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I have two sampling surjective techniques $\varphi_1,\varphi_2:[0,1)\to E$ mapping a random number $u\in[0,1)$ to a sample in a measurable space $(E,\mathcal E)$.

Say $u\in[0,1)$ and $x:=\varphi_1(u)$. Is there any chance to compute the random number $v\in[0,1)$ which would have produced the same sample $x$ under $\varphi_2$, i.e. $x=\varphi_2(v)$?

Clearly, if (for example) $\varphi_1$ is not injective, $\varphi_1$ might be constant on an interval and hence we're not able to identify a unique random number producing $x$. However, maybe we can somehow commit to a single value.

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The question is too vague in my opinion as there must be constraints on the transforms $\varphi_1$ and $\varphi_2$ for this to happen. Namely that the realised value of $X_1$ as $x=\varphi_1(u)$ must be a possible value of $X_2$ as well, namely that $x$ must belong to the support of $X_2$ for a $v$ such that $x=\varphi_2(v)$ to exist.

With this constraint in mind, an approach to the problem is to consider that $\varphi_1$ and $\varphi_2$ are the inverse cdfs of the random variables $X_1$ and $X_2$, namely$$\varphi_i(u)=F_i^-(u) = \sup \{x;\ F_i(x)\le u\}\qquad i=1,2$$(under the convention that cdf's are left-continuous). Either $x$ is a continuity point for $F_2$ and then $$v=F_2 \circ F_1^{-1}(u)$$ (since $F_2$ is then invertible at this point). For this is a discontinuity poiny, meaning it is an atom, in which case $$v\in \{\nu;\ \lim_{{y \to x}\\{y< x}} F_2(y)\le \nu\le \lim_{{y \to x}\\{y> x}} F_2(y)\}$$ This includes the special value $$v=F_2 \circ F_1^- (u)$$

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  • $\begingroup$ I will take a look at what you wrote. Just let me add that my question was motivated by section 6 here: cs.dartmouth.edu/~wjarosz/publications/bitterli18reversible.pdf. Maybe there is an additional assumption that I don't see for the moment. $\endgroup$ – 0xbadf00d Jan 2 at 17:59
  • $\begingroup$ Regarding your answer: I've forgot to mention that I want to assume that $\varphi_1,\varphi_2$ are surjective. $\endgroup$ – 0xbadf00d Jan 3 at 18:43
  • $\begingroup$ And, clearly, the problem with your solution is that it only works for $E=\mathbb R$, doesn't it? Actually, I'm willing to assume that $E$ is a Banach space, but I don't see how the notion of a CDF generalizes to this case. The case $E=\mathbb R^3$ is interesting for me as well. $\endgroup$ – 0xbadf00d Jan 3 at 19:15
  • $\begingroup$ You want to generate a three dimensional random variate from a single uniform?? $\endgroup$ – Xi'an Jan 3 at 19:22
  • $\begingroup$ Well, no. Sorry. I think I made some mistakes in my attempt to simplify my concern. For the case $E=\mathbb R^3$, I'm actually using a sample from a unit (hemi-)sphere. $\endgroup$ – 0xbadf00d Jan 3 at 19:40

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