Please let me use a specific example to illustrate the general title above.
(1) It is well known that if $X$ and $Y$ are independent and $X,Y\sim N(0,1)$ then $$ Z\equiv X^2+Y^2\sim\chi_2^2 $$ where $\chi_n^2$ denotes the $\chi^2$ distribution with $n\in\mathbb{R}_{++}$ degrees of freedom which is defined via its PDF (or, equivalently, CDF, MGF, etc.) without any mentioning of the normal distributions.
I'd like to say something like:
(2) If $Z\sim\chi_2^2$, then $\exists X,Y$ independent, $X,Y\sim N(0,1)$ such that $Z=X^2+Y^2$.
Are there some general techniques that lead from (1) to (2)? If not, are there some specific techniques that are applicable for this particular case?
Thank you.
There are two answers depending on how restrictive your setting is. As the OP asks for a general principle I will consider the following easier example:
Let $\Omega =\{0,1,2\}$, $P = (\delta_0+2\delta_1+\delta_2)/4$ and $Z:\Omega \to \mathbb{R}$ with $Z(\omega ) = \omega$. We know that we may obtain $Z \stackrel{d}{=} Z'$ with $Z' = X'+Y'$ where $X'$ and $Y'$ uniform on $\{0,1\}$ ($\stackrel{d}{=}$ for equal in distribution).
Restrictive setting: Do there exist $X$ and $Y$ on the same probability space as $Z$ such that $Z= X+Y$ almost surely and $X$ and $Y$ uniform on $\{0,1\}$ and independent of each other?
Answer: no. Quick: you cannot construct $X$ and $Y$ on the probability space with the correct probabilities and independence.
More elaborate: Suppose $X$ and $Y$ exist. Then $X: \Omega \to \mathbb{R}$ is measurable, so $X^{-1}(\{0\})=A$ and $X^{-1}(\{0\})=\Omega \backslash A$ for a certain $A \subset \Omega$. Moreover, to have $P \circ X^{-1}(\{0\})= 1/2$ we need to have $A= A_1=\{0,2\}$ or $A=A_2=\{1\}$. Then we (need to) define $Y= Z-X$. In the case $A=A_1$ this leads to $P(Y=2) = P(Z=2,X=0) = P(Z=2) = 1/4$, which means that $Y$ does not have uniform distribution on $\{0,1\}$. Furthermore $Y$ and $X$ won't be independent. If we have $A=A_2$ then we find that $P(Y=0)$ with the same problems. So to subsume, one cannot construct $X$ and $Y$ on the same probability space.
Less restrictive setting: Do there exist $X$ and $Y$ on an extension of the probability space as $Z$, which are independent and such that $Z= X+Y$ almost surely?
Answer: yes. Quick: define $X$ via regular conditional probabilities and set $Y:=Z-X$.
More elaborate: Let $\kappa(z,dx) = P(X' \in dx | X'+Y'=z)$ be a regular conditional probability for two independent, uniform $\{0,1\}$ random variables $X'$ and $Y'$. Use Lemma 3.22 in Kallenberg, Foundations of Modern Probability (this is for a generalization), to construct $X$ on a probability space $(\Omega \times [0,1],\mathcal{F}_1,P_1)$ with $P_1 (Z\in dz, X \in dx) = P(Z \in dz) \kappa(z,dx)$. Define $Y := Z-X$ and show that $X$ and $Y$ are independent and have uniform law on $\{0,1\}$. In our particular case, we can construct Kallenberg's function $f: S=\{0,1,2\}\times [0,1] \to T = \{0,1\}$ explicitely ($S$ is the set where $Z$ takes values and $T$ is the set where $X$ takes values): $f(0,\vartheta)=0$, $f(1,\vartheta)= \mathbb{1}_{(\vartheta \geq 1/2)}$ and $f(2,\vartheta)=1$. Then the pair $(Z,f(Z,\vartheta))$ has law $P(Z \in dz) \kappa(z,dx)$ by the aforementioned lemma.