Suppose I want to find a \(\beta\) which satisfies the following sort of equation:
\[k = \int e^{\beta h(\boldsymbol{\alpha})} d \boldsymbol{\alpha}\]where \(k >0\) is a constant and \(h\) is a known real function of the parameters \(\boldsymbol{\alpha} \in \mathbb{R}^n\). How might I do so? Here we present one possible approximate route, and briefly analyse the resulting equation. If the exponent has a suitably steep minima which we can expand around, then we might try a saddle point approximation
\[\begin{equation} \begin{aligned} k &= e^{\beta h({\boldsymbol{\alpha}*)}} \int e^{(\alpha - \alpha*)_i(\alpha - \alpha*)_j \frac{\partial^2 \beta h}{\partial \alpha_i \partial \alpha_j} } d \boldsymbol{\alpha} \\ &\approx e^{\beta h({\boldsymbol{\alpha}*)}} \Big(\frac{(2\pi)^n}{det \frac{\partial^2 \beta h}{\partial\alpha_i \partial\alpha_j} } \Big)^{1/2} \\ \end{aligned} \end{equation}\]Upon rearranging
\[\begin{equation} \label{eq:monge-ampere} \tag{1} \implies det \frac{\partial^2 \beta h}{\partial\alpha_i \partial\alpha_j} \vert_{\boldsymbol{\alpha = \alpha*}} = \frac{1}{k^2} e^{2 \beta h({\boldsymbol{\alpha}*)}} (2 \pi)^n \end{equation}\]This is similar to a nonlinear PDE called the Monge-Ampere Equation, which has uses in diffeential geometry and optimal transport (apparently). The similarity is superficial - remember \(h\) is known, and the expressions hold at a minima.
\(\beta\) is the solution to the non-linear algebraic equation which now involves no integrals. It still involves a determinant of the hessian, so it would be interesting to see if we could simplify this expression. This could be simplified exactly in a number of special cases. Some of the resulting equations for \(\beta\) are nonsensical, and thus betray the crassness of the saddle point approximation. We list some cases below.
\(h = h(\boldsymbol{c \cdot \alpha})\) (const \(\boldsymbol{c}\))
In this case the LHS becomes
\begin{equation}
\begin{aligned}
det \frac{\partial^2 h}{\partial\alpha_i \partial\alpha_j} &= det(c_i c_j h’’) = 0
\end{aligned}
\end{equation}
where the last equality comes from the fact that the tensor is of rank 1. Clearly Eq \eqref{eq:monge-ampere} makes little sense and therefore another approach is needed.
\(h =h(\vert \boldsymbol{\alpha}\vert)\) (rotational symmetry)
Denote \(\vert \boldsymbol{\alpha}\vert = \alpha\). In this case the LHS becomes
\[\begin{equation} \begin{aligned} det \frac{\partial^2 h}{\partial\alpha_i \partial\alpha_j} &= det\Big[ \delta_{ij} \frac{h'}{\alpha} + \alpha_i \alpha_j \big( \frac{h''}{\alpha^2} - \frac{h'}{\alpha^3} \big) \Big] \\ &= \Big( \frac{h'}{\alpha} \Big)^{n-1} h'' = \frac{1}{n\alpha} \big((h')^n \big)' \end{aligned} \end{equation}\]The last line comes from considering the product of eigenvalues of the derived second rank tensor.
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