# How to find a special random variable? [closed]

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.

## closed as off-topic by David Roberts, Sean Lawton, user44191, Jan-Christoph Schlage-Puchta, Mark WildonMar 19 at 20:07

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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.