# diffusions corresponding to estimators

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 will aprriciate it if someone can explain this to me.

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Regarding you first question, you might find this link useful: en.wikipedia.org/wiki/… – passerby51 Aug 30 '13 at 1:14

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.)