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.)


3 Answers 3


$\arg\min$ is argument of the minimum so it is in general the set of values where the function attains the minimum.

The simplest example is

$\arg\min _{x} f(x)$ is the value of $x$ for which $f(x)$ attains its minimum.

for your example

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

hopefully that helps.

  • $\begingroup$ 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 at 8:56

$arg min$ (or $arg max$) return the input for minimum (or maximum) output.

For example:

The graph illustrat $f(x)=(sin(x-0.5)+cos(x)^2)*2$

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

enter image description here

  • 2
    $\begingroup$ The arrow is approximately pointing to the point: (4.9, -2) $\endgroup$
    – Brandon
    Sep 23, 2021 at 4:24

$\operatorname{argmin}(f(x))$ 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).

  • 1
    $\begingroup$ 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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