Density of a diffeomorphic image of a random vector












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Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.




  1. If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?

  2. When does $f(X)$ have no density w.r.t. the $n$-dimensional Lebesgue measure?


In 1., it looks like integration by substitution should do the trick. Still, integration by substitution will give us integration over the wrong set, i.e. $f^{-1}(A)$ instead of $A$.



In 2., a sufficient condition is that at every point the derivative is degenerate, then by Sard's theorem the image of $f$ has measure zero, so it clearly has no density.










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    0












    $begingroup$


    Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.




    1. If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?

    2. When does $f(X)$ have no density w.r.t. the $n$-dimensional Lebesgue measure?


    In 1., it looks like integration by substitution should do the trick. Still, integration by substitution will give us integration over the wrong set, i.e. $f^{-1}(A)$ instead of $A$.



    In 2., a sufficient condition is that at every point the derivative is degenerate, then by Sard's theorem the image of $f$ has measure zero, so it clearly has no density.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.




      1. If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?

      2. When does $f(X)$ have no density w.r.t. the $n$-dimensional Lebesgue measure?


      In 1., it looks like integration by substitution should do the trick. Still, integration by substitution will give us integration over the wrong set, i.e. $f^{-1}(A)$ instead of $A$.



      In 2., a sufficient condition is that at every point the derivative is degenerate, then by Sard's theorem the image of $f$ has measure zero, so it clearly has no density.










      share|cite|improve this question









      $endgroup$




      Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.




      1. If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?

      2. When does $f(X)$ have no density w.r.t. the $n$-dimensional Lebesgue measure?


      In 1., it looks like integration by substitution should do the trick. Still, integration by substitution will give us integration over the wrong set, i.e. $f^{-1}(A)$ instead of $A$.



      In 2., a sufficient condition is that at every point the derivative is degenerate, then by Sard's theorem the image of $f$ has measure zero, so it clearly has no density.







      probability probability-distributions density-function






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      asked Jan 10 at 19:52









      Abraham SlavitskyAbraham Slavitsky

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