Silhouette value after normalizing a variable and re-clustering I have a non-normalized variable and other normalized variables and I make a clustering with k medoids (or k means).
If I let the first variable non-normalized, I get better results in terms of average silhouette coefficient.
If I normalize it, I get worse results. I know that if the variable doesn't get normalized then it's like to give a "weight" to the variable.
I'd ask if an high average silhouette coefficient means a better clustering whether I normalize or not.
Thanks.
 A: Majority of internal clustering criterions (including here average Silhouette value, I suppose) are suited to compare different cluster solutions (different number of clusters and/or clustering method) of the same data.
Some clustering criterions will tolerate arbitrary (dis)similarity measure which a clustering method operates with or which it implies behind the scene (if any); but others require just specific measures - typically euclidean distance. And a few criterions may impose their own (dis)similarity measure at their computation.
Very few, if any, clustering criterions have been so wisely standardized that they allow direct comparison among cluster solutions obtained on different datasets (be that different objects to cluster or different features). So, don't decide which cluster partition is better, A or B, if different data are involved in them.
Standardizing/normalizing of features, of all or selected, as well as other transformations with data, change data. Dimensions get "distorted" unequally. It is like another dataset what appears then. So, see the previous paragraph.

Here is an immediate and untested, one possible heuristic approach to conclude if a partition c1 based on cluster analysis of data first transformed t1 (or, maybe it means no transformation: raw data) is better or worse than a partition c2 based on cluster analysis of the data first transformed t2.
You have clustering index (e.g. average Silhouette) values: $I_{t1c1}$, $I_{t2c2}$. Instead of direct comparing $I_{t1c1} - I_{t2c2}$, calculate two more values, now without doing cluster analyses:


*

*$I_{c1t2}$ - t2-transtorm the original data and compute the index
value for c1 against it.

*$I_{c2t1}$ - t1-transtorm the original data and compute the index value for c2 against it.


Note that we can directly compare $I_{t1c1}$ with $I_{c2t1}$, and $I_{t2c2}$ with $I_{c1t2}$ - because transform t (i.e. the data values) within both cases is the same. Compute product $(I_{t1c1}-I_{c2t1})(I_{c1t2}-I_{t2c2})$. If solution c1 pretends to be definitely better than solution c2, it should show itself to be better both under t1 and t2, its overall advantage being the product of the two local advantages. Assuming that index I is "the greater is the better", - if both multipliers are positive c1 is better, if both are negative c2 is better. If the product is negative, no definite conclusion.
