# Rubin Causal Model and Selection Bias

In the Rubin Causal Model, with a binary treatment $$T \in \{0,1\}$$, the selection bias is expressed as: $$$$E(y_0|T=1) - E(y_0|T=0)$$$$ where $$E(y_0|T=1)$$ denotes the expected value of unit $$i$$ in the control group if it were treated (not observed), and $$E(y_0|T=0)$$ is the expected value of unit $$i$$ in the control group (observed).

Then, the slides suggest that by assuming the treatment is orthogonal to $$(y_0, y_1)$$, and therefore the treatment is randomly assigned to a treatment and a control group, this implies that $$E(y_0|T=1) = E(y_0|T=0)$$. Does this mean that unit $$i$$ in the control group receives, on average, the same effect whether treated or not? This would make the treatment, on average, uneffective for the control group. This seems very strange to me.

Lastly, the slides mention that we can eliminate the selection bias by assuming $$T_i \perp (Y_0, Y_1) | X$$. What does this mean?

I recommend you read my answer here and answers to linked questions and remember to consider $$y_0$$ as an unmeasured pre-treatment confounder. Then the quantity $$E[y_0|T=1] - E[y_0|T=0]$$ refers to pre-existing differences between the groups on this confounder.
It's also important to note that "in the control group" refers to those with $$T=0$$, and has nothing to do with $$y_0$$. All units, both treated and control units, have a value of $$y_0$$; remember, $$y_0$$ is a pre-treatment confounder. So your statement "unit $$i$$ in the control group receives, on average, the same effect whether treated or not" is getting these two backward; $$E[y_0|T=1] = E[y_0|T=0]$$ means that the mean of the pre-treatment confounder $$y_0$$ is the same in the treated group ($$T=1$$) and the control group ($$T=0$$).
I mention this a bit in my answer linked above, but the assumption $$T \perp(y_0, y_1)|X$$ means that conditional on the covariates $$X$$, the potential outcomes (which, again, should be thought of as unmeasured pre-treatment confounders) do not affect selection into treatment.