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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).

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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.

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  • $\begingroup$ Thanks for the response but I am a bit confused about the ordering. You say "$\rho$ is exogenous so we cannot include it as a regressor in the first stage." But then you process to investigate whether it is exogenous for the purposes of IV. But you just said it is endogenous. So in the derivation of rho in terms of e1, are you just proving the statement that it is endogenous? $\endgroup$ Mar 5 at 2:28
  • $\begingroup$ By "first equation" I meant $y_1 = f(y_2,\mathbf{x}_1) + e_1$, not the first-stage regression equation. Sorry for the confusion. I added "(i.e., the $y_1$ equation)". $\endgroup$
    – chan1142
    Mar 5 at 3:22
  • $\begingroup$ Thanks for the clarification. But what I am asking is, doesn't the reduced form equation for $\rho$ show both why $\rho$ cannot be used as a regressor in $y_1 = f() + e_1$ and why it cannot be used as an instrument? $\endgroup$ Mar 5 at 3:35
  • $\begingroup$ Hmm, that looks like a different issue. If we keep the linear model, the question is whether $E(y_1|y_2, \rho) = \alpha y_2 + \delta \rho$ for some $\delta$. Can this be true? Let me work a bit. I'll write a comment if I find an answer. $\endgroup$
    – chan1142
    Mar 5 at 8:08

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