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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 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 which arose my interest in reading this paper.

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

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Regarding you first question, you might find this link useful:… – passerby51 Aug 30 '13 at 1:14
up vote 2 down vote accepted

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.)

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