Is this an an endogeneity/simultaneity problem?

Question

I would like to know if the logic in these two situations is correct.

Situation 1: Let's say we have a continuous dependent variable, $y_1$, that then has a causal impact on an unobserved variable, $\rho$. This unobserved variable then has a causal impact on a variable, $y_2$, which has a causal impact on $y_1$. We want to estimate $ \frac{\partial y_1}{\partial y_2}$. So, we have a set of structural equations as follows:

$$y_1 = f(y_2, \mathbf{x_1})+ e_1$$ $$\rho = f(y_1, \mathbf{x_2})+ e_2$$ $$y_2 = f(\rho, \mathbf{x_3})+ e_3$$

where the $\mathbf{x_i}$ terms are exogenous and the $e_i$ terms are errors. By substituting the second equation into the third, we can see that we would have simultaneity and our estimates of the impact of $y_2$ on $y_1$ would be biased if we could not control for $\rho$ or some proxy for it in our estimate of the structural equation for $y_1$.

I am particularly unsure about this last part in italics. Could we use the proxy for $\rho$ in the estimate of the structural equation for $y_1$, or would we have to do 2SLS, with the proxy for $\rho$ being included in the first stage but excluded in the second?

Situation 2: Let's say we have a percentage dependent variable, $s_1$. Let's say the complement of $s_1$ is made up of two other percentages, $s_2$ and $s_3$. Furthermore, let's say that $s_3$ has a causal impact on an unobserved factor, $\rho$, and that $\rho$ has a causal impact on $y_2$, which has a causal impact on $s_1$. We want to estimate $\frac{\partial s_1}{\partial y_2}$. Thus, we have the following structural equations:

$$s_1 = f(y_2, \mathbf{x_1})+ e_1$$ $$\rho = f(s_3, \mathbf{x_2})+ e_2$$ $$y_2 = f(\rho, \mathbf{x_3})+ e_3$$

Let's now say that $s_2$ is more or less constant across observations. Thus, there is generally an inverse relation between $s_1$ and $s_3$. This implies that we can rewrite $s_3$ in the second structural equations in terms of $s_1$:

$$\rho = f(1 - (s_1 + \bar{s}_2), \mathbf{x_2})+ e_2$$

Then, just as in situation 1, by substituting this equation into the structural equation for $y_2$, we can see there would be simultaneity and our estimates of the impact of $y_2$ on $y_1$ would be biased (once again, I am unsure about whether controlling for $\rho$ or a proxy for it in our estimate of the structural equation for $y_1$ would solve this).

Answer 1

Let us consider Situation 1.

Let us assume that $\rho$ is observed. If it does not work when $\rho$ is observed, there is no reason why it (using a proxy of $\rho$ as instrument) should work when $\rho$ is not observed.

$\rho$ is endogenous so we can't just include it as a regressor in an equation. Thus, let us consider IV estimation.

Obvious instruments for the first equation are $\mathbf{x}_1$, $\mathbf{x}_2$, and $\mathbf{x}_3$. Can we use $\rho$ as an extra instrument? It depends on whether $\rho$ is relevant and whether $\rho$ is exogenous in the first equation (i.e., the $y_1$ equation). First, $\rho$ is relevant as it is correlated with $y_2$ (unless the last equation is degenerate). Next, is it exogenous?

To check it, let's go maths and consider the following simple model (without intercepts and exogenous regressors, for simplicity): $$y_1=\alpha y_2 + e_1,\;\; \rho = \beta y_1+e_2,\;\; y_2=\gamma \rho + e_3.$$ Write them in matrix form: $$ \begin{pmatrix} 1 & 0 & -\alpha\\ -\beta & 1 & 0\\ 0 & -\gamma & 1 \end{pmatrix} \begin{pmatrix} y_1\\ \rho\\ y_2 \end{pmatrix} = \begin{pmatrix} e_1\\ e_2\\ e_3 \end{pmatrix} . $$ Using Cramer's rule, we get $$\rho = (\beta e_1 + e_2 + \alpha \beta e_3 ) / (1-\alpha\beta\gamma).$$ $\rho$ is correlated with $e_1$ unless $\beta=0$, so $\rho$ can't be used as an instrument.

Edit

What happens if $y_1$ is regressed on $y_2$ and $\rho$? Then will $\alpha$ be consistently estimated? For simplicity, suppose that $e_1, e_2, e_3$ are $iid$ standard normal. Then the OLS estimator vector converges in probability to $$\begin{bmatrix} \alpha\\ 0 \end{bmatrix} + \begin{bmatrix} E(y_2^2) & E(y_2 \rho)\\ E(y_2\rho) & E(\rho^2) \end{bmatrix} ^{-1} \begin{bmatrix} E(y_2 e_1)\\ E(\rho e_1) \end{bmatrix} .$$ Letting $e=(e_1,e_2,e_3)'$, write $\rho = e'g$ and $y_2 = e'h$ for some $3\times 1$ nonrandom vectors $g$ and $h$. Then the second term is $$ \begin{pmatrix} h'h & h'g\\ g'h & g'g \end{pmatrix} ^{-1} \begin{pmatrix} h_1\\ g_1 \end{pmatrix} . $$ The first element of the above is the determinant inverse times $g'gh_1 - g'h g_1 = g'(gh_1 - hg_1)$. Let us calculate it. We have already obtained $g$. We now need $h$: $$y_2 = (-\beta\gamma e_1 - \gamma e_2 + e_3) / (1-\alpha\beta\gamma).$$

Let us ignore the determinant part (common). We can work out with $g=(\beta, 1, \alpha\beta)'$ and $h=(-\beta\gamma, -\gamma, 1)'$, ignoring the determinant. Then $$ gh_1 - hg_1 = [0, 0, -\beta (1+\alpha \beta\gamma) ]' \text{ times a constant}.$$ Thus, $$ g'(gh_1 - hg_1) = -\alpha \beta^2 (1+\alpha\beta\gamma) \text{ times another constant}.$$ The OLS regression of $y_1$ on $y_2$ and $\rho$ gives an inconsistent estimator in general, but gives a consistent estimator of $\alpha$ if $\alpha=0$ or $\beta=0$. Especially, if $\alpha=0$, then the OLS estimator is consistent. That is, we can regress $y_1$ on $y_2$ and $\rho$ if we want to test $H_0: \alpha=0$. This is unexpected. Not sure if my algebra is correct.