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, $$\sigma=1$$.

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

• Hmm. That wasn't my answer. I just took the second equation and put in either expectations or given values, and solved for $X.$ The solutions manual solution makes no sense to me, either. Where did the $1/9$ come from? Why divide $1/3$ by $10/9?$ Dec 2, 2021 at 14:59

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 $$2\times\beta_{XY} = .6$$.