I have a question regarding the OLS estimation, in the case of an estimation with instrumental variables:

We assume the linear model $π’š= 𝑿\beta+𝒖$ with $Z$ = instrumental variables.

Multiplying the model $π’š= 𝑿\beta+𝒖$ with $𝒁^{\prime}$ results in $𝒁^{\prime} y = 𝒁^{\prime} X\beta +𝒁^{\prime} u$

Estimating $\beta$ using OLS would lead to:

$\min{u^{\prime}u = (𝒁^{\prime}y - 𝒁^{\prime}X\beta)^{\prime} (𝒁'y - 𝒁^{\prime}X\beta)}$

FOC: $-2 X^{\prime}ZZ^{\prime}Y + 2X^{\prime}ZZ^{\prime}ß = 0$

that would lead to an estimate of $ß$: $ß = (X^{\prime}ZZ^{\prime}X)^{-1} * XZZ^{\prime}Y$

But I have never seen this form of the instrumental OLS estimator whether online nor in the literature. Can anyone help? What is my mistake?


1 Answer 1


You are minimizing $u'ZZ'u$ not $u'u$ like in original OLS. Therefore, yours is not IV OLS but something else, whatever you want to name it.

It's probably closest to weighted least squares in terms of the results, judging by the form of the equation.

  • $\begingroup$ Okay, so the IV OLS estimator would be minimizing u'u = (𝒁'y - 𝒁'𝜷)' (𝒁'y - 𝒁'𝜷) ? $\endgroup$
    – Joe94
    Jan 4, 2022 at 21:49
  • $\begingroup$ IV OLS minimizes $u'u$, of course, but it does so - conceptually - in two steps, that's why sometimes it's called two step least squares. First, it predicts $X$ as a linear combination of Z, then it run regression as usual but on predicted $\hat X$ instead of original X. you can work it out the equations to the original X in one shot, of course. $\endgroup$
    – Aksakal
    Jan 4, 2022 at 21:54
  • $\begingroup$ Yes, I understood how SLS works. But I wondered, if an IV estimator can be obtained by only using OLS. $\endgroup$
    – Joe94
    Jan 4, 2022 at 21:57

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.