# Is a continuous random variable always dependent on a linear combination which includes itself?

For example, where $Z$ may or may not be independent of $X$, is $X$ ever independent of $X+Z$? Intuitively, it seems absurd to suggest that a random variable may be independent of a function of itself. Indeed, I can think of cases which would provide zero covariance, however I can't think of a way to prove or disprove the claim that $X$ can be independent of $X+Z$, aside from citing the obviously degenerate case of $Z=-X$. Is there an approach which excludes this case and any other cases like it which boil down to just subtracting away the X variable?

• How about $Z = Y-X$ for some independent $Y$? Feb 5 '18 at 16:52
• Good point, would it be unfair of me to also exclude that case? Is there a way to exclude cases that do little more than totally remove the X term without also excluding all cases which would give independence? Feb 5 '18 at 16:54
• Some food for thought: if $(U,V)$ is i.i.d. standard normal, then $U/V$ and $U^2+V^2$ are independent. Thus, $X=U/V$ and $Z=(1+X^2)V^2-X$ are such that $X$ and $X+Z$ are independent.
– Did
Feb 5 '18 at 17:14

## 1 Answer

@Arthur suggests $Z = Y - X$, where $Y$ is independent of $X$, and OP asks "well, sure, but are there other examples?"

No, there are not. Here's why.

Suppose we have some "other" answer, i.e., suppose that $X$ is independent of $X + Z$.

Then we can let $Y = X + Z$, and we have a $Y$ that is independent of $X$, and the $Z$ we found is then $Z = Y - X$.

In short: no, there is no other possible answer, but this "simple" answer really covers quite a lot.