Regression specification: what if one regressor is function of another

Question

Consider the following regression model

$$ Y_i=D_{i}(\alpha_1+\beta_1 X_i)+(1-D_i)(\alpha_2+\beta_2 X_i)+\epsilon_i $$ where $D_i$ is a binary variable. Suppose the researcher has an i.i.d. sample $\{Y_i, X_i, D_i, Z_i\}_{i=1}^n$ where $Z_i$ is an instrument such that $E(Z_i^\top \epsilon_i)=0$.

One way to estimate $\theta\equiv (\alpha_1,\beta_1,\alpha_2,\beta_2)$ is by GMM. In particular, note that we can write $$ E(Z_i^\top\underbrace{(Y_i-D_{i}(\alpha_1+\beta_1 X_i)+(1-D_i)(\alpha_2+\beta_2 X_i))}_{g(Y_i, D_i, X_i; \theta)})=0 $$ We assume identification, i.e., the true parameter $\theta_0$ is the unique minimizer of $$ E(Z_i^\top g(Y_i, D_i, X_i; \theta) )^\top W E(Z_i^\top\ g(Y_i, D_i, X_i; \theta)) $$ where $W$ is some positive definite matrix. The GMM estimator $\hat{\theta}$ is the solution of $$ (1) \quad \min_{\theta}(\frac{1}{n}\sum_{i=1}^nZ_i^\top g(Y_i, D_i, X_i; \theta) )^\top W_n \frac{1}{n}\sum_{i=1}^n Z_i^\top\ g(Y_i, D_i, X_i; \theta) $$

Question: Now, suppose that each $D_i = f_i(\theta)$ is a parametric function of $\theta$ and I denote it as $D_i(\theta)$. Does the fact that I essentially observe $D_i(\theta_0)$ in the data (under correct model specification) allow me to solve (1) instead of $$ (2) \quad \min_{\theta}(\frac{1}{n}\sum_{i=1}^nZ_i^\top g(Y_i, D_i(\theta), X_i; \theta) )^\top W_n \frac{1}{n}\sum_{i=1}^n Z_i^\top\ g(Y_i, D_i(\theta), X_i; \theta) $$ ? If not, which are assumptions are needed for being justified to solve simply (1)?

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Further comments, related to the answer below:

Consider the problem at the population level. Let $\theta_0\in \mathbb{R}^4$ be the true parameter value.

(i) The orthogonality condition is $$ (3) \quad E\Big[Z_i^\top \Big(Y_i- f_i(\theta_0)(\alpha_{1,0}+\beta_{1,0} X_i)+(1-f_i(\theta_0))(\alpha_{2,0}+\beta_{2,0} X_i)\Big)\Big]=0 $$ Given that $D_i$ that we observe is indeed $f_i(\theta_0)$ (under correct model specification), (3) is equivalent to $$ (4) \quad E\Big[Z_i^\top \Big(Y_i- D_i(\alpha_{1,0}+\beta_{1,0} X_i)+(1-D_i)(\alpha_{2,0}+\beta_{2,0} X_i)\Big)\Big]=0 $$ Correct?

(ii) The identification condition is $$ (5) \quad E\Big[Z_i^\top \Big(Y_i- f_i(\theta)(\alpha_{1}+\beta_{1} X_i)+(1-f_i(\theta))(\alpha_{2}+\beta_{2} X_i)\Big)\Big]=0 \quad \text{ only if $\theta=\theta_0$} $$ From reading the answer below, I understand that (5) is equivalent to $$ (6) \quad \theta_0\underbrace{=}_{\text{unique}}\text{argmin}_{\theta\in \mathbb{R}^4} E\Big[Z_i^\top \Big(Y_i- f_i(\theta)(\alpha_{1}+\beta_{1} X_i)+(1-f_i(\theta))(\alpha_{2}+\beta_{2} X_i)\Big)\Big]^\top W E\Big[Z_i^\top \Big(Y_i- f_i(\theta)(\alpha_{1}+\beta_{1} X_i)+(1-f_i(\theta))(\alpha_{2}+\beta_{2} X_i)\Big)\Big]\\ \text{ s.t. $f_i(\theta)=D_i$} $$ where $W$ is some positive definite matrix. Correct?

(iii) Now, suppose I solve the unconstrained problem, where I simply plug in $f_i(\theta)=D_i$ in the objective function and obtain $$ (7) \quad \min_{\theta\in \mathbb{R}^4} E\Big[Z_i^\top \Big(Y_i- D_i(\alpha_{1}+\beta_{1} X_i)+(1-D_i)(\alpha_{2}+\beta_{2} X_i)\Big)\Big]^\top W E\Big[Z_i^\top \Big(Y_i- D_i(\alpha_{1}+\beta_{1} X_i)+(1-D_i)(\alpha_{2}+\beta_{2} X_i)\Big)\Big] $$ Can I get a solution for (7) that is different from the unique solution $\theta_0$ of (6)? I think I will get the same unique solution $\theta_0$. Therefore, at the population level, the constrained and unconstrained problems are equivalent. Could you confirm?

(iv) Let us now move to the sample level, where we replace $E$ with the sample average $\frac{1}{n}\sum_{i=1}^n$. Is the minimiser of the sample analogue of (6) the same as the minimiser of the sample analogue of (7)? If I solve (7) instead of (6), do I need to take care of some extra variation when computing standard errors (and this is perhaps what you refer to when saying "sensible" and "valid" estimator)?

Answer 1

So long as the vector $\mathbf{D}$ is observed, you could still estimate by solving $(1)$. This estimate would not take account of the information in your functions, but it would still be "valid" in the sense that it is a sensible estimator in the absence of that information.

If you want to incorporate the new information in your functions, you wouldn't do it in the way you're proposing --- the $D_i$ values are observed, so they are fixed values that should not vary in the minimisation. Instead, you would just take account of the fact that when you observe $\mathbf{D}=\mathbf{d}$ your parameters are now known to fall within the subspace:

$$\Theta(\mathbf{d}) \equiv \{ (\alpha_1, \beta_1, \alpha_2, \beta_2) | (\forall i=1,...,n): d_i = f_i(\alpha_1, \beta_1, \alpha_2, \beta_2) \}.$$

Consequently, your new minimisation would be just like $(1)$, except that you should only search over the parameters $\theta \in \Theta(\mathbf{d})$. In your comments you note that you only have access to your functions $f_1,...,f_n$ "in principle" but they are not in closed form and you would need to simulate them. This is likely to make it extremely difficult to determine the set $\Theta(\mathbf{d})$ for your observed $\mathbf{d}$. As a fallback if this is impossible, you could instead obtain roughly the same result by using constrained minimisation via penalty methods.$^\dagger$

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$^\dagger$ To do the constrained minimisation via penalty method, you would formulate a sequence of smooth penalty functions $p_1,p_2,p_3,...$ that satisfies the limiting requirement:

$$\lim_{k \rightarrow \infty} p_k(\theta) = \begin{cases} 0 & & & \text{if } \theta \in \Theta(\mathbf{d}), \\[6pt] \infty & & & \text{if } \theta \notin \Theta(\mathbf{d}). \\[6pt] \end{cases}$$

You would then solve a sequence of unconstrained minimisation problems where you minimise $F(\theta) - p_k(\theta)$ for $k=1,2,3,...$ (where $F$ is your objective function) and estimate using the result for some sufficiently large value of $k$.