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In Mostly Harmless Econometrics, the author uses the following identity to derive an estimator for the causal effect:

$$E \left[ \frac{Y_i D_i} {p(X_i)} \right] = E \left[Y_{1i} \right]$$

where:

  • $D_i$ is the dummy treatment,
  • $Y_{1i}$ is potential outcome when $D=1$,
  • $Y_i$ is the observed outcome, and
  • $p(X_i)$ is the propensity score

but I'm having a hard time trying to come up with a proof for it.

This question is related, saying that this "can be shown using Law of Iterated Expections and conditional independence assumption".


I tried using the law of iterated expectations twice (on $D_i$ and $p(X_i)$) but I end up with this:

$E(D_i) E \left[ \frac{1}{E(D_i|X_i)} \right] E(Y_{1i})$

but due to non-linearity of the function inside of the expectation it doesn't simplify to the expectation of potential outcome.

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    $\begingroup$ I think the question you are asking here is not very clear. Are you asking for a proof of the identity or is it something else that you are after? If you want a proof, I think people would appreciate if you show what you've tried so far. $\endgroup$ – baruuum Jun 30 at 2:19
  • $\begingroup$ Yes, I'm after a proof for the identity. Added a couple lines on what I tried. $\endgroup$ – gsmafra Jun 30 at 2:42
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    $\begingroup$ I think that the trick is to condition on $x_i$. Here are a few tips: 1) notice that conditional on $x_i$, the propensity score,$p(x_i)$ (which is just a function of $x_i$) is a constant, so you can pull it out of the conditional expectation. 2) Next, $D_iy_{i} = D_i y_{i1}$ (notice the change in subscripts). You can show this by expanding $y_i$ as $y_i = D_i y_{i1} + (1-D_i)y_{0i}$. 3) Lastly, think what the conditional independece (ignorability) condition implies on the conditional expectation $E[y_iD_i|x_i]$. $\endgroup$ – baruuum Jun 30 at 3:10
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    $\begingroup$ Oh I see. The trick I didn't realize was that $D_i Y_i = D_i Y_{1i}$. I was hopelessly trying to condition on $D_i$ instead. Do you want to write an answer for that? $\endgroup$ – gsmafra Jun 30 at 3:25
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I assume that by "conditional independence assumption" you mean

$$ Y_{1i}, Y_{0i} \perp D_i | X_i.$$

The trick is simply to condition on $X_i$. Conditional on $X_i$, the propoensity score $p(X_i)$ is just a constant. Also, note that $$Y_iD_i = [Y_{1i}D_i + Y_{0i}(1-D_i)]D_i = Y_{i1}D_i$$ since $D_i^2 = D_i$ and $D_i(1-D_i) = 0$ and that, by definition, $p(X_i) = E[D_i|X_i]$. So, you have $$\begin{aligned} E\left[\frac{Y_i D_i}{p(X_i)}\right] &=E\left[E\left[\frac{Y_i D_i}{p(X_i)}\Bigg| X_i\right]\right] \\ &= E\left[\frac{1}{p(X_i)}E[Y_{1i}D_i|X_i]\right] \\ &= E\left[\frac{1}{p(X_i)}E[Y_{1i}|X_i]E[D_i|X_i]\right] \\ &= E\left[\frac{1}{p(X_i)}E[Y_{1i}|X_i]p(X_i)\right]\\ &=E[E[Y_{i1}|X_i]]\\ &=E[Y_{i1}]. \end{aligned}$$

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