I have two sampling techniques $\varphi_1,\varphi_2$. Given $x=\varphi_1(u)$ can I compute a $v$ with $x=\varphi_2(v)$?

Question

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.

Answer 1

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 $$v=F_2 \circ F_1^- (u)$$