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While reading Jennifer Hill (2011), p. 220, in the context of conditional average treatment effect, the author is able to calculate $E[Y|Z]$ from $f(Y|X)$ and $f(X|Z)$.

My attempt to replicate the author's calculation is as follows: $$ \begin{align} E[Y|Z] &= \int y f(y|z) dy \\ &= \int y \int f(y, x| z) dx dy \\ &= \int y \int f(y|x, z) f(x|z) dx dy\\ &= \int y \int f(y|x) f(x|z) dx dy \end{align} $$

where the last equation is true because $y$ is independent from $z$ once conditional on $x$.

However, this integration is rather nightmarish even with the Gaussian distributions, i.e. $f(x|z=1) \sim N(40, 10^2)$ and $f(y|x) = N(72 + 3 \sqrt x, 1)$. I tried Mathematica, which couldn't give the numerical result either.

Is this the correct approach to calculate $E[Y|Z]$, or there is a more sensible approach?

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  • $\begingroup$ So you occasionally get complex means for $y$? $\endgroup$ – Taylor Jun 19 '16 at 13:20
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Continuing from your last statement, we have $$ \int_y y \int_x f(y|x) f(x|z) dx dy $$ $$ =\int_x\left[\int_y y f(y|x) dy \right] f(x|z) dx $$ $$ =\int_x E[Y|X] f(x|z) dx $$ and that's it.

But, if by any chance everything is Gaussian, then $$E[Y|X]=E[Y]+\alpha (X-E[X])$$ where $\alpha=\frac{cov(X,Y)}{var(X)}$. So we have $$ = \int_x \left[ \mu_Y + \alpha x -\alpha \mu_X \right] f(x|z) dx $$ $$ = \mu_Y -\alpha \mu_X + \alpha \int_x x f(x|z) dx $$ $$ = \mu_Y -\alpha \mu_X + \alpha E[X|Z]$$ which can be expressed easily using the same method, if it is Gaussian.

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  • $\begingroup$ How does $X$ and $Y$ being normal lead to that way of calculating $E[Y|X]$? I think that's only true when $X, Y$ are bivariate normal. $\endgroup$ – Heisenberg Jun 19 '16 at 15:54
  • $\begingroup$ You're right, by 'everything' I meant they're jointly normal. $\endgroup$ – yoki Jun 19 '16 at 16:28
  • $\begingroup$ Thanks a lot! Using $\int_x E[Y|X] f(x|z) dx$ Mathematica did manage to give the definite integral this time. Is it because we've simplified the integral? $\endgroup$ – Heisenberg Jun 19 '16 at 16:42
  • $\begingroup$ Well, I don't know how Mathematica works, but consider that switching the integral order does require some assumptions, so the algorithm might not have wanted to do that on its own. $\endgroup$ – yoki Jun 19 '16 at 16:46

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