3
$\begingroup$

I was reading Andrew Ng's ML lecture notes on K-mean clustering, in which the distortion function is defined as follow $$J(c,\mu) = \sum^m_{i=1} || x^{(i)} - \mu_{c^{(i)}}||^2$$

I am puzzled about the $L_2$ norm, since $|| x^{(i)} - \mu_{c^{(i)}}||^2 $ would imply $\sum^m_{i=1} (x^{(i)} - \mu_{c^{(i)}})^2$ and this means that there would be two summations $\sum_{i=1}^m$ in the entire expression.

I am sensing that I have misunderstood something crucial here. Please point out the error. Thanks.

UPDATE: the problem has a given a training set $\{x^{(1)}, ..., x^{(m)}\}$, where $x^{(i)} \in \mathbb{R}^n$ and the cluster centroids are $\mu_1, \mu_2,...\mu_k \in \mathbb{R}^n$

$\endgroup$
2
  • $\begingroup$ What are $x^{(i)}$ and $\mu_{c^{(i)}}$ ? Vectors ? If so, what is their dimension ? What is $m$ ? $\endgroup$
    – Pop
    Commented Feb 26, 2014 at 12:20
  • $\begingroup$ maybe the inner summation is summing over the $n$. But I am not sure about this $\endgroup$ Commented Feb 26, 2014 at 12:27

1 Answer 1

3
$\begingroup$

As the norm is applied to vectors in dimension $n$, $$|| x^{(i)} - \mu_{c^{(i)}}||^2 = \sum^n_{j=1} (x^{(i)}_j - \mu_{c^{(i)},j})^2$$

with $x^{(i)} = (x^{(i)}_1,..., x^{(i)}_n)$ and $\mu_{c^{(i)}}=(\mu_{c^{(i)},1},..., \mu_{c^{(i)},n})$.

The summation is first on the $m$ points of the sample ($x^{(i)}$) and then on their $n$ components, sothat: $$J(c,\mu) =\sum^m_{i=1}\sum^n_{j=1} (x^{(i)}_j - \mu_{c^{(i)},j})^2$$

$\endgroup$
1
  • $\begingroup$ There should also be a square root over the sum in the equation (for L2 norm) $\endgroup$
    – Oleg
    Commented Sep 8, 2022 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.