If $X_1,X_2$ are indep. and $Y_1,Y_2$ are indep. with $(Y_1,Y_2)∼N((x_1,x_2),σ^2I)$ if $(X_1,X_2)=(x_1,x_2)$, are $X_1(Y_1-X_1),X_2(Y_2-X_2)$ indep.? Let $n\in\mathbb N$, $\sigma>0$ and $$Q(x,\;\cdot\;):=\mathcal N_n(x,\sigma^2I_n)\;\;\;\text{for }x\in\mathbb R^n.$$ Note that $Q$ is a Markov kernel on $(\mathbb R^n,\mathcal B(\mathbb R^n))$.
Now, let $X,Y$ be $\mathbb R^n$ random variables on a common probability space $(\Omega,\mathcal A,\operatorname P)$. Let $X_\ast\operatorname P$ and $Y_\ast\operatorname P$ denote the distribution of $X$ and $Y$, respectively, and assume that $$Y_\ast\operatorname P=(X_\ast\operatorname P)Q,$$ where the right-hand side denotes the composition of $X_\ast\operatorname P$ and $Q$. Assume there is a continuous $f:\mathbb R\to\mathbb R$ such that $X$ has density $$\mathbb R^d\ni x\mapsto\prod_{i=1}^df(x_i)$$ with respect to the Lebesgue measure on $\mathcal B(\mathbb R^d)$

Are we able to show that $X_1(Y_1-X_1),\ldots,X_n(Y_n-X_n)$ are mutually independent?

By the product form of the densities of $X$ and $Y$, we easily see that $X_1,\ldots,X_n$ are mutually independent and $Y_1,\ldots,Y_n$ are mutually independent. Moreover, $$(Y-X)_\ast=\mathcal N_d(0,\sigma^2I_n)$$ and hence $Y_1-X_1,\ldots,Y_n-X_n$ are mutually independent as well. While this generally isn't enough, are we able to conclude in this specific situation?
 A: I think the "trick" is to observe that each selection of components of $X$ is independent of the selection of the same components of $Y-X$. This yields, for example, \begin{equation}\begin{split}\operatorname P\left[(X_1,Y_1-X_1)\in B_1\times C_1,(X_2,Y_2-X_2)\in B_2\times C_2\right]\\=\operatorname P\left[(X_1,X_2)\in B_1\times B_2,(Y_1-X_1,Y_2-X_2)\in C_1\times C_2\right]\\=\operatorname P\left[(X_1,X_2)\in B_1\times B_2\right]\operatorname P\left(Y_1-X_1,Y_2-X_2)\in C_1\times C_2\right]\\=\operatorname P\left[X_1\in B_1\right]\operatorname P\left[X_2\in B_2\right]\operatorname P\left[Y_1-X_1\in C_1\right]\operatorname P\left[Y_2-X_2\in C_2\right]\\=\operatorname P\left[(X_1,Y_1-X_1)\in B_1\times C_1\right]\operatorname P\left[(X_2,Y_2-X_2)\in B_2\times C_2\right]\end{split}\tag1\end{equation} for all $B_1,B_2,C_1,C_2\in\mathcal B(\mathbb R)$. Since $$f(x,y):=xy$$ is $(\mathcal B(\mathbb R^2),\mathbb R)$-measurable, we are able to conclude that $f(X_1,Y_1-X_1)$ and $f(X_2,Y_2-X_2)$ are independent. This should work in the same way for any selection of components.
If I'm missing something, please let me know.
