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.
- If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?
- 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
$endgroup$
add a comment |
$begingroup$
Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.
- If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?
- 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
$endgroup$
add a comment |
$begingroup$
Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.
- If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?
- 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
$endgroup$
Let $X = (X_1, dots, X_n)$ be a random vector and $fcolon mathbb{R}^nrightarrow mathbb{R}^n$ an a.e.-differentiable mapping.
- If $f$ is a diffeomorphism, is it possible to derive the density of $f(X)$?
- 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
probability probability-distributions density-function
asked Jan 10 at 19:52
Abraham SlavitskyAbraham Slavitsky
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