Likelihood modification in Metropolis Hastings ratio for transformed parameter












5












$begingroup$


I want to use MH to get samples from $p(theta mid y) approx p(y mid theta) p(theta)$. Let's assume $theta$ is heavily constrained and I transform $theta$ to $f(theta)$ so I can sample from an unconstrained space.



The new posterior becomes $p(f(theta) mid y) approx p(y mid f(theta) ) p(f(theta)) ,times, |det(J_{f^{-1}}(y)) |$. Note that I only changed the prior term (Pushforward measure) and left the likelihood term unchanged as it is a probability distribution on $y$, not on $theta$.



(1) My question now is: can I - in the Metropolis Hastings acceptance ratio - just evaluate



$$frac{p(y mid theta^star) }{ p(y mid theta) } ,times, frac{p(f(theta^star)) mid det{ J_{f^{-1}}( theta^star)} mid }{ p(f(theta)) mid det{ J_{f^{-1}}( theta )} mid }$$



? This term makes me nervous, because I transformed theta, evaluate the pdf of the transformed prior, but then transform it back and evaluate the likelihood of the parameter in the original space. However, I cannot evaluate the first term of this equation:



$$frac{p(y mid f(theta^star)) }{ p(y mid f(theta)) } ,times, frac{p(f(theta^star)) leftlvert det{ J_{f^{-1}}( theta^star)} rightrvert }{ p(f(theta)) leftlvert det{ J_{f^{-1}}( theta )} rightrvert }.$$



I could somehow reverse engineer this problem, i.e. define priors on $f(theta)$ and then map $f(theta)$ to $theta$. The Jacobian of the inverse transform then becomes the Jacobian of the transform of my original problem. That way I could evaluate all terms. However, I originally wanted to give some meaning to my priors for $theta$, not for some unconstrained $f(theta)$.




EDIT: Problem solved and clarified - thank you, I should have seen this myself! Please also see
linked stackexchange thread to this post for further
clarification.











share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    I want to use MH to get samples from $p(theta mid y) approx p(y mid theta) p(theta)$. Let's assume $theta$ is heavily constrained and I transform $theta$ to $f(theta)$ so I can sample from an unconstrained space.



    The new posterior becomes $p(f(theta) mid y) approx p(y mid f(theta) ) p(f(theta)) ,times, |det(J_{f^{-1}}(y)) |$. Note that I only changed the prior term (Pushforward measure) and left the likelihood term unchanged as it is a probability distribution on $y$, not on $theta$.



    (1) My question now is: can I - in the Metropolis Hastings acceptance ratio - just evaluate



    $$frac{p(y mid theta^star) }{ p(y mid theta) } ,times, frac{p(f(theta^star)) mid det{ J_{f^{-1}}( theta^star)} mid }{ p(f(theta)) mid det{ J_{f^{-1}}( theta )} mid }$$



    ? This term makes me nervous, because I transformed theta, evaluate the pdf of the transformed prior, but then transform it back and evaluate the likelihood of the parameter in the original space. However, I cannot evaluate the first term of this equation:



    $$frac{p(y mid f(theta^star)) }{ p(y mid f(theta)) } ,times, frac{p(f(theta^star)) leftlvert det{ J_{f^{-1}}( theta^star)} rightrvert }{ p(f(theta)) leftlvert det{ J_{f^{-1}}( theta )} rightrvert }.$$



    I could somehow reverse engineer this problem, i.e. define priors on $f(theta)$ and then map $f(theta)$ to $theta$. The Jacobian of the inverse transform then becomes the Jacobian of the transform of my original problem. That way I could evaluate all terms. However, I originally wanted to give some meaning to my priors for $theta$, not for some unconstrained $f(theta)$.




    EDIT: Problem solved and clarified - thank you, I should have seen this myself! Please also see
    linked stackexchange thread to this post for further
    clarification.











    share|cite|improve this question











    $endgroup$















      5












      5








      5


      3



      $begingroup$


      I want to use MH to get samples from $p(theta mid y) approx p(y mid theta) p(theta)$. Let's assume $theta$ is heavily constrained and I transform $theta$ to $f(theta)$ so I can sample from an unconstrained space.



      The new posterior becomes $p(f(theta) mid y) approx p(y mid f(theta) ) p(f(theta)) ,times, |det(J_{f^{-1}}(y)) |$. Note that I only changed the prior term (Pushforward measure) and left the likelihood term unchanged as it is a probability distribution on $y$, not on $theta$.



      (1) My question now is: can I - in the Metropolis Hastings acceptance ratio - just evaluate



      $$frac{p(y mid theta^star) }{ p(y mid theta) } ,times, frac{p(f(theta^star)) mid det{ J_{f^{-1}}( theta^star)} mid }{ p(f(theta)) mid det{ J_{f^{-1}}( theta )} mid }$$



      ? This term makes me nervous, because I transformed theta, evaluate the pdf of the transformed prior, but then transform it back and evaluate the likelihood of the parameter in the original space. However, I cannot evaluate the first term of this equation:



      $$frac{p(y mid f(theta^star)) }{ p(y mid f(theta)) } ,times, frac{p(f(theta^star)) leftlvert det{ J_{f^{-1}}( theta^star)} rightrvert }{ p(f(theta)) leftlvert det{ J_{f^{-1}}( theta )} rightrvert }.$$



      I could somehow reverse engineer this problem, i.e. define priors on $f(theta)$ and then map $f(theta)$ to $theta$. The Jacobian of the inverse transform then becomes the Jacobian of the transform of my original problem. That way I could evaluate all terms. However, I originally wanted to give some meaning to my priors for $theta$, not for some unconstrained $f(theta)$.




      EDIT: Problem solved and clarified - thank you, I should have seen this myself! Please also see
      linked stackexchange thread to this post for further
      clarification.











      share|cite|improve this question











      $endgroup$




      I want to use MH to get samples from $p(theta mid y) approx p(y mid theta) p(theta)$. Let's assume $theta$ is heavily constrained and I transform $theta$ to $f(theta)$ so I can sample from an unconstrained space.



      The new posterior becomes $p(f(theta) mid y) approx p(y mid f(theta) ) p(f(theta)) ,times, |det(J_{f^{-1}}(y)) |$. Note that I only changed the prior term (Pushforward measure) and left the likelihood term unchanged as it is a probability distribution on $y$, not on $theta$.



      (1) My question now is: can I - in the Metropolis Hastings acceptance ratio - just evaluate



      $$frac{p(y mid theta^star) }{ p(y mid theta) } ,times, frac{p(f(theta^star)) mid det{ J_{f^{-1}}( theta^star)} mid }{ p(f(theta)) mid det{ J_{f^{-1}}( theta )} mid }$$



      ? This term makes me nervous, because I transformed theta, evaluate the pdf of the transformed prior, but then transform it back and evaluate the likelihood of the parameter in the original space. However, I cannot evaluate the first term of this equation:



      $$frac{p(y mid f(theta^star)) }{ p(y mid f(theta)) } ,times, frac{p(f(theta^star)) leftlvert det{ J_{f^{-1}}( theta^star)} rightrvert }{ p(f(theta)) leftlvert det{ J_{f^{-1}}( theta )} rightrvert }.$$



      I could somehow reverse engineer this problem, i.e. define priors on $f(theta)$ and then map $f(theta)$ to $theta$. The Jacobian of the inverse transform then becomes the Jacobian of the transform of my original problem. That way I could evaluate all terms. However, I originally wanted to give some meaning to my priors for $theta$, not for some unconstrained $f(theta)$.




      EDIT: Problem solved and clarified - thank you, I should have seen this myself! Please also see
      linked stackexchange thread to this post for further
      clarification.








      sampling mcmc likelihood likelihood-ratio metropolis-hastings






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      edited Feb 16 at 14:52









      Xi'an

      57.3k895360




      57.3k895360










      asked Feb 15 at 14:37









      MrVengeanZeMrVengeanZe

      285




      285






















          1 Answer
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          5












          $begingroup$

          You should notice that what you denote $p(y|f(theta))$ is actually the same as $p(y|theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.



          With proper notations, if
          begin{align*}
          theta &sim pi(theta)qquadqquad&text{prior}\
          y|theta &sim f(y|theta)qquadqquad&text{sampling}\
          xi &= h(theta) qquadqquad&text{reparameterisation}\
          dfrac{text{d}theta}{text{d}xi}(xi) &= J(xi)qquadqquad&text{Jacobian}\
          y|xi &sim g(y|xi)qquadqquad&text{reparameterised density}\
          xi^{(t+1)}|xi^{(t)} &sim q(xi^{(t+1)}|xi^{(t)}) qquadqquad&text{proposal}
          end{align*}

          the Metropolis-Hastings ratio associated with the proposal $xi'sim q(xi'|xi)$ in the $xi$ parameterisation is
          $$
          underbrace{dfrac{pi(theta(xi'))J(xi')}{pi(theta(xi))J(xi)}}_text{ratio of priors}times
          underbrace{dfrac{f(y|theta(xi')}{f(y|theta(xi))}}_text{likelihood ratio}times
          underbrace{dfrac{q(xi|xi')}{q(xi'|xi)}}_text{proposal ratio}
          $$

          which also writes as
          $$dfrac{pi(h^{-1}(xi'))J(xi')}{pi(h^{-1}(xi))J(xi)}times
          dfrac{g(y|xi')}{g(y|xi)}times
          dfrac{q(xi|xi')}{q(xi'|xi)}
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Dear Xi'an, thank you for your answer! I will mark it as solved!
            $endgroup$
            – MrVengeanZe
            Feb 15 at 16:12











          Your Answer





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          1 Answer
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          active

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          5












          $begingroup$

          You should notice that what you denote $p(y|f(theta))$ is actually the same as $p(y|theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.



          With proper notations, if
          begin{align*}
          theta &sim pi(theta)qquadqquad&text{prior}\
          y|theta &sim f(y|theta)qquadqquad&text{sampling}\
          xi &= h(theta) qquadqquad&text{reparameterisation}\
          dfrac{text{d}theta}{text{d}xi}(xi) &= J(xi)qquadqquad&text{Jacobian}\
          y|xi &sim g(y|xi)qquadqquad&text{reparameterised density}\
          xi^{(t+1)}|xi^{(t)} &sim q(xi^{(t+1)}|xi^{(t)}) qquadqquad&text{proposal}
          end{align*}

          the Metropolis-Hastings ratio associated with the proposal $xi'sim q(xi'|xi)$ in the $xi$ parameterisation is
          $$
          underbrace{dfrac{pi(theta(xi'))J(xi')}{pi(theta(xi))J(xi)}}_text{ratio of priors}times
          underbrace{dfrac{f(y|theta(xi')}{f(y|theta(xi))}}_text{likelihood ratio}times
          underbrace{dfrac{q(xi|xi')}{q(xi'|xi)}}_text{proposal ratio}
          $$

          which also writes as
          $$dfrac{pi(h^{-1}(xi'))J(xi')}{pi(h^{-1}(xi))J(xi)}times
          dfrac{g(y|xi')}{g(y|xi)}times
          dfrac{q(xi|xi')}{q(xi'|xi)}
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Dear Xi'an, thank you for your answer! I will mark it as solved!
            $endgroup$
            – MrVengeanZe
            Feb 15 at 16:12
















          5












          $begingroup$

          You should notice that what you denote $p(y|f(theta))$ is actually the same as $p(y|theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.



          With proper notations, if
          begin{align*}
          theta &sim pi(theta)qquadqquad&text{prior}\
          y|theta &sim f(y|theta)qquadqquad&text{sampling}\
          xi &= h(theta) qquadqquad&text{reparameterisation}\
          dfrac{text{d}theta}{text{d}xi}(xi) &= J(xi)qquadqquad&text{Jacobian}\
          y|xi &sim g(y|xi)qquadqquad&text{reparameterised density}\
          xi^{(t+1)}|xi^{(t)} &sim q(xi^{(t+1)}|xi^{(t)}) qquadqquad&text{proposal}
          end{align*}

          the Metropolis-Hastings ratio associated with the proposal $xi'sim q(xi'|xi)$ in the $xi$ parameterisation is
          $$
          underbrace{dfrac{pi(theta(xi'))J(xi')}{pi(theta(xi))J(xi)}}_text{ratio of priors}times
          underbrace{dfrac{f(y|theta(xi')}{f(y|theta(xi))}}_text{likelihood ratio}times
          underbrace{dfrac{q(xi|xi')}{q(xi'|xi)}}_text{proposal ratio}
          $$

          which also writes as
          $$dfrac{pi(h^{-1}(xi'))J(xi')}{pi(h^{-1}(xi))J(xi)}times
          dfrac{g(y|xi')}{g(y|xi)}times
          dfrac{q(xi|xi')}{q(xi'|xi)}
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Dear Xi'an, thank you for your answer! I will mark it as solved!
            $endgroup$
            – MrVengeanZe
            Feb 15 at 16:12














          5












          5








          5





          $begingroup$

          You should notice that what you denote $p(y|f(theta))$ is actually the same as $p(y|theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.



          With proper notations, if
          begin{align*}
          theta &sim pi(theta)qquadqquad&text{prior}\
          y|theta &sim f(y|theta)qquadqquad&text{sampling}\
          xi &= h(theta) qquadqquad&text{reparameterisation}\
          dfrac{text{d}theta}{text{d}xi}(xi) &= J(xi)qquadqquad&text{Jacobian}\
          y|xi &sim g(y|xi)qquadqquad&text{reparameterised density}\
          xi^{(t+1)}|xi^{(t)} &sim q(xi^{(t+1)}|xi^{(t)}) qquadqquad&text{proposal}
          end{align*}

          the Metropolis-Hastings ratio associated with the proposal $xi'sim q(xi'|xi)$ in the $xi$ parameterisation is
          $$
          underbrace{dfrac{pi(theta(xi'))J(xi')}{pi(theta(xi))J(xi)}}_text{ratio of priors}times
          underbrace{dfrac{f(y|theta(xi')}{f(y|theta(xi))}}_text{likelihood ratio}times
          underbrace{dfrac{q(xi|xi')}{q(xi'|xi)}}_text{proposal ratio}
          $$

          which also writes as
          $$dfrac{pi(h^{-1}(xi'))J(xi')}{pi(h^{-1}(xi))J(xi)}times
          dfrac{g(y|xi')}{g(y|xi)}times
          dfrac{q(xi|xi')}{q(xi'|xi)}
          $$






          share|cite|improve this answer











          $endgroup$



          You should notice that what you denote $p(y|f(theta))$ is actually the same as $p(y|theta)$ [if you overlook the terrible abuse of notations]. As you mention, changing the parameterisation does not modify the density of the random variable at the observed value $y$ and there is no Jacobian associated with that part.



          With proper notations, if
          begin{align*}
          theta &sim pi(theta)qquadqquad&text{prior}\
          y|theta &sim f(y|theta)qquadqquad&text{sampling}\
          xi &= h(theta) qquadqquad&text{reparameterisation}\
          dfrac{text{d}theta}{text{d}xi}(xi) &= J(xi)qquadqquad&text{Jacobian}\
          y|xi &sim g(y|xi)qquadqquad&text{reparameterised density}\
          xi^{(t+1)}|xi^{(t)} &sim q(xi^{(t+1)}|xi^{(t)}) qquadqquad&text{proposal}
          end{align*}

          the Metropolis-Hastings ratio associated with the proposal $xi'sim q(xi'|xi)$ in the $xi$ parameterisation is
          $$
          underbrace{dfrac{pi(theta(xi'))J(xi')}{pi(theta(xi))J(xi)}}_text{ratio of priors}times
          underbrace{dfrac{f(y|theta(xi')}{f(y|theta(xi))}}_text{likelihood ratio}times
          underbrace{dfrac{q(xi|xi')}{q(xi'|xi)}}_text{proposal ratio}
          $$

          which also writes as
          $$dfrac{pi(h^{-1}(xi'))J(xi')}{pi(h^{-1}(xi))J(xi)}times
          dfrac{g(y|xi')}{g(y|xi)}times
          dfrac{q(xi|xi')}{q(xi'|xi)}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 15 at 15:52

























          answered Feb 15 at 14:52









          Xi'anXi'an

          57.3k895360




          57.3k895360








          • 1




            $begingroup$
            Dear Xi'an, thank you for your answer! I will mark it as solved!
            $endgroup$
            – MrVengeanZe
            Feb 15 at 16:12














          • 1




            $begingroup$
            Dear Xi'an, thank you for your answer! I will mark it as solved!
            $endgroup$
            – MrVengeanZe
            Feb 15 at 16:12








          1




          1




          $begingroup$
          Dear Xi'an, thank you for your answer! I will mark it as solved!
          $endgroup$
          – MrVengeanZe
          Feb 15 at 16:12




          $begingroup$
          Dear Xi'an, thank you for your answer! I will mark it as solved!
          $endgroup$
          – MrVengeanZe
          Feb 15 at 16:12


















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