How to derive the form of the posterior for regression?












1














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







share|cite|improve this question



























    1














    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







    share|cite|improve this question

























      1












      1








      1


      1





      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      basicidea

      212




      212






















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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*}






          share|cite|improve this 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*}






            share|cite|improve this 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*}






              share|cite|improve this 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*}






                share|cite|improve this answer












                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*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Taylor

                11.6k11745




                11.6k11745






























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