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.

## **put on hold** as off-topic by David Roberts, Sean Lawton, user44191, Jan-Christoph Schlage-Puchta, Mark Wildon 18 hours ago

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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.