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Oct 16, 2021 at 1:02 history edited Fcold CC BY-SA 4.0
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Oct 16, 2021 at 0:52 comment added Fcold Absolutely, the effect of A on P(y) will change depending on how large is Var(U). Reduction to the absurd. IF Var(U) goes to infinity, A has no effect. However, you can do the following. 1) y* = x1 + x2 + u. with u~N(0,1). then dy = y*>0 . Here you can compare the results modeling P(y=1|X) = F(x1) and F(x1,x2). Coefficients will change, but marginal effects will be almost the same.
Oct 15, 2021 at 19:26 comment added Pburg What calculation are you suggesting? I think $\Pr(Y=1|A)$ will depend on the distribution of $U$. Suppose $U$ has, compared to $A$, very extreme values in our sample so that it basically determines $Y$. Then we might see almost no effect from $A$. But if instead $U$ is low variance, then we will get a substantial effect from $A$. In either case $U$ and $A$ are drawn independently. I've simulated this with $A$ a Bernouli(0.5) rv, $U$ being normal with high/low variance, then checking the cross tabs for $A$ and $Y$. Similar results with $A$ normal.
Oct 15, 2021 at 17:33 history answered Fcold CC BY-SA 4.0