What does it mean to write X=Y+W where W is gaussian with mean 0.

Question

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.

Michael Hardy · Answer

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 $Xmid y simtext{some distribution}$ is something I would express either by saying $Xmid (Y=y) simtext{some distribution depending on $y$}$ (thus $Y$ is a random variable and $y$ is not) or by saying $Xmid Y sim text{some distribution depending on $Y$}$ (so that things like $operatorname E(Xmid Y)$ or $operatorname{var}(Xmid Y)$ would themselves be random variables that are functions of $Y$).
Notice that if you say that $Xmid (Y=y) sim operatorname N(y,sigma^2),$ that will entail that $X-Y$ is independent of $Y,$ as follows:
begin{align}
Xmid (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.

What does it mean to write X=Y+W where W is gaussian with mean 0.

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