I'm looking at this paper "An Alternative Method for Estimating and Simulating Maximum Entropy Densities" (https://editorialexpress.com/cgi-bin/conference/download.cgi?db_name=FEMES09&paper_id=780) that presents an alternative method to solve the maxent problem.
The problem is as follows: $p(x)=argmin\int p(x)\log\frac{p(x)}{q(x)}dx$ s.t.
$\int p(x)dx=1$
$\forall i\int c_{j}(x)p(x)=\mathbb{E}\left[c_{j}(X)\right]=d_{j}$
The author of the paper shows that this optimization proble is equivalent to:
$\min_{\lambda}\int\exp\left\{ \sum_{j}\lambda_{j}\left(c_{j}(x)-d_{j}\right)\right\} q(x)dx$
with solution:
$p(x)=\frac{1}{\mu}\exp(\lambda^{'}c(x))q(x)$ where $\mu=\int\exp(\lambda'c(x))q(x)dx$
In his proof (p. 4), he constructs the following lagrangian to transform the constrained optimization problem into an unconstrained problem:
$L(p)=\int p(x)\log\frac{p(x)}{q(x)}dx+(1+\lambda_{0})(1-\int p(x)dx)+\sum_{j}\lambda_{j}(d_{j}-\int c_{j}(x)p(x)dx)$
By stating that $L(p)$ is maximized when its Frechet derivative $\delta L$ equals zero. This way, the author shows that the objective function can be rewritten as follows:
$\int p(x)\log\frac{p(x)}{q(x)}dx=-\log\mu+\sum_{j}\lambda_{j}d_{j}$
From that result, he concludes "the problem is therefore equivalent to minimizing $\min_{\lambda}\int\exp\left\{ \sum_{j}\lambda_{j}\left(c_{j}(x)-d_{j}\right)\right\} q(x)dx$"
Since, he does not provide more details, it must a very easy step. but i can't seem to find why he can make such statement.
I have tried taking $\exp()$ (monotonic increasing function, should preserve the order) of $-\log\mu+\sum_{j}\lambda_{j}d_{j}$ but that leads me to a different result:
$argmin\exp\left(-\log\mu+\sum_{j}\lambda_{j}d_{j}\right)=argmin\frac{1}{\int\exp\left\{ \sum_{j}\lambda_{j}\left(c_{j}(x)-d_{j}\right)\right\} q(x)dx}$
Any idea what i am missing?
Thanks a lot for your help
Lamp'