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I got this example and I was wondering about a certain statement: $$ \begin{aligned} (I) \ y_1 &= \alpha_{12}y_2 + \alpha_{13}y_3 + \beta_{11}z_1 + u_1 \\ (II) \ y_2 &= \alpha_{21}y_1 + \beta_{21}z_1 + \beta_{22}z_2 + \beta_{23}z_3 u_2 \\ (III) \ y_3 &= \alpha_{32}y_2 + \beta_{31}z_1 + \beta_{32}z_2 + \beta_{33}z_3 + \beta_{34}z_4 + u_3 \end{aligned} $$

It is written that for equation (I), we could use all the excluded exogenous variables $z_2, z_3, z_4$ as instruments for the two endogenous regressors $y_2,y_3$.

But as far as I remembered, I can't use $z_2$ nor $z_3$ for $y_2$ because these variables already appear in (II). In the same sense, I cannot use $z_2, z_3, z_4$ for $y_3$.

In my understanding, I could use $z_2, z_3, z_4$ for $y_1$ but not for $y_2, y_3$. Is that right?

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Instruments need to fulfill two conditions. First, they need to be informative which I take as given for the sake of this answer. However serious issues arise if they're not very but that's another discussion.

The second condition is that they are valid, which means they must be uncorrelated with your dependent variable (or the error term on the LHS). So say you want to estimate the first equation, which means your are performing LIML estimation, you can use $z_2$, $z_3$, and $z_4$ as instruments. However, you may run the risk of weak identification if two instruments appear in the same way twice for different endogenous regressors that you are instrumenting for.

In order to eyeball valid instruments, you need to make sure that instruments do not affect your dependent variable other than through instrumented endogenous regressors.

Moreover your question is not quite clear when you say using an instrument "for" an endogenous regressor like $y_2$. I am assuming you mean to say "to instrument for".

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    $\begingroup$ That is what I meant. By now, I would say that the system defined by equation (I,II,III) is not identifiable as a whole but equation (I) and (II) may be identified if the necessary requirements for the instrument $z_2,z_3,z_4$ are met for (I) and for $z_4$ for (II). $\endgroup$ – Druss2k Jun 14 '16 at 19:32

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