5
$\begingroup$

Suppose that $X_1$ and $X_2$ are independent. I wonder if $X_1$ and $X_2$ conditioning on $X_1+X_2$ can be independent as well.

If $X_1$ and $X_2$ are normally distributed, then the above statement is wrong. I wonder if the statement can be true for some random variables.

$\endgroup$
1
  • 2
    $\begingroup$ It will be helpful if you type this explicitly mathematically as a theorem. $\endgroup$ Commented Jul 3, 2021 at 10:15

2 Answers 2

6
$\begingroup$

$X_1 ~|~ X_1+X_2$ and $X_2 ~|~ X_1+X_2$ are not independent. They are perfectly negatively correlated distributions.

$\endgroup$
2
  • $\begingroup$ They can be independent - see the other answers. $\endgroup$
    – fblundun
    Commented Jul 4, 2021 at 8:18
  • 1
    $\begingroup$ $X_1 + X_2$ is an instance of a "collider" for $X_1$ and $X_2$. This is discussed in: Day, Felix R., et al. "A robust example of collider bias in a genetic association study." The American Journal of Human Genetics 98.2 (2016): 392-393. See also en.wikipedia.org/wiki/Collider_(statistics) . $\endgroup$
    – krkeane
    Commented Jul 6, 2021 at 11:51
3
$\begingroup$

It's possible if one of them is constant - for example if $X_1$ has a Bernoulli distribution and $X_2$ is always equal to zero.

$\endgroup$
4
  • $\begingroup$ OP: _ I wonder if the statement can be true for some random variables_ $\endgroup$
    – krkeane
    Commented Jul 3, 2021 at 18:39
  • $\begingroup$ @krkeane I don't understand your comment. $\endgroup$
    – fblundun
    Commented Jul 3, 2021 at 21:44
  • $\begingroup$ Plural: "random variables", so constant ruled out. $\endgroup$
    – krkeane
    Commented Jul 3, 2021 at 22:26
  • $\begingroup$ @krkeane no, a constant random variable is still a random variable. See the "Constant random variable" section on this Wikipedia page: en.wikipedia.org/wiki/Degenerate_distribution $\endgroup$
    – fblundun
    Commented Jul 4, 2021 at 10:09

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.