$\begingroup$

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.)

$\endgroup$

3 Answers

$\begingroup$

$\arg\min$ 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.

$\endgroup$ $\begingroup$

$arg min$ (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 .

$\endgroup$ 1 $\begingroup$

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

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy