N(x|μ1,σ1^2)N(x|μ1,σ2^2) ∝ N(x|μ1+μ2,σ1^2+σ2^2)











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N is d-th dimensional Gaussian distribution. In case of d=1



$ N(x|mu,sigma^2)=frac{1}{sqrt{2pisigma^2}}e^{-frac{(x-mu)^2}{2sigma^2}} $



in case of $d neq 1$,



$N({bf x}|{boldsymbol mu},{bf Sigma})=frac{1}{sqrt{(2pi)^d|{bf Sigma}|}}{rm exp}{-frac{1}{2}(bf{x}-boldsymbol mu)^TSigma^{-1}(x-boldsymbol mu)}$



Here,$bf{x}$ and $boldsymbol mu$ are d dimensional vertical vectors and $bf Sigma$ is d dimensional square matrix.



We can proof $N(x|mu_1,sigma_1)N(x|mu,sigma_2)propto N(x|mu_1+mu_2,sigma_1^2+sigma_2^2)$ easily.



How about in case of production of multi dimensional and single dimensional Gaussian distribution??



I mean, what is $N(x|mu,sigma^2)N({bf W}|{bf m},bfSigma^2) $ proportional to? I want to know what kind of probability distribution will be appeared.










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    up vote
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    down vote

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    N is d-th dimensional Gaussian distribution. In case of d=1



    $ N(x|mu,sigma^2)=frac{1}{sqrt{2pisigma^2}}e^{-frac{(x-mu)^2}{2sigma^2}} $



    in case of $d neq 1$,



    $N({bf x}|{boldsymbol mu},{bf Sigma})=frac{1}{sqrt{(2pi)^d|{bf Sigma}|}}{rm exp}{-frac{1}{2}(bf{x}-boldsymbol mu)^TSigma^{-1}(x-boldsymbol mu)}$



    Here,$bf{x}$ and $boldsymbol mu$ are d dimensional vertical vectors and $bf Sigma$ is d dimensional square matrix.



    We can proof $N(x|mu_1,sigma_1)N(x|mu,sigma_2)propto N(x|mu_1+mu_2,sigma_1^2+sigma_2^2)$ easily.



    How about in case of production of multi dimensional and single dimensional Gaussian distribution??



    I mean, what is $N(x|mu,sigma^2)N({bf W}|{bf m},bfSigma^2) $ proportional to? I want to know what kind of probability distribution will be appeared.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      N is d-th dimensional Gaussian distribution. In case of d=1



      $ N(x|mu,sigma^2)=frac{1}{sqrt{2pisigma^2}}e^{-frac{(x-mu)^2}{2sigma^2}} $



      in case of $d neq 1$,



      $N({bf x}|{boldsymbol mu},{bf Sigma})=frac{1}{sqrt{(2pi)^d|{bf Sigma}|}}{rm exp}{-frac{1}{2}(bf{x}-boldsymbol mu)^TSigma^{-1}(x-boldsymbol mu)}$



      Here,$bf{x}$ and $boldsymbol mu$ are d dimensional vertical vectors and $bf Sigma$ is d dimensional square matrix.



      We can proof $N(x|mu_1,sigma_1)N(x|mu,sigma_2)propto N(x|mu_1+mu_2,sigma_1^2+sigma_2^2)$ easily.



      How about in case of production of multi dimensional and single dimensional Gaussian distribution??



      I mean, what is $N(x|mu,sigma^2)N({bf W}|{bf m},bfSigma^2) $ proportional to? I want to know what kind of probability distribution will be appeared.










      share|cite|improve this question













      N is d-th dimensional Gaussian distribution. In case of d=1



      $ N(x|mu,sigma^2)=frac{1}{sqrt{2pisigma^2}}e^{-frac{(x-mu)^2}{2sigma^2}} $



      in case of $d neq 1$,



      $N({bf x}|{boldsymbol mu},{bf Sigma})=frac{1}{sqrt{(2pi)^d|{bf Sigma}|}}{rm exp}{-frac{1}{2}(bf{x}-boldsymbol mu)^TSigma^{-1}(x-boldsymbol mu)}$



      Here,$bf{x}$ and $boldsymbol mu$ are d dimensional vertical vectors and $bf Sigma$ is d dimensional square matrix.



      We can proof $N(x|mu_1,sigma_1)N(x|mu,sigma_2)propto N(x|mu_1+mu_2,sigma_1^2+sigma_2^2)$ easily.



      How about in case of production of multi dimensional and single dimensional Gaussian distribution??



      I mean, what is $N(x|mu,sigma^2)N({bf W}|{bf m},bfSigma^2) $ proportional to? I want to know what kind of probability distribution will be appeared.







      probability statistics






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      asked Nov 15 at 4:54









      Sakurai.JJ

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