The average treatment effect and the difference in means

Question

Hi I have a question related to the treatment effect.

Recently, I am reading literatures on treatment effect and have a question.

In the literatures, we denote the counterfactual outcomes as $Y_1$ and $Y_0$ where $Y_1$ is for the treated and $Y_0$ is for the untreated. Then, the observed outcome is $Y=W\cdot Y_1+(1-W)\cdot Y_0$ where $W$ is the indicator of the treatment.

Here, my first question is whether or not $E(Y_1|W=1)$ and $E(Y|W=1)$ are different?

Second, I found an equation that is as follows: $$ \begin{aligned} E(Y|W=1)-E(Y|W=0) &= E(Y_1-Y_0)\\ &+\{E(Y_1|W=1)-E(Y_1|W=0)\}P(W=0)\\ &+\{E(Y_0|W=1)-E(Y_0|W=0)\}P(W=1) \end{aligned} $$ where P() is the probability function. But, I can't derive the equation.

Please, Help Me! Thank you for your time spent to read this question.

Answer 1

To answer your first question, we have $$E(Y | W = 1) = E(W\cdot Y_1 + (1 - W) \cdot Y_0 | W = 1) = E(Y_1 | W = 1)$$ where the last equality follows by the fact that we are allowed to plug in the conditioned value of $W$ when evaluating the conditional expectation, so yes, they are equal. Of course, a similar argument shows that $E(Y | W = 0) = E(Y_0 | W = 0)$. I end up using both of these facts below:

To answer your second question, it seems the expression you wrote down is off by a negative sign. To get the correct expression, the only real trick is to use the law of iterated expectations prudently and to do some light algebra afterwards:

$$ \begin{aligned} E(Y_1) &= E(Y_1 | W = 1)P(W = 1) + E(Y_1 | W = 0) P(W = 0)\\ &= E(Y_1 | W = 1)[1 - P(W = 0)] + E(Y_1 | W = 0) P(W = 0)\\ &= E(Y_1 | W = 1) - \{E(Y_1 | W = 1) - E(Y_1 | W = 0)\} P(W = 0)\\ &= E(Y | W = 1) - \{E(Y_1 | W = 1) - E(Y_1 | W = 0)\} P(W = 0) \end{aligned} $$

$$ \begin{aligned} E(Y_0) &= E(Y_0 | W = 1)P(W = 1) + E(Y_0 | W = 0) P(W = 0)\\ &= E(Y_0 | W = 1)P(W = 1) + E(Y_0 | W = 0) [1 - P(W = 1)]\\ &= E(Y_0 | W = 0) + \{E(Y_0 | W = 1) - E(Y_0 | W = 0)\} P(W = 1)\\ &= E(Y | W = 0) + \{E(Y_0 | W = 1) - E(Y_0 | W = 0)\} P(W = 1) \end{aligned} $$

Now, adding these two equalities together and rearranging, we get

$$ \begin{aligned} E(Y|W=1)-E(Y|W=0) &= E(Y_1-Y_0)\\ &+\{E(Y_1|W=1)-E(Y_1|W=0)\}P(W=0)\\ &-\{E(Y_0|W=1)-E(Y_0|W=0)\}P(W=1) \end{aligned} $$