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The generalized f-mean is a generalization of multiple estimators, and even generalizes the generalized mean.

For some invertible function $f$, and $k$-dimensional vector, it is given as:

$$M_f(\vec{x}) \triangleq f^{-1} \left( \frac{1}{n} \sum_{k=1}^{n} f(x_k) \right)$$

It would serve a similar purpose to define a (possibly more robust) generalized $f$-median.

$$\mathcal{M}_f(\vec{x}) \triangleq f^{-1} \left( \text{median} \left( f(x_k) \right) \right)$$

What are the tradeoffs of using the generalized $f$-median vs the generalized $f$-mean?

DifferentialPleiometry
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  • Isn't your "generalized $f$-median" identical to the median? If not, then please explain what your notation "$\mathcal{M}(f(x_k))$" is intended to signify. – whuber Jul 26 '21 at 17:41
  • @whuber If $f$ is the identity function then they are the same, but I see what you mean about the notation. I will improve it. – DifferentialPleiometry Jul 26 '21 at 17:42
  • When $f$ is strictly invertible they appear to be identical, apart from the usual issues with defining medians when $n$ is an even number. – whuber Jul 26 '21 at 17:43
  • I tried a quick Monte Carlo of $\sqrt(\text{median} (x_k^2))$ where $k=100$ and $X$ was drawn from a standard normal, and they turned out to be different values. I think you're on to something about choice of $f$ that will preserve the order of the values leads to an invariant. – DifferentialPleiometry Jul 26 '21 at 17:52
  • Of course they will be different values, because the square is not invertible on the support of a Normal variable. But when $f$ is not invertible, it's not terribly useful for defining any measure of central tendency, is it? – whuber Jul 26 '21 at 18:09
  • @whuber Not for all choices of distribution, no. I think the [generalized mean](https://en.wikipedia.org/wiki/Generalized_mean) motivates a general case of specific aggregation functions that are sometimes useful. – DifferentialPleiometry Jul 26 '21 at 18:22

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