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Assume that $\sqrt{n} (\boldsymbol{Z}_n - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, as $n\to \infty$, for some $\boldsymbol{\mu}\in \mathbb{R}^d$ and $\Sigma$ a symmetric positive definite matrix (here, $\stackrel{\mathcal{D}}{\longrightarrow}$ denotes the convergence in distribution). The delta method says that for a $\mathcal{C}^1$ function $h: \mathbb{R}^d \rightarrow \mathbb{R}$, we have $$ \sqrt{n} (h(\boldsymbol{Z}_n) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z}_n)$ (with appropriate conditions on $h$)? We know that $$ f_{h(\boldsymbol{Z}_n)}(t) = f_{\boldsymbol{Z}_n}(h^{-1}(t)) \, \left|\frac{d}{d t} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$ where the last term denotes the Jacobian of the transformation. If we let $W\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z}_n)}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ ? I found these papers:

but they don't quite answer my question.

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    $\begingroup$ Presumably $Z$ should be $Z_n$, but they don't need to have a density at all under your assumption... $\endgroup$ Sep 6, 2021 at 9:17
  • $\begingroup$ I understand, but there could be a class of $h$ for which we could say something about the ratio $f_{h(\boldsymbol{Z}_n)}(t) / f_{W}(t)$. In Boos (1985), the first paper linked above, his theorems 1 and 2 relate to translation and scale statistics for example. $\endgroup$ Sep 6, 2021 at 9:25
  • $\begingroup$ also at math.stackexchange.com/questions/4243094/… $\endgroup$
    – Henry
    Sep 6, 2021 at 22:28

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