1

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 $P_{Y|r(X)}(y|\{x_1,x_2\})=P_{Y}(y)$.

TEX
  • 266
  • 3
  • 19

1 Answers1

6

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 $X$.

Stefan
  • 368
  • 1
  • 5