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Aftermath 12345
Aftermath 12345
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Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$$\sqrt{n} (\boldsymbol{Z}_n - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, as $n\to \infty$, for some $\boldsymbol{\mu}\in \mathbb{R}^d$ and $\Sigma$ a symmetric positive definite matrix (here, $\stackrel{\mathcal{D}}{\longrightarrow}$ denotes the convergence in distribution). The delta method says that for a $\mathcal{C}^1$ function $h: \mathbb{R}^d \rightarrow \mathbb{R}$, we have $$ \sqrt{n} (h(\boldsymbol{Z}) - 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. $$$$ \sqrt{n} (h(\boldsymbol{Z}_n) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$$h(\boldsymbol{Z}_n)$ (with appropriate conditions on $h$)? We know that $$ f_{h(\boldsymbol{Z})}(t) = f_{\boldsymbol{Z}}(h^{-1}(t) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$$$ f_{h(\boldsymbol{Z}_n)}(t) = f_{\boldsymbol{Z}_n}(h^{-1}(t)) \, \left|\frac{d}{d t} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$ 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: $$ \frac{f_{h(\boldsymbol{Z})}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$$$ \frac{f_{h(\boldsymbol{Z}_n)}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(t) = f_{\boldsymbol{Z}}(h^{-1}(t) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$ 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: $$ \frac{f_{h(\boldsymbol{Z})}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

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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Aftermath 12345
Aftermath 12345
  • 219
  • 1
  • 8

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(\boldsymbol{z}) = f_{\boldsymbol{Z}}(h^{-1}(\boldsymbol{z})) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(\boldsymbol{z})\right|, $$$$ f_{h(\boldsymbol{Z})}(t) = f_{\boldsymbol{Z}}(h^{-1}(t) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$ where the last term denotes the Jacobian of the transformation. If we let $\boldsymbol{W}\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$$W\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty, $$$$ \frac{f_{h(\boldsymbol{Z})}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(\boldsymbol{z}) = f_{\boldsymbol{Z}}(h^{-1}(\boldsymbol{z})) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(\boldsymbol{z})\right|, $$ where the last term denotes the Jacobian of the transformation. If we let $\boldsymbol{W}\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(t) = f_{\boldsymbol{Z}}(h^{-1}(t) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(t)\right|, \quad t\in \mathbb{R}, $$ 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: $$ \frac{f_{h(\boldsymbol{Z})}(t)}{f_{W}(t)} = 1 + \frac{\text{error}_1(t)}{\sqrt{n}} + \frac{\text{error}_2(t)}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

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Aftermath 12345
Aftermath 12345
  • 219
  • 1
  • 8

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(\boldsymbol{z}) = f_{\boldsymbol{Z}}(h^{-1}(\boldsymbol{z})) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(\boldsymbol{z})\right|, $$ where the last term denotes the Jacobian of the transformation. If we let $\boldsymbol{W}\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty. $$$$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course. ? I found these papers:

but they don't quite answer my question.

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(\boldsymbol{z}) = f_{\boldsymbol{Z}}(h^{-1}(\boldsymbol{z})) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(\boldsymbol{z})\right|, $$ where the last term denotes the Jacobian of the transformation. If we let $\boldsymbol{W}\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty. $$ with appropriate restrictions on $h$ of course. I found these papers:

but they don't quite answer my question.

Assume that $\sqrt{n} (\boldsymbol{Z} - \boldsymbol{\mu}) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0},\Sigma)$, 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 $$ \sqrt{n} (h(\boldsymbol{Z}) - h(\boldsymbol{\mu})) \stackrel{\mathcal{D}}{\longrightarrow} \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu})), \quad \text{as } n\to \infty. $$

Now, my question is: Are there more precise results in the literature involving asymptotic expansion(s) for the DENSITY function of $h(\boldsymbol{Z})$ ? We know that $$ f_{h(\boldsymbol{Z})}(\boldsymbol{z}) = f_{\boldsymbol{Z}}(h^{-1}(\boldsymbol{z})) \, \left|\frac{d}{d \boldsymbol{z}} h^{-1}(\boldsymbol{z})\right|, $$ where the last term denotes the Jacobian of the transformation. If we let $\boldsymbol{W}\sim \mathcal{N}(\boldsymbol{0}, \nabla h (\boldsymbol{\mu})^{\top} \Sigma \, \nabla h (\boldsymbol{\mu}))$, would it be possible to obtain a result of the form: $$ \frac{f_{h(\boldsymbol{Z})}(\boldsymbol{z})}{f_{\boldsymbol{W}}(\boldsymbol{z})} = 1 + \frac{\text{error 1}}{\sqrt{n}} + \frac{\text{error 2}}{n} + ~..., \quad \text{as } n\to \infty, $$ with appropriate restrictions on $h$ of course ? I found these papers:

but they don't quite answer my question.

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Aftermath 12345
Aftermath 12345
  • 219
  • 1
  • 8