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my main problem is that, i don't understand what argmax means in this equation (page 134., figure 4., the output part)

I want to write a code, but i don't understand this equation. Is there any simpler form of this equation ? (maybe an example)

This is how i understand it: 1. i count the sum first 2. i find the argmax

Sorry for my dummy question, but i am not realy good in this area. Thanks for your answers !

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    $\begingroup$ Did you do any research? $\endgroup$
    – user296602
    Aug 25, 2016 at 20:47
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    $\begingroup$ You are reading an exceptionally complicated paper for someone who hasn't gotten acquainted with the term "argmax"... $\endgroup$
    – Ian
    Aug 25, 2016 at 20:48
  • $\begingroup$ Note: I fully agree with the two comments above. Now, if you want an algorithm-version of this equation: "loop over all elements $y\in Y$, compute for each of them the value $\sum_{i=1}^T (\log\frac{1}{\beta_t}) h_t(x,y)$; return the $y$ for which this quantity is maximum." $\endgroup$
    – Clement C.
    Aug 25, 2016 at 21:00

2 Answers 2

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Lets say we have a funktion $f(x) = -(x-2)^2 + 3$, which has a global maximum at $x_0=2$ with $f(x_0)=3$.

This means that $\max(f) = 3$. $\arg\!\max$ answers not how high the maximum is, but where it occurs: $\arg\!\max(f) = 2$.

Note that $f(\arg\!\max(f)) = \max(f)$.

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    $\begingroup$ Strictly speaking, $f(\arg \max(f))=\{ \max(f) \}$, because the argmax is a set. $\endgroup$
    – Ian
    Aug 29, 2016 at 20:26
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In general $$x^*=\arg \max_x f(x) $$ means return $x$ that maximizes $f(x)$.

How to find $x^*$? 1) Find $f(x)$ for all possible values of $x$ as $\{x_n,\,f(x_n)\}$. 2) Find the maximum value of $f(x)$ in the set $\{f(x_n)\}$, let's denote it by $f(x_\max)$. 3) return $x_\max$.

You don't have to store all values. You can start by saying that $x_\max=x_1,\,f_\max=f(x_1)$. Then if $f(x_2)>f_\max$ update the values as $x_\max=x_2,\,f_\max=f(x_2)$. If not, keep the previous values. When all the possible values of $x$ are considered, your $x_\max$ at the end will be the value you are looking for.

Apply this to your situation.

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