0

I know only a little about Fisher information and optimal experimental design, but I'm trying to better understand the subject. If I have an experiment composed of a single detector and my detector location is along a single axis and is parameterized by a scalar $\theta$ (my explanatory variable), then my D-Criterion would be $D=\text{det}(I^T I)$, where $I =I(\theta)$ is the Fisher information matrix. Since the experiment depends on a single parameter it would just be $D=I^2(\theta)$. If the goal for the optimal experiment is to maximize $D$, then what is this doing to the probability distribution for an observable $X$? It seems like $f(X;\theta)$ is tuned to have some characteristic property to it, but what would that be?

If there's anything off about my example/reasoning please correct me.

David G.
  • 117
  • 5
  • See also https://stats.stackexchange.com/questions/114039/why-people-often-optimize-the-determinant-of-x-sigma-x-1/114048#114048 – kjetil b halvorsen Feb 27 '24 at 21:29
  • Please see https://stats.stackexchange.com/a/564349/919 for a discussion of an important and intuitive application of d-optimality. – whuber Feb 27 '24 at 21:35
  • @whuber Thank you for the suggestions. I suppose my question ends up being, if D-optimality gives me a choice of parameter for my experimental design, what does that chosen parameter do to the shape of the probability distribution function? It seems like it makes the PDF flat for distributions I've played around with. – David G. Feb 28 '24 at 16:33
  • I don't understand what you mean by "choice of parameter." If you're doing an experiment, presumably there is at least one parameter (and it would seem to be $\theta$ itself) that you are attempting to discover: you don't get to choose it. Nature does. Where you do have a choice lies in how you conduct the experiment: how many observations you make and what experimental settings (such as values of explanatory variables) you choose. – whuber Feb 28 '24 at 17:47
  • $\theta$ in this example is meant to be the location of a detector (so an explanatory variable). I was attempting to create an example where the optimal design is based on the choice of just one parameter. I suppose I might be incorrect, but does there not need to be at least one explanatory variable for OED to be considered? If not, what exactly is being designed? Detector placement/location seems like the most intuitive example to use. – David G. Feb 28 '24 at 21:09
  • You appear to use the term "parameter" as a synonym for "explanatory variable," which in this forum will confuse most readers. Thus, offering a clearer, more concrete description of your problem will be helpful. – whuber Mar 01 '24 at 14:11

0 Answers0