Can $X_1$ and $X_2$ be independent conditioning on $X_1+X_2$?

Question

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.

It will be helpful if you type this explicitly mathematically as a theorem. — msuzen, Jul 03 '21 at 10:15

score 5 · Answer 1 · answered Jul 03 '21 at 14:05

5

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

answered Jul 03 '21 at 14:05

krkeane

They can be independent - see the other answers. – fblundun Jul 04 '21 at 08:18
1

$X_1 + X_2$ is an instance of a "collider" for $X_1$ and $X_2$. This is discussed in: [Day, Felix R., et al.](https://www.cell.com/ajhg/pdf/S0002-9297(15)00514-5.pdf) "A robust example of collider bias in a genetic association study." The American Journal of Human Genetics 98.2 (2016): 392-393. See also https://en.wikipedia.org/wiki/Collider_(statistics) . – krkeane Jul 06 '21 at 11:51

score 3 · Answer 2 · answered Jul 03 '21 at 14:37

3

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.

answered Jul 03 '21 at 14:37

fblundun

OP: _ I wonder if the statement can be true for some random variables_ – krkeane Jul 03 '21 at 18:39
@krkeane I don't understand your comment. – fblundun Jul 03 '21 at 21:44
Plural: "random variables", so constant ruled out. – krkeane Jul 03 '21 at 22:26
@krkeane no, a constant random variable is still a random variable. See the "Constant random variable" section on this Wikipedia page: https://en.wikipedia.org/wiki/Degenerate_distribution – fblundun Jul 04 '21 at 10:09

2 Answers2