If $X$ and $Y-\mathbb E[Y|X]$ are independent, and $(X_1,Y_1) \stackrel{d}= (X,Y)$, are $X_1$ and $Y_1 -\mathbb E[Y_1|X_1]$ independent?

Question

Suppose $X$ and $Y$ are random variables such that $\mathbb E|Y| \lt \infty$. If $X$ and $Y-\mathbb E[Y|X]$ are independent, and $(X_1,Y_1) \stackrel{d}= (X,Y)$, are $X_1$ and $Y_1-\mathbb E[Y_1|X_1]$ independent?

Attempt:

Let $m$ be Borel measurable function such that $m(X)=\mathbb E[Y|X]$. For $B\subseteq \mathbb R$ any Borel measurable set, it follows that $$\mathbb E[Y_11_{\{X_1 \in B\}}] = \int_{\mathbb R^2}y1_{B}(x)~d\mathbb P_{(X_1,Y_1)} = \int_{\mathbb R^2} y 1_B(x)d \mathbb P_{(X,Y)}(x,y) = \int_{\mathbb R^2} m(x)1_B(x)d\mathbb P_{(X,Y)}(x,y) = \int_{\mathbb R^2} m(x)1_B(x)d\mathbb P_{(X_1,Y_1)}(x,y) = \mathbb E[m(X_1)1_{\{X_1\in B\}}].$$ Therefore, $\mathbb E[Y_1|X_1] = m(X_1)$.

For Borel measurable $B\subseteq \mathbb R$ it follows $$\mathbb P_{Y-m(X)}(B) = \int_{\mathbb R^2} 1_B(y-m(x)) ~d\mathbb P_{(X,Y)}(x,y) = \int_{\mathbb R^2} 1_B(y-m(x)) ~d\mathbb P_{(X_1,Y_1)}(x,y) = \mathbb P_{Y_1-m(X_1)}(B)$$ Therefore $Y-m(X) \stackrel{d}= Y_1-m(X_1).$

For Borel measurable $C \subseteq \mathbb R^2$ we have $$\mathbb P_{(X,Y-m(X))}(C) = \int_{\mathbb R^2} 1_C(x,y-m(x))~d\mathbb P_{(X,Y)}(x,y) = \int_{\mathbb R^2} 1_C(x,y-m(x))~d\mathbb P_{(X_1,Y_1)}(x,y) = \mathbb P_{(X_1,Y_1-m(X_1))}(C) $$ which implies $(X,Y-m(X)) \stackrel{d}=(X_1,Y_1-m(X_1))$.

Finally, we have $$\mathbb P_{(X_1,Y_1-m(X_1))} = \mathbb P_{(X,Y-m(X))} = \mathbb P_{X}\mathbb P_{Y-m(X)} = \mathbb P_{X_1}\mathbb P_{Y_1-m(X_1)}$$ which implies that $X_1$ and $Y_1-\mathbb E[Y_1|X_1]$ are independent.

Is my proof correct?

Answer 1

If $(X_1, Y_1)\stackrel{d}=(X,Y)$ that means their join distributions are the same. That then applies to all marginal and conditional distributions, i.e. $\Bbb P_{X_1} = \Bbb P_X$, $\Bbb P_{Y_1} = \Bbb P_Y$ and $\Bbb P(Y_1|X_1)=\Bbb P(Y|X)$ and $\Bbb P_{(X, Y, Y|X)} = \Bbb P_{(X_1, Y_1, Y_1|X_1)}$ which in particular means that if $X\perp Y - \Bbb E[Y|X]$ then $X_1\perp Y_1 - \Bbb E[Y_1|X_1]$.

Your proof looks fine to me, given the fact that you just had to write every time that all things to be checked involve $\Bbb P$ and $\Bbb P_1$, which are said to be equal.