Let's say we have two random discrete variables. Given an arbitrary element $y$ in $[0, 2^{n}-1]$, we compute the probability of $y$, where $y = x + b \mod 2^{n}$ by:
$$P(y) = \sum_{x=0}^{2^{n}-1} (P(x) \cdot P(b))$$
Am I right? But can we do the same if we consider $y = x \oplus b$? Namely, given an arbitrary element $y$ in $[0, 2^{n}-1]$, we compute the probability of $y$, where $y = x \oplus b$ by:
$$P(y) = \sum_{x=0}^{2^{n}-1} (P(x) \cdot P(b))$$
Or some other way?