Can anyone help explain this basic example of posterior

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I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"



I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.



It is,



Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$



Then (From the book, page 24).



enter image description here






bayesian normal-distribution independence posterior identifiability







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    up vote
    3
    down vote

    favorite












    I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"



    I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.



    It is,



    Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$



    Then (From the book, page 24).



    enter image description here






    bayesian normal-distribution independence posterior identifiability







    share|cite|improve this question

























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"



      I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.



      It is,



      Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$



      Then (From the book, page 24).



      enter image description here






      bayesian normal-distribution independence posterior identifiability







      share|cite|improve this question















      I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"



      I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.



      It is,



      Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$



      Then (From the book, page 24).



      enter image description here






      bayesian normal-distribution independence posterior identifiability




      bayesian normal-distribution independence posterior identifiability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 8 mins ago









      Xi'an

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          Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
          $$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
          Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
          $$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
          which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.






          share|cite




















          • Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
            – Learning
            2 mins ago










          Your Answer





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          Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
          $$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
          Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
          $$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
          which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.






          share|cite




















          • Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
            – Learning
            2 mins ago














          up vote
          3
          down vote



          accepted










          Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
          $$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
          Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
          $$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
          which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.






          share|cite




















          • Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
            – Learning
            2 mins ago












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
          $$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
          Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
          $$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
          which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.






          share|cite












          Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
          $$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
          Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
          $$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
          which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.







          share|cite












          share|cite



          share|cite










          answered 9 mins ago









          Xi'an

          51.4k686337




          51.4k686337











          • Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
            – Learning
            2 mins ago
















          • Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
            – Learning
            2 mins ago















          Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
          – Learning
          2 mins ago




          Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
          – Learning
          2 mins ago

















           

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