Independence of random variables $X$ and $X+Y$

Question

Let $X,Y$ be two random variables on a probability space $(\Omega, \mathcal{F},\mathbb{P})$.

I want to prove the fact that \begin{align} X \text{ is indep}&\text{endent of } X+Y \\ &\iff \\ \exists c \in \mathbb{R}&: \mathbb{P}(X=c) =1. \end{align} First, we consider the implication $(\implies)$. We know that \begin{align} \mathbb{P}(X=c, X+Y=c+y) = \mathbb{P}(X=c)\mathbb{P}(X+Y=c+y). \end{align} Which we could interpret as the conditional expectation \begin{align} (*) \qquad \mathbb{P}(X=c|X+Y=c+y) = \frac{\mathbb{P}(X=c, X+Y=c+y)}{\mathbb{P}(X+Y=c+y)}= \mathbb{P}(X=c). \end{align} How to conclude from $(*)$ that $\exists c \in \mathbb{R}: \mathbb{P}(X=c) =1 $?

Answer 1

I believe the equivalence you are trying to prove is wrong. A simple counter example is given by taking $X\thicksim N(0,1)$ and setting $Y=-X$. Since $X+Y=0$ is a constant, $X+Y$ is independent of any random variable.

Note that constant random variables (or almost surely constant) are always independent of anything, which is what you have essentially proven above. However the reverse implication is not true.

Best regards

Answer 2

The premise is clearly wrong. For example, let $X$ and $Z$ be independent random variables, and let $Y=Z-X$. Then $X$ and $X+Y=Z$ are independent, regardless of what $X$'s distribution looks like.