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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!

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  • $\begingroup$ 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 $10/9?$ $\endgroup$ Dec 2, 2021 at 14:59

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

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