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Problem Statement: Suppose we have the following Structural Causal Model (SCM). Assume all exogenous variables ($U$) are independent identically distributed standard normals. \begin{align*} V&=\{X,Y,Z\},\qquad U=\{U_X, U_Y, U_Z\},\qquad F=\{f_X, f_Y, f_Z\}\\ f_X: X&=U_X\\ f_Y: Y&=\frac{X}{3}+U_Y\\ f_Z: Z&=\frac{Y}{16}+U_Z. \end{align*} 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 $E[Y|X=x,Z=z]=R_{Y\!X\cdot Z}x+R_{Y\!Z\cdot X}z.$]

My Answer: We assume the model $$Y=\alpha+\beta_XX+\beta_ZZ+\varepsilon,$$ with $\beta_X$ and $\beta_Z$ given by \begin{align*} \beta_X=R_{Y\!X\cdot Z} &=\frac{\sigma_Z^2\sigma_{Y\!X}-\sigma_{Y\!Z}\sigma_{Z\!X}}{\sigma_X^2\sigma_Z^2-\sigma_{X\!Z}^2}\\ \beta_Z=R_{Y\!Z\cdot X} &=\frac{\sigma_X^2\sigma_{Y\!Z}-\sigma_{Y\!X}\sigma_{X\!Z}}{\sigma_Z^2\sigma_X^2-\sigma_{Z\!X}^2}. \end{align*} Now here's where things get a little hazy. Going off the model, and the idea that if $C=aA+bB,$ then $\sigma_C^2=a^2\sigma_A^2+b^2\sigma_B^2,$ we have the following equations: \begin{align*} \sigma_X^2&=\sigma_{U_X}^2=1\\ \sigma_Y^2&=\frac19 \sigma_X^2+\sigma_{U_Y}^2=\frac{10}{9}\\ \sigma_Z^2&=\frac{1}{256}\sigma_Y^2+\sigma_{U_Z}^2=\frac{1157}{1152}. \end{align*} Here's where I have trouble: computing the covariances. I know that, for example, $$\sigma_{XY}=E(XY)-\underbrace{E(X)}_{=0}E(Y)=E(XY)$$ in this case, since $E(X)=0.$ Then I substitute in to obtain \begin{align*} E(XY) &=E(X(X/3+U_Y))\\ &=\frac13 E\big(X^2\big)+\underbrace{E(XU_Y)}_{=0}\\ &=\frac13\left(\sigma_X^2-(E(X))^2\right)\\ &=\frac13. \end{align*} A similar calculation reveals that \begin{align*} \sigma_{XZ}&=\frac{1}{48}\\ \sigma_{YZ}&=\frac{5}{72}. \end{align*}

My question is: are these calculations correct so far? If so, I think I can make my way to the end by plugging into the expressions above for the regression coefficients, and then the model.

Thank you for your time!

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  • $\begingroup$ Just for my edification, what does the notation $R_{YX \cdot Z}x$ means? $\endgroup$ Jan 8, 2021 at 15:57
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    $\begingroup$ Sure! $R_{YX\cdot Z}$ means the slope of $Y$ on $X$ when we hold $Z$ constant; it's also known as a partial regression coefficient. See Causal Inference in Statistics: A Primer, p. 23. $\endgroup$ Jan 8, 2021 at 16:19
  • $\begingroup$ Got it, thanks! So, question then. Shouldn't the formula be $E[Y|X,Z]=y_0+R_{YX \cdot Z}x+R_{YZ \cdot X}z$ where $y_0=E[Y|X=0,Z=0]$? $\endgroup$ Jan 8, 2021 at 16:29
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    $\begingroup$ Well, the formula I typed up in the hint is exactly what Pearl has. If you notice in my answer, I allow an offset such as you suggest. In the context of this problem, I doubt it matters, as I think the $y_0$ you've got there would be zero, given that the exogenous variables are all standard normals. $\endgroup$ Jan 8, 2021 at 20:24
  • $\begingroup$ Ah, yes, good point. I did not consider that - I suppose, in general, its just an offset. Thanks! $\endgroup$ Jan 8, 2021 at 20:50

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I believe that given that $X=1$, then $E(X)$ should be 1, not 0. I understand the confusion once $X=U_x$ and $E(U_x)=0$. But the previous only state that the value of $X$ is caused by the value of $U_x$, in this case the value of X is already observed to be 1, hence a constant $c$, and because $E(c)=c$ then $E(X)=1$. $E(Y|X=1)=E(1/3 +U_y)=1/3$.

Note that $\sigma_{xy}=0$ does not imply that X and Y are independent.

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I was, indeed, on the right track. To finish the calculations: \begin{align*} \sigma_X^2&=1\\ \sigma_Y^2&=10/9\\ \sigma_Z^2&=1157/1152\\ \sigma_{XY}&=1/3\\ \sigma_{YZ}&=5/72\\ \sigma_{XZ}&=1/48\\ \\ E[Y|X=1,Z=3]&=400/771. \end{align*}

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    $\begingroup$ Might a bit late, but this result also concurs with mine. Cheers! $\endgroup$ Dec 4, 2021 at 3:55

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