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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 $$ \text{if} \quad p(A) = P(B)P(C)\cdots \quad\text{then}\quad p(A|X) = P(B|X)P(C|X)\cdots $$ Does such a rule exist?

What is the right approach?

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    $\begingroup$ 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 '19 at 7:53
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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*}

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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:

$$ p(\theta\vert v)=p(\theta\vert x,y) =\frac{p(v\vert\theta)p(\theta)}{p(v)} =\frac{p(y\vert x,\theta)p(x\vert\theta)p(\theta)}{p(y\vert x)p(x)}=\frac{p(y\vert x,\theta)p(\theta)}{p(y\vert x)} $$

Where we used the fact that $p(x\vert\theta)=p(x)$

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