Suppose random variables $X\_1$ and $X\_2$ have the same distribution under P, $Y\_1$ is an arbitrary random variable,let $Z\_1:=X\_1+Y\_1$.Can we find a r.v. $Y\_2$ which has same distribution as $Y\_1$,such that $Z\_2:=X\_2+Y\_2$ having same distribution as $Z\_1$?
I gave much reflection on the problem but still had no answer，so I ask for help here.Can you give me a proof of the existence or a counterexample?Thanks a lot.



How to find a special random variable?
Yes, if the underlying probability space is "rich enough." In particular, assume there is a uniformly distributed random variable $W$ that is independent of $X\_2.$ Let $\mu$ be the joint distribution of $X\_1$ and $Y\_1$ and let $\kappa:\mathbb{R}\times\mathcal{B} \to [0,1]$ be a regular conditional probability of $\mu$, so that for every measurable rectangle $A\times B$,
$$\mu(A\times B)=\int\_A\kappa(x,B)~\mathrm dP\circ X\_1^{-1}=\int\_A\kappa(x,B)~\mathrm dP\circ X\_2^{-1}.$$
Now, there will be a measurable function $g:\mathbb{R}\times[0,1]\to\mathbb{R}$ such that for all $x\in\mathbb{R}$ and every Borel set $B$,
$$\kappa(x,B)=\lambda\big(\{y\in[0,1]\mid g(x,y)\in B\}\big),$$
where $\lambda$ is the uniform distribution. Indeed, one can take
$$g(x,y)=\sup\big\{r\in [0,1]\mid \kappa(x,[0,r])< y\big\},$$
 (see Proposition 10.7.6. in Bogachev's book on measure theory for the details.) Then you can define $Z\_2$ by $Z\_2(\omega)=g\big(X\_2(\omega),W\_2(\omega)\big)$. One can verify that the distribution of $(X\_2,Z\_2)$ is equal to the distribution of $(X\_1,Z\_1)$ and that implies that the sum and the second coordinate have the same distribution. This construction is basically taken from [this paper](https://link.springer.com/article/10.1007%2Fs00199-014-0815-1).


