Marginalizing over product of same probability distribution












0












$begingroup$


Let's say I have a probability distribution defined as a product of probability distributions:



begin{equation}tag{1}label{1}
p(boldsymbolmu)=
prod_{r=1}^{R}prod_{c=1}^{C}p(mu_{r-1,c})p(mu_{r,c})p(mu_{r+1,c})p(mu_{r,c+1})p(mu_{r,c-1})
end{equation}

where $boldsymbolmu$ is a matrix with dimensions $R, C$.



Let's say I want to marginalize out all variables except two particular ones:



$$
sum_{boldsymbolmu notin {mu_{alpha, beta}, mu_{alpha, beta+1} }} p(boldsymbolmu) = p(mu_{alpha, beta}, mu_{alpha, beta+1})
$$



I want to perform this marginalization on (ref{1}). How do I do that? Let's for example say that $R = 1$ and $C = 2$ and we want $p(mu_{2,1}, mu_{2,2})$. Trying to marginalize, we get:



$$
p(mu_{2,1}, mu_{2,2}) = sum_{mu_{0,1}} sum_{mu_{1,1}} sum_{mu_{1,0}} sum_{mu_{1,2}} sum_{mu_{0,2}} p(mu_{0,1})p(mu_{1,1})p(mu_{2,1})p(mu_{1,0})p(mu_{1,2})p(mu_{0,2})p(mu_{1,2})p(mu_{2,2})p(mu_{1,1})p(mu_{1,2})
$$



Three of the variables are easy to eliminate since they only show up once in the product. We're then left with:



$$
sum_{mu_{1,1}} sum_{mu_{1,2}} p(mu_{1,1}) p(mu_{1,1}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{2,1}) p(mu_{2,2})
$$



where there are many duplicates of the distributions we want to marginalize out. Is it correct to simply write



$$
p(mu_{2,1}, mu_{2,2}) = p(mu_{2,1}) p(mu_{2,2})?
$$



And why?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let's say I have a probability distribution defined as a product of probability distributions:



    begin{equation}tag{1}label{1}
    p(boldsymbolmu)=
    prod_{r=1}^{R}prod_{c=1}^{C}p(mu_{r-1,c})p(mu_{r,c})p(mu_{r+1,c})p(mu_{r,c+1})p(mu_{r,c-1})
    end{equation}

    where $boldsymbolmu$ is a matrix with dimensions $R, C$.



    Let's say I want to marginalize out all variables except two particular ones:



    $$
    sum_{boldsymbolmu notin {mu_{alpha, beta}, mu_{alpha, beta+1} }} p(boldsymbolmu) = p(mu_{alpha, beta}, mu_{alpha, beta+1})
    $$



    I want to perform this marginalization on (ref{1}). How do I do that? Let's for example say that $R = 1$ and $C = 2$ and we want $p(mu_{2,1}, mu_{2,2})$. Trying to marginalize, we get:



    $$
    p(mu_{2,1}, mu_{2,2}) = sum_{mu_{0,1}} sum_{mu_{1,1}} sum_{mu_{1,0}} sum_{mu_{1,2}} sum_{mu_{0,2}} p(mu_{0,1})p(mu_{1,1})p(mu_{2,1})p(mu_{1,0})p(mu_{1,2})p(mu_{0,2})p(mu_{1,2})p(mu_{2,2})p(mu_{1,1})p(mu_{1,2})
    $$



    Three of the variables are easy to eliminate since they only show up once in the product. We're then left with:



    $$
    sum_{mu_{1,1}} sum_{mu_{1,2}} p(mu_{1,1}) p(mu_{1,1}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{2,1}) p(mu_{2,2})
    $$



    where there are many duplicates of the distributions we want to marginalize out. Is it correct to simply write



    $$
    p(mu_{2,1}, mu_{2,2}) = p(mu_{2,1}) p(mu_{2,2})?
    $$



    And why?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let's say I have a probability distribution defined as a product of probability distributions:



      begin{equation}tag{1}label{1}
      p(boldsymbolmu)=
      prod_{r=1}^{R}prod_{c=1}^{C}p(mu_{r-1,c})p(mu_{r,c})p(mu_{r+1,c})p(mu_{r,c+1})p(mu_{r,c-1})
      end{equation}

      where $boldsymbolmu$ is a matrix with dimensions $R, C$.



      Let's say I want to marginalize out all variables except two particular ones:



      $$
      sum_{boldsymbolmu notin {mu_{alpha, beta}, mu_{alpha, beta+1} }} p(boldsymbolmu) = p(mu_{alpha, beta}, mu_{alpha, beta+1})
      $$



      I want to perform this marginalization on (ref{1}). How do I do that? Let's for example say that $R = 1$ and $C = 2$ and we want $p(mu_{2,1}, mu_{2,2})$. Trying to marginalize, we get:



      $$
      p(mu_{2,1}, mu_{2,2}) = sum_{mu_{0,1}} sum_{mu_{1,1}} sum_{mu_{1,0}} sum_{mu_{1,2}} sum_{mu_{0,2}} p(mu_{0,1})p(mu_{1,1})p(mu_{2,1})p(mu_{1,0})p(mu_{1,2})p(mu_{0,2})p(mu_{1,2})p(mu_{2,2})p(mu_{1,1})p(mu_{1,2})
      $$



      Three of the variables are easy to eliminate since they only show up once in the product. We're then left with:



      $$
      sum_{mu_{1,1}} sum_{mu_{1,2}} p(mu_{1,1}) p(mu_{1,1}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{2,1}) p(mu_{2,2})
      $$



      where there are many duplicates of the distributions we want to marginalize out. Is it correct to simply write



      $$
      p(mu_{2,1}, mu_{2,2}) = p(mu_{2,1}) p(mu_{2,2})?
      $$



      And why?










      share|cite|improve this question









      $endgroup$




      Let's say I have a probability distribution defined as a product of probability distributions:



      begin{equation}tag{1}label{1}
      p(boldsymbolmu)=
      prod_{r=1}^{R}prod_{c=1}^{C}p(mu_{r-1,c})p(mu_{r,c})p(mu_{r+1,c})p(mu_{r,c+1})p(mu_{r,c-1})
      end{equation}

      where $boldsymbolmu$ is a matrix with dimensions $R, C$.



      Let's say I want to marginalize out all variables except two particular ones:



      $$
      sum_{boldsymbolmu notin {mu_{alpha, beta}, mu_{alpha, beta+1} }} p(boldsymbolmu) = p(mu_{alpha, beta}, mu_{alpha, beta+1})
      $$



      I want to perform this marginalization on (ref{1}). How do I do that? Let's for example say that $R = 1$ and $C = 2$ and we want $p(mu_{2,1}, mu_{2,2})$. Trying to marginalize, we get:



      $$
      p(mu_{2,1}, mu_{2,2}) = sum_{mu_{0,1}} sum_{mu_{1,1}} sum_{mu_{1,0}} sum_{mu_{1,2}} sum_{mu_{0,2}} p(mu_{0,1})p(mu_{1,1})p(mu_{2,1})p(mu_{1,0})p(mu_{1,2})p(mu_{0,2})p(mu_{1,2})p(mu_{2,2})p(mu_{1,1})p(mu_{1,2})
      $$



      Three of the variables are easy to eliminate since they only show up once in the product. We're then left with:



      $$
      sum_{mu_{1,1}} sum_{mu_{1,2}} p(mu_{1,1}) p(mu_{1,1}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{1,2}) p(mu_{2,1}) p(mu_{2,2})
      $$



      where there are many duplicates of the distributions we want to marginalize out. Is it correct to simply write



      $$
      p(mu_{2,1}, mu_{2,2}) = p(mu_{2,1}) p(mu_{2,2})?
      $$



      And why?







      probability-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 5 '18 at 15:45









      SandiSandi

      262112




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