Conditional Distribution Deconstructing a Random Variable

Question

If I have a R.V $$X|Y=y\sim N(y,9) \text{ and } Y\sim N(10,8)$$ then $(X|y)=y+3Z$ where $Z\sim N(0,1).$ Is it possible to write $X$ as $X=Y+3Z$ to find the marginal distribution of $X$ i.e can I construct a random variable $Z$ that is independent of $Y$. Then will $$X=Y+3Z$$ have the same conditional distribution of $X|Y$. If this is true can I just get the marginal of $X$ through this form. $$E[X]=E[Y]+E[3Z]$$ $$Var(X)=Var(Y)+Var(3Z)$$ Then $X\sim N(10,17)$

Answer 1

The specific way your $X$ depends on $Y$ makes this work. We have

$$\forall y: (X \mid Y=y) \sim N(y, 9) = y + N(0, 9)$$

so $X = Y + N(0, 9)$, where the $N(0, 9)$ is independent of $Y$. Your results then follow.

Note that if the dependence were different, e.g. $(X \mid Y=y) \sim Uniform(0, |y|)$ then this kind of "decomposition" may not work.