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In Angrist's book Mostly Harmless Econometrics, section 3.4.2, it says $$ E[Y_i|D_i]=E[Y_i|Y_i>0,D_i]P[Y_i>0|D_i] $$, where $Y_i$ is a non-negative variable ($Y_i$ can be $0$) and $D_i$ is a dummy variable ($0$ and $1$).

I can somehow understand it intuitively, but I want a strict proof.

A discrete case is okay. Please avoid notations about measure theory since I'm a beginner.

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    $\begingroup$ Do you know the Law of Total Expectation? en.wikipedia.org/wiki/Law_of_total_expectation $\endgroup$
    – sudeep5221
    Commented Mar 29 at 15:22
  • $\begingroup$ I know the law of total expectation and iterated expectation basically. But I'm a little confused about the application of these laws on conditional expectation, especially the case that the condition contains the random variable too. Sorry for my silly question. $\endgroup$
    – Kozack51
    Commented Mar 30 at 2:55
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    $\begingroup$ $E[Y_i|Y_i>0,D_i]=\frac{E[Y_iI_{(0,\infty)}(Y_i)|D_i]}{P[Y_i>0|D_i]}$, but in this case $Y_iI_{(0,\infty)}(Y_i)=Y_i$ $\endgroup$
    – Speltzu
    Commented Mar 30 at 16:59
  • $\begingroup$ At a high level, you can think of conditioning on a variable as simply inducing a new probability law. So you can do whatever you know of without conditioning, you can do with conditioning. For example, in the standard case, you know that the law of total expectation looks like $\mathbb{E}[X] = \mathbb{E}[X|A]\Pr(A) + \mathbb{E}[X|A^c] \Pr(A^c)$. Here $A$ is some event and $A^c$ is its complement and the expectation is with respect to the probability law under which the probability of $A$ is calculated. $\endgroup$
    – sudeep5221
    Commented Mar 30 at 20:40
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    $\begingroup$ Thus, $\mathbb{E}[Y_i|D_i] = \mathbb{E}[Y_i|Y_i > 0, D_i] \Pr(Y_i > 0|D_i) + \mathbb{E}[Y_i|Y_i = 0, D_i] \Pr(Y_i = 0|D_i)$. Now, conditioning on $Y_i = 0$, the expected value of $Y_i$ is $0$, so the second term vanishes giving the result. Note that all this explanation is intuitive like you wanted (so I have skipped some rigor). However, if you continue studying probability, I would suggest to eventually understand these ideas in measure theoretic terms. $\endgroup$
    – sudeep5221
    Commented Mar 30 at 20:46

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