How to derive the form of the posterior for regression?
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 2 hours ago
basicidea
212
add a comment |
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 2 hours ago
basicidea
212
add a comment |
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 2 hours ago
basicidea
212
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 2 hours ago
basicidea
212
asked 2 hours ago
basicidea
212
asked 2 hours ago
basicidea
212
asked 2 hours ago
basicidea
212
asked 2 hours ago
basicidea
212
212
add a comment |
add a comment |
1 Answer
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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 2 hours ago
Taylor
11.6k11745
add a comment |
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1 Answer
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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 2 hours ago
Taylor
11.6k11745
add a comment |
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 2 hours ago
Taylor
11.6k11745
add a comment |
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 2 hours ago
Taylor
11.6k11745
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 2 hours ago
Taylor
11.6k11745
answered 2 hours ago
Taylor
11.6k11745
answered 2 hours ago
Taylor
11.6k11745
answered 2 hours ago
Taylor
11.6k11745
11.6k11745
add a comment |
add a comment |
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