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This fact appears in my statistics textbook (Pg 543, statistical decision theory and bayesian analysis). it says : for normal distribution the generalized bayes estimator becomes \begin{align*} \delta_F(x)&=\nabla \log[(2\pi)^{-p/2}e^{-|x|^2/2}\int e^{x'\theta}e^{-|\theta|^2/2}d(F\theta)]-\nabla \log[e^{-|x|^2/2}]\\ &=\nabla \log \int e^{x'\theta}e^{-|\theta|^2/2}F(d\theta) \end{align*} This integral is a Laplace transform of $exp\{-\frac{1}{2}|\theta|^2\}dF(\theta)$, which is known to be infinitely differentiable.

It seems like this is a well-known fact. However, I don't know how it was deduced since I don't know about laplace transform. I want to know why this is indeed the Laplace transform of $exp\{-\frac{1}{2}|\theta|^2\}dF(\theta)$ since the $log$ doesn't make much sense at first look. I also want to know why this is infinitely differentiable.

Thanks in advance for any help.

user st
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  • Seems to me the assertion is about the integral, not the log of the integral, so I don't think the log and the gradient should concern you regarding your confusion about the def. of the Laplace transform. – treble Sep 24 '13 at 16:00
  • @treble，thanks. In fact, I just found this http://math.stackexchange.com/questions/84382/show-that-laplace-transform-is-differentiable. Thought this can be directly applied to show the integral is infinitely differentiable by induction. – user st Sep 24 '13 at 16:05

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