I am an undergraduate math student preparing my thesis. Currently I am reading L.D Brown's (1971) paper Admissible Estimators, Recurrent Diffusions, and Insoluble Boundary Value Problems. Here is a link of the paper <http://www.stat.yale.edu/~hz68/619/Brown1971.pdf>. One of the main ideas of the paper is to transfer the necessary and sufficient condition for the admissibility of an estimator to the recurrence of the "corresponding" diffusions. I have some difficulty of understanding what "corresponding" means in this case.


On page 862, (1.3.10), this says if we write $\sum j''\_{ii}+\sum\frac{{f'\_i}^\*}{f^\*}j\_i'=0$, then left-hand side of is actually the generator of the diffusion with local variance matrix $2I$ and local mean $\nabla f^\*/f^\*$


I would like to know that why this operator generates this particular diffusion, since Brown didn't say this explicitly in his paper. 


I also want to know the reasoning of the statement in the following paragraph, which says if we choose $\delta$ as the usual estimator $x$, then the diffusion corresponding is a version of Brownian motion.


I read <https://stats.stackexchange.com/questions/13494/intuition-behind-why-steins-paradox-only-applies-in-dimensions-ge-3> which arose my interest in reading this paper.


I will aprriciate it if someone can explain this to me.



diffusions corresponding to estimators
Here is an attempt to answer your questions, admitting that I have not read the paper carefully.


The generator corresponding to a diffusion satisfying the SDE $dX\_t = b(X\_t) dt + \sigma(X\_t) dB\_t$ is the following partial differential operator $A$, acting on a function $f$ as follows
$$
Af(x) = \sum\_i b\_i(x) \,\partial\_i f(x) + \frac12 \sum\_{ij} \Sigma\_{ij}(x) \, \partial\_{ij}f(x) 
$$
where $\Sigma\_{ij}(x) = [\sigma(x) \sigma(x)^T]\_{ij}$. What the paper calls local mean is $b(x)$ and what it calls local variance-covariance matrix is $\Sigma(x)$. The PDE in the paper corresponds to $b\_i = \partial\_i f^\*/ f^\*$ or equivalently $b = \nabla f^\* / f^\*$ and $\Sigma(x) = 2I$.


According to the paper, apparently, $f^\*$ corresponding to an estimator $\delta\_F$ satisfies $\nabla f^\*(x)/f^\*(x) = \delta(x) - x$. (It seems so, I am not sure here. For this, one needs a bit more careful reading of the paper.) Thus, the $f^\*$ corresponding to estimator given by $\delta(x) = x$ gives $b$ which is identically zero, that is, $b(x) = \nabla f^\*(x)/f^\*(x) = x - x = 0$. Since $b(x) = 0$ in this case, the corresponding diffusion satisfies $dX\_t = \sqrt{2} \, d B\_t$ (which is a version of Brownian motion.)


