How to estimate the best guess given observed viariables in structural causal model? Here is a question in the book CAUSAL INFERENCE IN STATISTICS: A PRIMER.
Suppose we have the following SCM. Assume all exogenous variables are independent and that the expected value of each is 0.
$V=\{X, Y, Z\},\quad U=\{U_X, U_Y, U_Z\}, \quad F=\{f_X, f_Y, f_Z\}$
$f_X: X=U_X$
$f_Y: Y=X/3+U_Y$
$f_Z: Z=Y/16+U_Z$
(e) Assume that all exogenous variables are normally distributed with zero means and unit variance, that is, $\sigma=1$.

*

*Determine the best guess of X, given that we observed $Y=2$.​

*(Advanced) Determine the best guess of $Y$, given that we observed $X=1$ and $Z=3$.​
[Hint: You may wish to use the technique of multiple regression, together with the fact that, for every three normally distributed variables, say $X$, $Y$, and $Z$, we have $E[Y|X=x,Z=z]=R_{YX\cdot Z} x+R_{YZ\cdot X}z$.]

For 1, the solution provided says that the regression coefficient of $X\sim Y$ should be $\frac{1}{3}/(1+\frac{1}{9})=\frac{9}{30}=0.3$. So the best guess should be 0.6.
I totally can't understand where the idea of solution comes from. I can get the best guess of the value of $Y$, given that we observe $X=x$. But in reverse it makes me confused.
Could you please tell me how to calculate it and the basic idea? Thank you very much!

The question is essentially asking you to imagine running a regression of $X$ on $Y$ and then plugging in the value of $Y$ to get the predicted value of $X$. We known that $\beta_{YX}$, the coefficient on $X$ in $f_Y$, is equal to $\frac{\text{Cov}(Y, X)}{\text{Var}(X)}$. Our goal is to compute $\beta_{XY}$, the coefficient on $Y$ in a hypothetical regression of $X$ on $Y$.
We know that $\beta_{XY} = \frac{\text{Cov}(Y, X)}{\text{Var}(Y)}$, so we need to find the values of those components and plug them in to get $\beta_{XY}$.
We know $\text{Cov}(Y, X) = \beta_{YX}\text{Var}(X)$ by doing a little algebra. $\beta_{YX} = 1/3$ and $\text{Var}(X) = 1$ (because $\text{Var}(U_X) = 1$), so $\text{Cov}(Y, X) = 1/3$. We still need to find $\text{Var}(Y)$.
\begin{align}
\text{Var}(Y) &= \text{Var}(X/3 + U_Y) \\
&= \text{Var}(X)/3^2 +\text{Var}(U_Y) \\
&= 1/9 + 1 \\ &= 10/9
\end{align}
So, $\beta_{XY} = \frac{\text{Cov}(Y, X)}{\text{Var}(Y)}=\frac{1/3}{10/9}=3/10=.3$. Our best guess of $X$ when $Y=2$ is thus $2\times\beta_{XY} = .6$.
