$\arg\min_\theta [f(\theta x)+f(\theta y)]=\arg\min_\theta [\log f(\theta x)+\log f(\theta y)]$ for all $x,y$ and monotonic $f$?

Question

I would like to know whether the following statement is true for all $x,y \in \mathcal{R}$:

$\arg\min_\theta [f(\theta x)+f(\theta y)]=\arg\min_\theta [\log f(\theta x)+\log f(\theta y)]$

where $f$ is any monotonically increasing function.

The statement is a priori surprising. Did you look for counterexamples? If yes, which functions did you try? — Did, Dec 20 '13 at 17:03

score 1 · Accepted Answer · answered Dec 20 '13 at 19:22

It is not equivalent.

$$\arg\min_\theta [\log f(\theta x)+\log f(\theta y)] =\arg\min_\theta \log [f(\theta x) f(\theta y)]$$

Suppose $f(x)=x$, then $\log [f(\theta x) f(\theta y)]=log(xy\theta^2)$, where the argmin is $\theta=0 $ for all $ x,y$. [One might argue a minimum does not exist; I take the limiting case $\log x\to -\infty$ as $x\to 0$.]

On the other hand, $f(\theta x)+f(\theta y)=\theta x+\theta y$, which has argmin $\theta=\infty$ if $x+y<0$ and $\theta=-\infty$ if $x+y>0$.

$\arg\min_\theta [f(\theta x)+f(\theta y)]=\arg\min_\theta [\log f(\theta x)+\log f(\theta y)]$ for all $x,y$ and monotonic $f$?

1 Answers1