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

Mathematics Asked by zhongyuan chen on January 3, 2022

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.

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.

Answered by Michael Hardy on January 3, 2022

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