$
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 $p(xverttheta)=p(x)$
answered Jan 2 at 8:04
BlackBearBlackBear
1616
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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*}
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
$endgroup$
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*}
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
$endgroup$
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*}
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
$endgroup$
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
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
answered Jan 2 at 5:37
TaylorTaylor
11.8k11945
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 $p(xverttheta)=p(x)$
answered Jan 2 at 8:04
BlackBearBlackBear
1616
$endgroup$
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 $p(xverttheta)=p(x)$
answered Jan 2 at 8:04
BlackBearBlackBear
1616
$endgroup$
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 $p(xverttheta)=p(x)$
answered Jan 2 at 8:04
BlackBearBlackBear
1616
$endgroup$
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 $p(xverttheta)=p(x)$
answered Jan 2 at 8:04
BlackBearBlackBear
1616
edited Jan 2 at 8:16
answered Jan 2 at 8:04
BlackBearBlackBear
1616
answered Jan 2 at 8:04
BlackBearBlackBear
1616
answered Jan 2 at 8:04
BlackBearBlackBear
1616
1616
add a comment |
add a comment |
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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