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Let $x_1,...,x_n$ be a random sample from some population with $n\geq3$ and with at most $n-2$ sample points being equal. How can one show $$ \sum_1^n (x_i-\bar{x})^2 = \sum_1^{n-1} (x_i -\bar{x}^*)^2+\frac n{n-1}(x_n-\bar{x})^2, $$ where $ \bar{x}=\frac1n\sum_1^nx_i$ and $ \bar{x}^* = \frac 1{n-1} \sum_1^{n-1} x_i.$ This is a useful identity since it would allow me to have an upper bound on the deviation of $x_i$ from the mean. That is $$ \max_{1\leq i \leq n} | x_i - \bar{x}| < \frac {n-1} {\sqrt{n}} S, $$ where $ S^2 = \frac 1 {n-1} \sum_1^n (x_i - \bar{x})^2.$

Thank you very much for your help.

Source of the question

A Matrix Formulation on How Deviant an Observation Can Be
Author: Ingram Olkin.
Source: The American Statistician, Vol. 46, No. 3 (Aug., 1992), pp. 205-209 https://www.jstor.org/stable/2685215

I will add shortly what I've tried so far.

Fiori_Modena
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  • Your first term on the right hand side is $0$. – lightxbulb Jan 18 '19 at 02:05
  • Welcome to MSE. Please indicate what you've tried so far, and are possibly having difficulty with. Also, I haven't checked your equation carefully, but I'm fairly certain you are missing a square of the values being added in the first the term on the right. – John Omielan Jan 18 '19 at 02:06
  • I found the answer in another question. I just didn't find it prior to posting mine. https://math.stackexchange.com/a/2442224/512044 – Fiori_Modena Jan 18 '19 at 03:21
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    The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be?? – StubbornAtom Jan 18 '19 at 14:40
  • @StubbornAtom, Yes, I have seen that article. It's in that article that I have found this useful equality that I am trying to show. The proof of the equality is in this question https://math.stackexchange.com/a/2442224/512044. I used it and I managed to answer my question. Should I write my own answer or leave to somebody else the chance to answer? – Fiori_Modena Jan 22 '19 at 03:03
  • @Fiori_Modena You are welcome to post your own answers and accept them to mark the question as answered. – StubbornAtom Jan 22 '19 at 07:20

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