2
$\begingroup$

Suppose that I am working in a model containing a nuisance parameter $h$ and a finite dimensional parameter of interest $\theta$, whose true values are $h_0$ and $\theta_0$, respectively. Newey (1994) proves the asymptotic properties of a two-step GMM with a non parametric first step, and studies the properties of $\hat{\theta}= \arg \max_{\theta} m_{n}(\theta, \hat{h})^{T} \hat{W} m_{n}(\theta,\hat{h}) $ where $m_{n}(\theta,h)=\frac{1}{n} \sum\limits_{i=1}^n m(W_i,\theta,h)$ and $\mathbb{E}[m(W,\theta_0,h_0)]=0$.

In this setup, can one choose $\hat{W}$ (positive semi-definite) to depend on $\hat{h}$ and still expect the asymptotic results to hold, if all the regularity conditions are met and $\hat{W} \to W$ in probability ? There are no indications in the papers that $\hat{W}$ cannot depend on the nuisance parameter.

$\endgroup$

1 Answer 1

2
$\begingroup$

As long as $\hat{W}\rightarrow W$ where $W$ is the inverse of the variance-covariance matrix of moments, then all of the asymptotic properties hold. You seem to be asking if this applies in your case.

I think you would need to justify that $\hat{h}\rightarrow h$ and that $\hat{W}(h)$ is "continuous" in $h$. Given this, $\hat{W}(\hat{h})\rightarrow W$ and all is well.

$\endgroup$
3
  • 1
    $\begingroup$ Does it have to be the inverse variance-covariance matrix of moments? Or W hat could be a positive definite matrix, that converges to a positive definite matrix W. Not looking for efficiejxy. $\endgroup$
    – mich95
    Sep 12, 2023 at 11:02
  • 1
    $\begingroup$ If you aren't looking for efficiency, then you are correct, you just need convergence to a positive definite matrix. $\endgroup$ Sep 13, 2023 at 7:30
  • $\begingroup$ I am afraid that nuisance parameters are not satisfying the requirement $\hat{h}\rightarrow h_0$ when sample size goes to infinity. $\endgroup$
    – Bertrand
    Sep 14, 2023 at 11:24

Your Answer

By clicking “Post 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.