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 $
$begingroup$
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.
statistics
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
$endgroup$
|
show 1 more comment
$begingroup$
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.
statistics
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
statistics
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
$endgroup$
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
statistics
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
edited Jan 18 at 2:35
Fiori_Modena
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
asked Jan 18 at 1:58
Fiori_ModenaFiori_Modena
225
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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