Lets say that i have linear regression $Y= a +XB$, and in my $X$ matrix i have a dummy for the gender, lets call it $d_{g}$ which is 1 for male and 0 for female. The coefficient $b$ of $d_{g}$ shows the effect of being male relative to the effect for being female. How can i get the effect of being female ? Or the effect of being male, but not relative to the female.
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$\begingroup$ Could you elaborate on "the effect of being male, but not relative to the female"? Relative to what then? $\endgroup$– Scortchi ♦Commented Jun 7, 2017 at 8:47
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2 Answers
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You can reconstruct what you want from your fit using the population fractions of male and female $f_M$ and $f_F = 1 - f_M$.
The expectation of $Y$ for males and females is given by: $$ E[Y|M] = a + b \qquad E[Y|F] = a$$ so $$ E[Y|M] - E[Y|F] = b$$ $b$ gives the expected change in $Y$ going from female to male, as you said in your post.
Given the male and female population fractions we can construct the unconditional expectation: $$E[Y] = f_F E[Y|F] + f_M E[Y|M] = (1-f_M) a + f_M (a + b) = a + f_M b$$ and from that, we can get the expected change in $Y$ going from sex-unknown to male: $$E[Y|M] - E[Y] = (1 - f_M) b$$ and, similarly, the expected change in $Y$ going from sex-unknown to female: $$E[Y|F] - E[Y] = - f_M b$$ You should get the same results with this method as from the procedure suggested by Michael above, without having to re-run the regression.
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$\begingroup$ +1, this is the better solution once the regression is already done. $\endgroup$ Commented Jun 7, 2017 at 8:33
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$\begingroup$ (+1) Wonder if that's the desired interpretation of "the effect of being male, but not relative to the female". (I can't think of another that makes any sense) $\endgroup$– Scortchi ♦Commented Jun 7, 2017 at 8:45
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$\begingroup$ Thank you. @Scortchi I am sorry if it was not clear enough. $\endgroup$– quantCommented Jun 7, 2017 at 8:58
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1$\begingroup$ @quant: You can extend it, but the algebra gets pretty harry. Suppose your model is $Y = a + b u + c v$, where $u = \{0, 1\}$ and $v = \{0, 1\}$ are your dummy variables for female/male and poor/rich, respectively. You can write fully conditional expectations $E[Y|0 0] = a, E[Y|1 0] = a+b, E[Y|0 1] = a+c, E[Y|1 1] = a + b + c$. Given the population fractions of each subgroup $p_{0 0}, p_{1 0}, p_{0 1}, p_{1 1}$, you can compute semi-conditional probabilities like $E[Y|1 .] = (p_{1 0} E[Y|1 0] + p_{1 1} E[Y| 1 1])/(p_{1 0} + p_{1 1}) = a + b + \frac{p_{1 1}}{p_{1 0} + p_{1 1}} c$ $\endgroup$ Commented Jun 8, 2017 at 17:21
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1$\begingroup$ and unconditional probabilities like $E[Y|. .] = a + (p_{1 0} + p_{1 1}) b + (p_{0 1} + p_{1 1}) c$, and then take difference to get expected changes such as $E[Y|1 .] - E[Y|. .] = (p_{0 0} + p_{0 1}) b + \left[ \frac{p_{1 1}}{p_{1 0} + p_{1 1}} - (p_{0 1} + p_{1 1}) \right] c$. There may be some way to convert this to more compact matrix expressions; I don't know. If you are willing to re-do the regression, it would also be straightforward to fall back to Michael's suggested procedure. $\endgroup$ Commented Jun 8, 2017 at 17:27
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This can simply be achieved by leaving out the constant (other wise you run in the so called dummy-trap) and using a model like $$Y=d_{female} + d_{male} + X\beta + u, $$ where $d_{female}$ is one if female and $d_{male}$ is one if male. Then you will have $$E[Y|X,male]=d_{male} + E[X]\beta,$$ $$E[Y|X,female]=d_{female} + E[X]\beta$$
and $E[Y|X,male]-E[Y|X,female]=d_{male} -d_{female}$.
