I am using nonlinear optimizers such as BOBYQA to train a model with 10-20 parameters. It so happens that some of the parameters have high correlation. Roughly speaking, imagine that you are fitting parameters $a,b$ in a function like this:


The response to the change in parameters $a,b$ is much larger than the response in to the change of $a-b$. Would performance of Powell's nonlinear optimization algorithms such as BOBYQA, NEWUOA, etc., improve if I would "rotate parameters" to isolate gross responses from fine ones?

That is, would nonlinear optimizers perform better if instead of optimizing $a,b$ in the above function I would optimize for parameters $\alpha,\beta$ a function like this:


The real model is more complicated, of course, and I'd like to find out whether changing parameters to minimize correlations has a chance to improve anything.


1 Answer 1


If possible, you want to choose a well-scaled objective for your problem, which would mean de-correlating your parameters as much as possible.

For a quadratic approximation to an objective function, correlated parameters correspond to objective function isosurfaces that are ellipsoidal in shape; the ideal shape for isosurfaces would be roughly spherical. In algebraic terms, ellipsoidal isosurfaces correspond to ill-conditioned Hessian matrices (the relationship being that the magnitude of the eigenvalues of the Hessian corresponds to the lengths of the semiaxes of the ellipsoids), and for second-order methods, an ill-conditioned Hessian definitely degrades convergence. Empirically, this behavior would be seen by taking small "zig-zag" steps as one approaches a local optimum; from a numerical analysis standpoint, it follows by noting that second-order methods take a form that looks roughly like

\begin{align} H(x^{n})(x^{n+1} - x^{n}) = -\nabla{f}(x^{n}), \end{align}

where $H$ is the Hessian, $f$ is the objective function, and $\nabla{f}$ is the gradient of the objective function, yielding an ill-conditioned linear system that will be solved inaccurately with numerical methods. (Here, I've assumed a Newton method, but quasi-Newton methods operate similarly.)


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