What does it mean to write X=Y+W where W is gaussian with mean 0. Assuming X Y are two random variables. When we write X=Y+W with W~Gaussian(with 0 mean), Do we mean the random variable X-Y ~ Gaussian(with 0 mean)? Or does it mean X|y ~ Gaussian(with mean y)? I think the two are clearly not equivalent since the second case seems to imply the first but the first doesn't imply the second.
 A: Probably what was intended is that $Y$ and $W$ are independent of each other and $W$ is Gaussian with expectation $0.$ People often omit to mention such an assumption of independence.
Necessarily this would imply that $X-Y$ is Gaussian with mean $0,$ but people don't usually express the situation that way when they intend the hypothesis of independence of $W$ and $Y$ to be tacitly understood.
The thing that you express by saying $X\mid y \sim\text{some distribution}$ is something I would express either by saying $X\mid (Y=y) \sim\text{some distribution depending on $y$}$ (thus $Y$ is a random variable and $y$ is not) or by saying $X\mid Y \sim \text{some distribution depending on $Y$}$ (so that things like $\operatorname E(X\mid Y)$ or $\operatorname{var}(X\mid Y)$ would themselves be random variables that are functions of $Y$).
Notice that if you say that $X\mid (Y=y) \sim \operatorname N(y,\sigma^2),$ that will entail that $X-Y$ is independent of $Y,$ as follows:
\begin{align}
X\mid (Y=y) \sim \operatorname N(y,\sigma^2) \\[8pt]
X-Y \mid (Y=y) \sim \operatorname N(0,\sigma^2)
\end{align}
and the expression $\text{“}\operatorname N(0,\sigma^2)\text{''}$ has no $\text{“}y \text{''}$ in it.
