Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is regarding a clarification in page 384 of a paper published in Annals of Statistics by Amari.

In page no. 384, he defines $$R_i(t)=\frac{\partial}{\partial \theta_i} D_{\alpha}(q(x,t),p(x,\theta))$$ where for $|\alpha|\neq 1$,

\begin{eqnarray*} D_{\alpha}(q(x,t),p(x,\theta))=\frac{8}{1-\alpha^2}\left[1-\int q(x,t)^{\frac{1+\alpha}{2}}p(x,\theta)^{\frac{1-\alpha}{2}}~dx\right] \end{eqnarray*}

The one I am struggling to understand is that how did the author obtain the differential equation (A.19): $$\ddot{R_i}(t)=-\frac{1-\alpha^2}{4}i(t)R(t)$$ where, I think, $i(t)=\int \frac{1}{q(x,t)}(\dot{q}(x,t))^2~dx$ from bottom of page 382.

When I derived manually I obtained (by allowing interchange of differentiation and integral) $$\ddot{R_i}(t)=\int p(x,\theta)^{-\frac{1+\alpha}{2}}\left[q(x,t)^{\frac{\alpha-1}{2}}\ddot{q}(x,t)+\frac{\alpha-1}{2}q(x,t)^{\frac{\alpha-3}{2}}\dot{q}(x,t))^2\right]\frac{\partial}{\partial \theta_i}(p(x,\theta)~dx.$$ I am wondering how can the right hand side of the above be written as a product of two integrals as in (A.19). As this is appeared in Annals of Statistics, I am not in a position to doubt this as well.

It would be of great help if anyone who knows differential/Riemannian geometry can clarify why should (A.16) be true under the mentioned circumstances.

NB: I asked the above question in math.stackexchange where I couldn't get any response so far.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.