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Would someone be so kind to explain this to me:

$$\pi_nk=\left\{\begin{array}{cl}1&\textrm{if }k=\arg\min_j\left\Vert\mathbf x_n-\mu_j\right\Vert^2\\0&\textrm{otherwise}\end{array}\right..$$

Especially the $\arg\min$ part.

(It's from the $k$-means algorithm.)

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3 Answers 3

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Define $\arg\min_x f(x)$ as the set of values of $x$ for which the minimum of $f(x)$ is attained, so it is the set of values where the function attains the minimum. Thus, $\arg\min_x f(x)$ is a subset of the domain of $f(x)$.

For your example: $x_n$ is known and depends on $\pi_{nk}$; $k$ equals to the value of $j$ such that $\begin{Vmatrix} x_n-\mu_j \end{Vmatrix}^2$ attains a minimum among all values of $\mu_j$ and given $x_n$.

Hopefully that helps.

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    $\begingroup$ If you define arg min $f$ as the set of values where the function attains its minimum, then one should write $x\in $arg min $f$ whereas $x=$ arg min $f$ is not what most authors intend to say. $\endgroup$
    – Jochen
    Sep 18, 2023 at 8:56
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$\arg \min$ (or $\arg \max$) return the input(s) for which the output is minimum (or maximum).

For example:

The graph illustrates the function $f(x)=2 \sin(x-0.5)+\cos(x)^2$.

The global minimum of $f(x)$ is $\min(f(x)) \approx -2$, while $\arg \min f(x) \approx 4.9$.

enter image description here

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    $\begingroup$ The arrow is approximately pointing to the point: (4.9, -2) $\endgroup$
    – Brandon
    Sep 23, 2021 at 4:24
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$\operatorname{argmin}(f(x))$ simply returns the value of $x$ which minimizes $f(x)$ over the set of candidates for $x$ as opposed to the minimum value itself. This arises, of course, in all kinds of statistical estimates of parameters when building models (like the LS situation alluded to in your example).

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    $\begingroup$ Practical! yes it means argument that returns least not the least passed argument! $\endgroup$
    – Learner
    Apr 27, 2016 at 15:07

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