Let's say I have one endogenous variable $X_1$ in the linear model

$$ Y=X_1\beta $$

and two instrumental variables $Z_1$ and $Z_2$ (strongly correlated with $X_1$ but not $Y$).

I compute the two-stage least squares in the following way:

$$ \widehat{\beta}_{2SLS} = [X'Z(Z'Z)^{-1}Z'X]^{-1}[X'Z(Z'Z)^{-1}Z'y] $$

I'm trying to understand the number of degrees of freedom in this situation in order to correct the calculation of the sample variance of my final regression model. I have two options:

  1. According to Multiple linear regression degrees of freedom, I would have $N-2$ degrees of freedom.

  2. However, because during the first stage of 2SLS I regress $X_1$ on the $Zs$, i.e., I run OLS on the linear model

$$ X_1=\delta_1Z_1 + \delta_2Z_2 $$

and in this case we have two predictor variables ($Z_1$ and $Z_2$), so perhaps I have $N-3$ degrees of freedom.

Any hints about which one works here?


Either is fine, asymptotically. Recall that the main goal in the degrees of freedom corrections of the error variance estimate in OLS is to render the estimate $s^2$ an unbiased estimator of $\sigma^2$, and also a required ingredient to $t$- and $F$-finite-sample distribution theory in normal regression models.

Now, finite-sample properties for IV estimators are, outside toy models, either unwieldy or plain unavailable, such that asymptotic approximations are needed.

In particular, it can be shown that


is consistent for $\sigma^2$. Now, rescaling this expression by $\frac{n}{n-K}$ for any finite $K$ such as 2 or 3 will not matter asymptotically, as $\frac{n}{n-K}\to1$.

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