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 $












3












$begingroup$


Let $x_1,...,x_n$ be a random sample from some population with $ngeq3$ 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}=frac1nsum_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
$$ max_{1leq i leq n} | x_i - bar{x}| < frac {n-1} {sqrt{n}} S, $$ 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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Your first term on the right hand side is $0$.
    $endgroup$
    – lightxbulb
    Jan 18 at 2:05










  • $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 18 at 2:06










  • $begingroup$
    I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
    $endgroup$
    – Fiori_Modena
    Jan 18 at 3:21






  • 1




    $begingroup$
    The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
    $endgroup$
    – StubbornAtom
    Jan 18 at 14:40










  • $begingroup$
    @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 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?
    $endgroup$
    – Fiori_Modena
    Jan 22 at 3:03


















3












$begingroup$


Let $x_1,...,x_n$ be a random sample from some population with $ngeq3$ 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}=frac1nsum_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
$$ max_{1leq i leq n} | x_i - bar{x}| < frac {n-1} {sqrt{n}} S, $$ 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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Your first term on the right hand side is $0$.
    $endgroup$
    – lightxbulb
    Jan 18 at 2:05










  • $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 18 at 2:06










  • $begingroup$
    I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
    $endgroup$
    – Fiori_Modena
    Jan 18 at 3:21






  • 1




    $begingroup$
    The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
    $endgroup$
    – StubbornAtom
    Jan 18 at 14:40










  • $begingroup$
    @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 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?
    $endgroup$
    – Fiori_Modena
    Jan 22 at 3:03
















3












3








3





$begingroup$


Let $x_1,...,x_n$ be a random sample from some population with $ngeq3$ 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}=frac1nsum_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
$$ max_{1leq i leq n} | x_i - bar{x}| < frac {n-1} {sqrt{n}} S, $$ 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.










share|cite|improve this question











$endgroup$




Let $x_1,...,x_n$ be a random sample from some population with $ngeq3$ 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}=frac1nsum_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
$$ max_{1leq i leq n} | x_i - bar{x}| < frac {n-1} {sqrt{n}} S, $$ 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.







statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 2:35







Fiori_Modena

















asked Jan 18 at 1:58









Fiori_ModenaFiori_Modena

225




225












  • $begingroup$
    Your first term on the right hand side is $0$.
    $endgroup$
    – lightxbulb
    Jan 18 at 2:05










  • $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 18 at 2:06










  • $begingroup$
    I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
    $endgroup$
    – Fiori_Modena
    Jan 18 at 3:21






  • 1




    $begingroup$
    The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
    $endgroup$
    – StubbornAtom
    Jan 18 at 14:40










  • $begingroup$
    @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 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?
    $endgroup$
    – Fiori_Modena
    Jan 22 at 3:03




















  • $begingroup$
    Your first term on the right hand side is $0$.
    $endgroup$
    – lightxbulb
    Jan 18 at 2:05










  • $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 18 at 2:06










  • $begingroup$
    I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
    $endgroup$
    – Fiori_Modena
    Jan 18 at 3:21






  • 1




    $begingroup$
    The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
    $endgroup$
    – StubbornAtom
    Jan 18 at 14:40










  • $begingroup$
    @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 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?
    $endgroup$
    – Fiori_Modena
    Jan 22 at 3:03


















$begingroup$
Your first term on the right hand side is $0$.
$endgroup$
– lightxbulb
Jan 18 at 2:05




$begingroup$
Your first term on the right hand side is $0$.
$endgroup$
– lightxbulb
Jan 18 at 2:05












$begingroup$
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.
$endgroup$
– John Omielan
Jan 18 at 2:06




$begingroup$
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.
$endgroup$
– John Omielan
Jan 18 at 2:06












$begingroup$
I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
$endgroup$
– Fiori_Modena
Jan 18 at 3:21




$begingroup$
I found the answer in another question. I just didn't find it prior to posting mine. math.stackexchange.com/a/2442224/512044
$endgroup$
– Fiori_Modena
Jan 18 at 3:21




1




1




$begingroup$
The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
$endgroup$
– StubbornAtom
Jan 18 at 14:40




$begingroup$
The main result you are trying to prove is Samuelson's inequality. Did you see the original paper How Deviant Can You Be??
$endgroup$
– StubbornAtom
Jan 18 at 14:40












$begingroup$
@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 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?
$endgroup$
– Fiori_Modena
Jan 22 at 3:03






$begingroup$
@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 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?
$endgroup$
– Fiori_Modena
Jan 22 at 3:03












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