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Intuitively, I feel like the following should hold, but I fail to prove it:

$\lVert x^*- \frac 1k \Sigma_{i=1}^k w_ix_i\rVert_2 \overset ?= \frac 1k \Sigma_{i=1}^k w_i \lVert x^*- x_i\rVert_2$

Where $x_1,...,x_k, x^*\in\mathbb R^d$

I.e. is the distance to a (weighted) average point equal to the average of (weighted) distances to original points?

Shay BE
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  • Fails in most cases! Have you tried any examples? – Kavi Rama Murthy Mar 28 '19 at 09:04
  • Suppose $x^*$ is the average point. Then the left-hand side is zero but the right-hand side is generally not. –  Mar 28 '19 at 09:05
  • I don't want to sound rude, it's definitely not my intention, but have you even tried a single example (to check for counterexamples)? Just take a simple example, with $k\leq 2$. I tried once taking a few arbitrarily chosen numbers and the first example failed. – Eff Mar 28 '19 at 09:09
  • @Eff, no. I was working by geometric intuition. – Shay BE Mar 28 '19 at 09:11
  • @ShayBE That's fine to do. But it's always a good idea to check a variety of cases to test your intuition. If you try a few examples, then you will either find a counter-example or you will not. If you find a counter-example, great, you've disproven your intuition. If you don't find a counter-example, then either you just missed a counter-example or your intuition is really correct. If you don't find a counter-example, then you need proof. – Eff Mar 28 '19 at 09:15
  • @Eff, like in most things, hindsight is 20-20. I am working through some things that are a bit out of my element, appreciate the help. – Shay BE Mar 28 '19 at 09:15
  • @ShayBE This was mostly advice for the future. It's a very common thing for mathematicians (or people who are interested in mathematics) to have intuitions about what they think would have to be correct in some situation. In these cases, it is always a good idea to start out with a few cases to test your intuition. Because this can be an easy and quick way to falsify your intuition. If you don't find counter-examples, then it's inconclusive. – Eff Mar 28 '19 at 09:18
  • Again, appreciated. I will try to rephrase and post a new question better fitting to my actual problem. – Shay BE Mar 28 '19 at 09:20

1 Answers1

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A simple counterexample: $d=1,k=2,x_1=0,x_2=2,x^{*}=1,w_1=w_2=1$.

Kavi Rama Murthy
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