Consider the random variables $Y$ and $X$. Let $\mathcal{X}$ denote the support of $X$. Let $\mathcal{Y}$ denote the support of $Y$. Let $r:\mathcal{X}\rightarrow \mathbb{R}$. I have doubts about the following:

1) Let $X$ be independent of $Y$, i.e. $X\perp Y$. Does this imply $Y\perp r(X)$?

2) Let $Y\perp r(X)$. Does this imply $Y\perp X$?

My thoughts: I believe that the answer to both questions is "no", but I'm struggling to formalise arguments.

For example, regarding 1), suppose that the codomain of the function $r$ is $\{0,1\}$ and that $\{x\in \mathcal{X}: r(x)=1\}=\{x_1,x_2\}$. We assume that $X\perp Y$, that is $$ P_{Y|X}(y|x)=P_Y(y) \hspace{1cm} \forall y\in \mathcal{Y}, \forall x \in \mathcal{X} $$ We want to show that $$ P_{Y|r(X)}(y|r)=P_Y(y) \hspace{1cm} \forall y\in \mathcal{Y}, $\forall r\in \{0,1\} $$ Now, note that $$ P_{Y|r(X)}(y|1)=P_{Y|r(X)}(y|\{x_1,x_2\}) $$ I don't think that we can claim that $P_{Y|r(X)}(y|\{x_1,x_2\})=P_{Y}(y)$.


For 1) see that for any measurable sets $A,B$: $$\mathbf P (r(X) \in A, Y \in B) = \mathbf P (X \in r^{-1}(A), Y \in B) = \mathbf P (X \in r^{-1}(A))\mathbf P (Y \in B) = \mathbf P(r(X) \in A) \mathbf P (Y \in B)$$ and thus $r(X)$ is independent of $Y$

For 2) take $r(X) = 1$ constant and $Y = X$. Then $Y$ is independent of $1$ but certainly not of $X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.