Suppose $\phi, \Phi$ are PDF and CDF for a $1$-dimensional normal Gaussian, and $X\sim\mathcal{N}(\theta,1)$, 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 $m\to\infty$. Then the following estimator for $\theta$ can be obtained:
$$\Rightarrow \hat \theta(X)=\lim_{m\to\infty}\int_{-\infty}^\infty{\vartheta P(\vartheta|X)d\vartheta}$$ $$=\lim_{m\to\infty}\frac{\int_{-\infty}^\infty\vartheta\alpha_m(\vartheta)P(X\mid\vartheta)d\vartheta}{\int_{-\infty}^\infty \alpha_m(\vartheta)P(X\mid\vartheta)d\vartheta}$$ $$=\frac{\int_{0}^\infty\vartheta P(X\mid\vartheta)d\vartheta}{\int_{0}^\infty P(X\mid\vartheta)d\vartheta}$$ 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?