Mean estimator of a Gaussian variable with positive mean for quadratic loss
$begingroup$
Suppose $phi, Phi$ are PDF and CDF for a $1$-dimensional normal Gaussian, and $Xsimmathcal{N}(theta,1)$, in which $theta>0$ is positive but othrewise unknown. We want to estimate $theta$ with respect to quadratic loss. If one incorporates the positivity of $theta$ by introducing the extended Bayesian prior $alpha_m(theta)=text{Uniform}(0,m)$ and then taking the limit when $mtoinfty$. Then the following estimator for $theta$ can be obtained:
$$Rightarrow hat theta(X)=lim_{mtoinfty}int_{-infty}^infty{vartheta P(vartheta|X)dvartheta}$$
$$=lim_{mtoinfty}frac{int_{-infty}^inftyvarthetaalpha_m(vartheta)P(Xmidvartheta)dvartheta}{int_{-infty}^infty alpha_m(vartheta)P(Xmidvartheta)dvartheta}$$
$$=frac{int_{0}^inftyvartheta P(Xmidvartheta)dvartheta}{int_{0}^infty P(Xmidvartheta)dvartheta}$$
This is equivalent to computing the mean of a random variable $vartheta$ with truncated Gaussian PDF:
$$Rightarrow hat theta(X)=X + frac{phi(X)}{Phi(X)}.$$
Now my question is, is it possible to compute $mathbb{E}_theta[hat theta(X)]$ and $Var_theta[hat theta(X)]$ in order to compute its loss? If it is not possible to derive this explicitly, is it possible to somehow prove it's a minimax and/or admisible estimator?
Is it possible to introduce some other prior that makes calculations easier and hence makes it possible to assert its minimaxity/admissibility?
statistics bayesian estimation parameter-estimation gaussian-integral
asked Jan 18 at 13:57
kvphxgakvphxga
1568
$endgroup$
add a comment |
$begingroup$
Suppose $phi, Phi$ are PDF and CDF for a $1$-dimensional normal Gaussian, and $Xsimmathcal{N}(theta,1)$, in which $theta>0$ is positive but othrewise unknown. We want to estimate $theta$ with respect to quadratic loss. If one incorporates the positivity of $theta$ by introducing the extended Bayesian prior $alpha_m(theta)=text{Uniform}(0,m)$ and then taking the limit when $mtoinfty$. Then the following estimator for $theta$ can be obtained:
$$Rightarrow hat theta(X)=lim_{mtoinfty}int_{-infty}^infty{vartheta P(vartheta|X)dvartheta}$$
$$=lim_{mtoinfty}frac{int_{-infty}^inftyvarthetaalpha_m(vartheta)P(Xmidvartheta)dvartheta}{int_{-infty}^infty alpha_m(vartheta)P(Xmidvartheta)dvartheta}$$
$$=frac{int_{0}^inftyvartheta P(Xmidvartheta)dvartheta}{int_{0}^infty P(Xmidvartheta)dvartheta}$$
This is equivalent to computing the mean of a random variable $vartheta$ with truncated Gaussian PDF:
$$Rightarrow hat theta(X)=X + frac{phi(X)}{Phi(X)}.$$
Now my question is, is it possible to compute $mathbb{E}_theta[hat theta(X)]$ and $Var_theta[hat theta(X)]$ in order to compute its loss? If it is not possible to derive this explicitly, is it possible to somehow prove it's a minimax and/or admisible estimator?
Is it possible to introduce some other prior that makes calculations easier and hence makes it possible to assert its minimaxity/admissibility?
statistics bayesian estimation parameter-estimation gaussian-integral
asked Jan 18 at 13:57
kvphxgakvphxga
1568
$endgroup$
add a comment |
$begingroup$
Suppose $phi, Phi$ are PDF and CDF for a $1$-dimensional normal Gaussian, and $Xsimmathcal{N}(theta,1)$, in which $theta>0$ is positive but othrewise unknown. We want to estimate $theta$ with respect to quadratic loss. If one incorporates the positivity of $theta$ by introducing the extended Bayesian prior $alpha_m(theta)=text{Uniform}(0,m)$ and then taking the limit when $mtoinfty$. Then the following estimator for $theta$ can be obtained:
$$Rightarrow hat theta(X)=lim_{mtoinfty}int_{-infty}^infty{vartheta P(vartheta|X)dvartheta}$$
$$=lim_{mtoinfty}frac{int_{-infty}^inftyvarthetaalpha_m(vartheta)P(Xmidvartheta)dvartheta}{int_{-infty}^infty alpha_m(vartheta)P(Xmidvartheta)dvartheta}$$
$$=frac{int_{0}^inftyvartheta P(Xmidvartheta)dvartheta}{int_{0}^infty P(Xmidvartheta)dvartheta}$$
This is equivalent to computing the mean of a random variable $vartheta$ with truncated Gaussian PDF:
$$Rightarrow hat theta(X)=X + frac{phi(X)}{Phi(X)}.$$
Now my question is, is it possible to compute $mathbb{E}_theta[hat theta(X)]$ and $Var_theta[hat theta(X)]$ in order to compute its loss? If it is not possible to derive this explicitly, is it possible to somehow prove it's a minimax and/or admisible estimator?
Is it possible to introduce some other prior that makes calculations easier and hence makes it possible to assert its minimaxity/admissibility?
statistics bayesian estimation parameter-estimation gaussian-integral
asked Jan 18 at 13:57
kvphxgakvphxga
1568
$endgroup$
Suppose $phi, Phi$ are PDF and CDF for a $1$-dimensional normal Gaussian, and $Xsimmathcal{N}(theta,1)$, in which $theta>0$ is positive but othrewise unknown. We want to estimate $theta$ with respect to quadratic loss. If one incorporates the positivity of $theta$ by introducing the extended Bayesian prior $alpha_m(theta)=text{Uniform}(0,m)$ and then taking the limit when $mtoinfty$. Then the following estimator for $theta$ can be obtained:
$$Rightarrow hat theta(X)=lim_{mtoinfty}int_{-infty}^infty{vartheta P(vartheta|X)dvartheta}$$
$$=lim_{mtoinfty}frac{int_{-infty}^inftyvarthetaalpha_m(vartheta)P(Xmidvartheta)dvartheta}{int_{-infty}^infty alpha_m(vartheta)P(Xmidvartheta)dvartheta}$$
$$=frac{int_{0}^inftyvartheta P(Xmidvartheta)dvartheta}{int_{0}^infty P(Xmidvartheta)dvartheta}$$
This is equivalent to computing the mean of a random variable $vartheta$ with truncated Gaussian PDF:
$$Rightarrow hat theta(X)=X + frac{phi(X)}{Phi(X)}.$$
Now my question is, is it possible to compute $mathbb{E}_theta[hat theta(X)]$ and $Var_theta[hat theta(X)]$ in order to compute its loss? If it is not possible to derive this explicitly, is it possible to somehow prove it's a minimax and/or admisible estimator?
Is it possible to introduce some other prior that makes calculations easier and hence makes it possible to assert its minimaxity/admissibility?
statistics bayesian estimation parameter-estimation gaussian-integral
statistics bayesian estimation parameter-estimation gaussian-integral
asked Jan 18 at 13:57
kvphxgakvphxga
1568
asked Jan 18 at 13:57
kvphxgakvphxga
1568
edited Jan 18 at 15:21
kvphxga
asked Jan 18 at 13:57
kvphxgakvphxga
1568
asked Jan 18 at 13:57
kvphxgakvphxga
1568
asked Jan 18 at 13:57
kvphxgakvphxga
1568
1568
add a comment |
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