Suppose that $X,Y,Z \in L^2(\Omega,\mathcal{F},\mathbb{P};\mathbb{R}^d)$, are $d$-dimensional random-vectors and there exists functions $f,g\in L^2(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),Law(X);\mathbb{R}^d)$ satisfying $$ f(X)=Y \qquad g(X)=Z. $$

If $f$ is invertible, then the function $h\triangleq g\circ f^{-1}$ implies that $$ Z=h(Y). $$

However, if $f$ is not invertible, when does there exists a function $h$ satisfying $ h\circ f = g $ except possibly on a set of $Law(X)$-measure $0$? Hence $$ h(Y)=Z . $$

  • $\begingroup$ The law of $X$ is a measure on the Borel sets of $\Bbb R^d$. Something does not add up here. $\endgroup$ – Mars Plastic Mar 12 at 13:29
  • $\begingroup$ Thanks, made it clear now :) $\endgroup$ – AIM_BLB Mar 12 at 13:34
  • $\begingroup$ The space $L^2(\Omega,\mathcal{F},Law(X);\mathbb{R}^d)$ still doesn't make sense. ;-) $\endgroup$ – Mars Plastic Mar 12 at 13:37
  • $\begingroup$ Should be good now, thanks for noticing that; I fixed it to the Borel measure space on $d$-dimensional Euclidean space. :) $\endgroup$ – AIM_BLB Mar 12 at 13:47

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