# Explanation on arg min

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.)

$$\arg\min$$ is argument of the minimum.

The simplest example is

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

$$x_n$$ 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.

$arg min$ (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 . • The arrow is approximately pointing to the point: (4.9, -2) 13 hours ago

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

• Practical! yes it means argument that returns least not the least passed argument! Apr 27 '16 at 15:07