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Suppose I have an estimator $B\in\mathbb{R}^m$ converging to $\beta$, such that $$ \sqrt{n}(B-\beta)\rightarrow\mathcal{N}(0,\Sigma). $$

I am interested in a quantity $\mathbf{h}(B):\ \mathbb{R}^m\rightarrow\mathbb{R}^p$, and would like to use delta method to approximate the asymptotic distribution.

Let $\nabla\mathbf{h}(\beta)$ be a matrix $\left[\frac{\partial h_k(\beta)}{\partial \beta_j}\right]_{jk}$. When $p=1$, multivariate Delta method would provide $$ \sqrt{n}(\mathbf{h}(B)-\mathbf{h}(\beta))\rightarrow\mathcal{N}(0,\nabla\mathbf{h}(\beta)^T\cdot\Sigma\cdot\nabla\mathbf{h}(\beta)). $$ Can we say that it is the same for the case when $p>1$? When I actually tried this numerically, I find that the covariance $\nabla\mathbf{h}(\beta)^T\cdot\Sigma\cdot\nabla\mathbf{h}(\beta)$ is not invertible.

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  • $\begingroup$ not invertible is OK, given it is positive semidefined. $\endgroup$ – user158565 Jul 17 at 18:00
  • $\begingroup$ I think I found the problem. My $\mathbf{h}$ is a function whose image is a subset of $\mathbb{R}^p$, and probably that is why the covariance is singular. $\endgroup$ – user15988 Jul 17 at 18:44
  • $\begingroup$ If you can give out $h$ function, it will be easy for discussion. Another posibility is $\Sigma$ itself is singular. $\endgroup$ – user158565 Jul 17 at 18:52

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