Why weights are not squared in the weighted Euclidean distance formula?

Question

I am working on Data Mining and need distance between data points.

With two points $(x_1, y_1), (x_2, y_2)$ & weights $w_1$ and $w_2$ respectively. The euclidean distance is: $$ \sqrt{w_1\times(x_2-x_1)^2+w_2\times(y_2-y_1)^2}. $$ I think intuitively $w_1$ & $w_2$ were supposed to be squared as they are getting square rooted in the result.

mitsos · Accepted Answer · 2019-09-14T12:09:00.157

2

You are right. $w$ is just a name for $\frac{1}{s^2}$, where $s$ for standard deviation. Full answer here:

http://www.econ.upf.edu/~michael/stanford/maeb4.pdf

edited Sep 14 '19 at 12:09

answered Sep 14 '19 at 12:02

mitsos

93

+1 for the reference. But I think the answer to the OP's question is "no". The entire expression is a square root. The weights should not be squared (they are squares). – Ethan Bolker Sep 14 '19 at 12:14
@Ethan Bolker, I see your point. I found it correct because $w=(\frac{1}{s})^2$ – mitsos Sep 14 '19 at 12:18
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@mitsos what if I take my own arbitrary weights? – mrityunjoynath Sep 14 '19 at 12:31
@mrityunjoynath you choose $w$ as the inverse variance of each variable. Why did you use this formula in the first place? I think the pdf link was pretty clear in page 4-7. – mitsos Sep 14 '19 at 12:45
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@mitsos yeah now it's clear. Awesome explanation. Thanks. – mrityunjoynath Sep 14 '19 at 13:01
@mrityunjoynath good question, helped me clear it also. – mitsos Sep 14 '19 at 14:41

Why weights are not squared in the weighted Euclidean distance formula?

1 Answers1