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Can someone tell me how in equation (40) the two equal signs are attained here?

I know pictures and specific questions are not cool but I am so lost

enter image description here

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  • $\begingroup$ We will surely need more information $\endgroup$ – George Dewhirst Apr 17 at 17:19
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Although not specified, $d_{i}$ is $1$ if individual is treated, $0$ otherwise. By very definition of Expected Value:

$$E(d_{i} \mid z_{i} ) = 1 \times P(d_{i} = 1 \mid z_{i}) + 0 \times P(d_{i} = 0 \mid z_{i}) = 1 \times P(d_{i} = 1 \mid z_{i}) = P(z_{i})$$

the latter by definition of propensity score. Additionally, it is assumed (standard assumption, external validity of the instrument), $E(u_{i}) = E(u_{i} \mid z_{i}) = 0$, hence proving the second equality. For the third, last equality in your question, we use the fact that, given two random variables $X$ and $Y$, it holds that:

$$E(X\mid f(Y)) = E(E(X\mid Y)\mid f(Y))$$

hence, in our case:

\begin{align} E( y_{i} \mid P(z_{i}) ) &= E(E(y_{i} \mid z_{i}) \mid P(z_{i})) \\ &= E(\beta + \alpha P(z_{i}) + E(u_{i} \mid z_{i}) \mid P(z_{i})) \\ &= \beta + \alpha P(z_{i}) \\ &= E(y_{i} \mid z_{i}) \end{align} where the first equality follows from the aforementioned property, the second has been proved before, the third is immediate, the last by again what proved initially.

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  • $\begingroup$ For the third equality did you use the Law of iterated expectations? Is that what that is? E(X|f(y)) = E(E(X|Y)|f(Y)) ? $\endgroup$ – Emil Krabbe Apr 18 at 8:34
  • $\begingroup$ No, I just used the conditional mean independence $E(u_{i} \mid z_{i})$ $=$ $E(u_{i})$ and $E(u_{i})$ $=$ $0$. The LIE is $E(E(X \mid Y))$ $=$ $E(X)$, while the first property used is a derivation of it. $\endgroup$ – Nicg Apr 18 at 9:48

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