# A regression decomposition for dummy variables

Suppose that I have a regression model

\begin{align} Y = X\beta_X + Z_1 \beta_{Z_1} + Z_2 \beta_{Z_2} + \varepsilon \end{align}

where $X$ are some controls, and $Z_1$ and $Z_2$ are two sets of dummy ($\{0,1\}$-valued) variables. I assume that $Z_2$ does not depend upon $Z_1$, but $Z_1$ might depend upon $Z_2$ in the sense that the probability that an element of $Z_1$ might be affected by the value of $Z_2$. I would like to decompose the information (variation) in $Z_2$ into a component that affects $Y$ only through $Z_1$, and a residual that affects $Y$ directly.

If $Z_1$ were continuous, I could do something like regress (each coordinate of) $Z_1$ onto $Z_2$, then regress $Y$ on the fitted values. The natural analogue of this, say fitting a logistic (or similar) regression model for $Z_1$ gives up linearity, losing interpretability, and takes much longer to compute.

I would appreciate any insight on literature addressing this or similar problems.

In the approach you write for the case where the r.v.'s were continuous, you would regress $Z_1$ on $Z_2$ and use the fitted values $\hat Z_1$ in the $Y$-regression. But $\hat Z_1 = \hat E(Z_1\mid Z_2)$. So what you need is the conditional expectation function of $Z_1$ given $Z_2$.

This paper has a nice exposition for the bivariate Bernoulli distribution (as an introduction to the multivariate case).

Denoting the joint probability $P(Z_1 = 0, Z_2=0)=p(0,0)$ and analogously for the other three possible combinations, the conditional density of $Z_1\mid Z_2$ is (eq. 2.11 of paper) $$P(Z_1=z_1\mid Z_2=z_2) = \left(\frac {p(1,z_2)}{p(1,z_2)+p(0,z_2)}\right)^{z_1}\left(\frac {p(0,z_2)}{p(1,z_2)+p(0,z_2)}\right)^{1-z_1}$$

Then the conditional expectation function we are looking for is

$$E(Z_1\mid Z_2=z_2) = \frac {p(1,z_2)}{p(1,z_2)+p(0,z_2)} \equiv Z_1^*$$

You will then have to approximate the joint probabilities, creating a standard $2 \times 2$ contingency table of relative empirical frequencies -from which actually you can also test your dependence suspicion: theoretically, $Z_1$ and $Z_2$ will be independent if and only if $p(0,0)\cdot p(1,1) = p(0,1)\cdot p(1,0)$ and you can test this by a Fisher exact test.

Assuming that independence is rejected, then the contigency table will give you the estimated probabilities that you need and you will obtain the regressor

$$\hat Z_1^* = \frac {\hat p(1,z_2)}{\hat p(1,z_2)+\hat p(0,z_2)}$$

..which will produce a series of numbers, as for each observation $j$ of the sample we will use the ratio that corresponds to the value $z_{2j}$ of $Z_2$, i.e. if $z_{2j}=1$ we will have $$\hat z_{1j}^* = \frac {\hat p(1,1)}{\hat p(1,1)+\hat p(0,1)}$$ etc. $\hat Z_1^*$ is still a binary variable, but not taking the $\{0,1\}$ values any more.

• I agree, this is the way to go if $Z_1$ and $Z_2$ are one-dimensional. This approach suffers some rather severe curse of dimensionality, however. Since I care only about modeling the mean of $Y$, it seems that there should be better approaches? It seems that the paper you reference gives a way of parameterizing the correlation structure. Would you mind making it clearer how I could estimate the conditional means of $Z_1$ without estimating exponentially many parameters in the dimension of $Z_2$? – user39430 Sep 19 '13 at 5:28
• Multidimensionality needs some clarifications: you mean that $Z_1$ is a vector of r,v,'s, $\beta_{Z_1}$ is a vector of coefficients and $Z_1\beta_{Z_1}$ is their inner product? (the same question applies to $Z_2$). Also, in such a case, all elements of the random vector $Z_2$ have a dependence relationship with all elements of the random vector $Z_1$? – Alecos Papadopoulos Sep 19 '13 at 9:41
• Yes, sorry. I said "...$Z_1$ and $Z_2$ are two sets of...", but I could have been clearer. All elements of $Z_2$ could depend upon $Z_1$ (and hopefully whatever procedure I use would give me some information on this). – user39430 Sep 19 '13 at 14:56