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Throughout this question assume $f_i \ge 0 \forall i $. I know that for any (single) function the following is true

$$f(x^\star) \ge f(x) \text{ }\forall x\in X$$

iff

$$\log(f(x^\star)) \ge \log(f(x)) \text{ } x\in X$$

Now suppose fundamentally we are interested in maximizing a function of the form

$$F = \sum_i f_i$$

Is this clearly equivalent to maximizing

$$\displaystyle \hat{F} = \prod_i f_i$$ ?

Clearly, by the same line of reasoning we have that maximizing the products is equivalent to maximizing the following

$$\log(\hat{F}) = \sum_i \log(f_i)$$

So ultimately, the question boils down to, is maximizing $F$ equivalent to maximizing $\log (\hat{F} )$?

Squirtle
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1 Answers1

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No to both questions.

Maximizing $f$ is equivalent to maximizing $g\circ f$, where $g$ is an increasing function. The key idea is that $g$ can be "canceled" from inequalities: $$f(x_1) < f(x_2) \Leftrightarrow g(f(x_1)) < g(f(x_2))$$ and so a maximum of one must be a maximum of the other.

Unless $g$ is linear, the above does not generalize to sums: it's not generally true for increasing functions $g$ that $$x_1 + x_2 < x_3 \Leftrightarrow g(x_1)+g(x_2) < g(x_3)$$ and so you cannot apply $g$ (in your case $\log$) separately to terms of a sum. As a concrete counterexample for $\log$:

$$\log 2 + \log 4 \not < \log 7.$$

user7530
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