0
$\begingroup$

Let $X_1$ and $X_2$ be random variables, and $R(X_1)$ be a function of $X_1$. Here are two statements:

(a) $X_1\perp\!\!\!\!\perp (X_2, Y) \mid R(X_1) $

(b) $X_1\perp\!\!\!\!\perp Y \mid \{R(X_1),X_2\} $

Is it true that (a) implies (b)?

This conclusion appears in the Proposition 1 of this paper: https://doi.org/10.1111/rssb.12093

This paper says this conclusion is based on Proposition 4.6 in the book Regression Graphics: Ideas for Studying Regressions through Graphics by R. Dennis Cook. Here is how Proposition 4.6 looks alike: enter image description here

Statement (a) corresponds to condition (c) and statement (b) corresponds to condition (a1).

Proposition 4.6 says that condition (c) is equivalent to condition (a1) and condition (a2) holding at the same time.

It seems like the paper interprets this as condition (c) implies condition (a1). Is it really okay to drop condition (a2) like this?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is just a simple logic problem: \begin{align*} (c) \iff (a_1) + (a_2) \text{ implies } (c) \implies (a_1) + (a_2) \text{ implies } (c) \implies (a_1) \end{align*} It is of course "OK" to drop $(a_2)$ from left to right. What is not OK is trying to deduce $(c)$ based on $(a_1)$ alone (i.e., dropping $(a_2)$ from right to left).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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