I encountered this problem while reading a statistic text. Since I am not quite familar with the background knowledge. Wonder can someone help me to explain the details of the following proof?
Suppose X follows a spherically symmetric distribution which has a density $f(||x-\theta||^2)$, and let $F(t)=2^{-1}\int_{t}^{\infty}f(u)du$ and $Q(t)=F(t)/f(t)$.
I want to evaluate the risk of a general function $X+g(X)$:$R(\theta,X+g(X))=E_{\theta}[||X+g(X)-\theta||^2]=E_{\theta}[||X-\theta||^2]+E_{\theta}[||g(X)||^2]+2E[(X-\theta)'g(X)]$, where all values here are vector values of dimension p.
Now $E[(x-\theta)'g(X)]=\int_{R^p}(x-\theta)'g(X)f(||x-\theta||^2)dx=\int_{R^p}g(X)'\nabla F(||x-\theta||^2)dx=\int_{R^p}\nabla g(X)' F(||x-\theta||^2)dx=E[Q(||X-\theta||^2)\nabla 'g(X)]$
It says the above whole thing follows from Green's theorem. However, I can't see how to get third and fourth equality and where Green's theorem comes from . I also can't see why the last step comes to $E[Q(||X-\theta||^2)\nabla 'g(X)]$
Some things that are useful:
First, there is the identity $\int_\Omega \nabla f dx = \int_{\partial \Omega} f n dA$. This, along with the identity $\nabla(fg) = f\nabla g + g \nabla f$, lets you turn integrals over a region into integrals over the boundary. It's basically a higher-dimensional integration by parts:
$$\int_\Omega g\nabla f = \int_{\partial \Omega} fndA - \int_\Omega f \nabla g.$$ What you can often do is show that the boundary integral goes to zero as the region gets larger and larger, and show these two integrals are equal. It looks like something like this might have been used for the third equality. (But it's missing a negative...)
I'm not sure what you're using $'$ for, as in the expression $\nabla'g$.