Does $ arg min_{x_{2n}} sum_{i = 1}^{N} left( {s}_{i} - x_{2n} right)^{2n} $, for $n>1$, have a name...
$begingroup$
Given a list of $N$ real numbers $s_i$:
the median $x_0$ is
$$ arg min_{x_0} sum_{i = 1}^{N} left| {s}_{i} - x_0 right| $$
the mean $x_2$ is
$$ arg min_{x_2} sum_{i = 1}^{N} left( {s}_{i} - x_2 right)^2 $$
What about $x_4$?
$$ arg min_{x_4} sum_{i = 1}^{N} left( {s}_{i} - x_4 right)^4 $$
does have it a name?
And what about the generic $x_{2n}$
$$ arg min_{x_{2n}} sum_{i = 1}^{N} left( {s}_{i} - x_{2n} right)^{2n} $$
Are there any known application for $n>1$?
means median
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
$endgroup$
add a comment |
$begingroup$
Given a list of $N$ real numbers $s_i$:
the median $x_0$ is
$$ arg min_{x_0} sum_{i = 1}^{N} left| {s}_{i} - x_0 right| $$
the mean $x_2$ is
$$ arg min_{x_2} sum_{i = 1}^{N} left( {s}_{i} - x_2 right)^2 $$
What about $x_4$?
$$ arg min_{x_4} sum_{i = 1}^{N} left( {s}_{i} - x_4 right)^4 $$
does have it a name?
And what about the generic $x_{2n}$
$$ arg min_{x_{2n}} sum_{i = 1}^{N} left( {s}_{i} - x_{2n} right)^{2n} $$
Are there any known application for $n>1$?
means median
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
$endgroup$
add a comment |
$begingroup$
Given a list of $N$ real numbers $s_i$:
the median $x_0$ is
$$ arg min_{x_0} sum_{i = 1}^{N} left| {s}_{i} - x_0 right| $$
the mean $x_2$ is
$$ arg min_{x_2} sum_{i = 1}^{N} left( {s}_{i} - x_2 right)^2 $$
What about $x_4$?
$$ arg min_{x_4} sum_{i = 1}^{N} left( {s}_{i} - x_4 right)^4 $$
does have it a name?
And what about the generic $x_{2n}$
$$ arg min_{x_{2n}} sum_{i = 1}^{N} left( {s}_{i} - x_{2n} right)^{2n} $$
Are there any known application for $n>1$?
means median
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
$endgroup$
Given a list of $N$ real numbers $s_i$:
the median $x_0$ is
$$ arg min_{x_0} sum_{i = 1}^{N} left| {s}_{i} - x_0 right| $$
the mean $x_2$ is
$$ arg min_{x_2} sum_{i = 1}^{N} left( {s}_{i} - x_2 right)^2 $$
What about $x_4$?
$$ arg min_{x_4} sum_{i = 1}^{N} left( {s}_{i} - x_4 right)^4 $$
does have it a name?
And what about the generic $x_{2n}$
$$ arg min_{x_{2n}} sum_{i = 1}^{N} left( {s}_{i} - x_{2n} right)^{2n} $$
Are there any known application for $n>1$?
means median
means median
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
edited Dec 7 '18 at 8:32
Alessandro Jacopson
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
asked Dec 6 '18 at 20:16
Alessandro JacopsonAlessandro Jacopson
4261522
4261522
add a comment |
add a comment |
1 Answer
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$begingroup$
They are partial moments with respect to the minimizing reference point of an un-normalized uniformly distributed probability function.
From wikipedia:
$$mu_n^-(r) + mu_n^+(r)$$
$$r_o = min_{r}left{int_{-infty}^infty (r-x)^n f(x)dxright}$$
Where $f(x)$ is uniform discrete probability density $[0,N]$. (Since we are working with continuous case and integral instead of sum this is actually a sequence of Paul Dirac's delta distribution at integer values.)
Edit I first misread as $(x_k-i)^k$, that would have given the uniform discrete distribution mentioned above. Now what instead happens is we have sum of Dirac distributions like this: $$f(x) = sum_{i=0}^N delta(x-s_i)$$
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
$endgroup$
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
add a comment |
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1 Answer
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$begingroup$
They are partial moments with respect to the minimizing reference point of an un-normalized uniformly distributed probability function.
From wikipedia:
$$mu_n^-(r) + mu_n^+(r)$$
$$r_o = min_{r}left{int_{-infty}^infty (r-x)^n f(x)dxright}$$
Where $f(x)$ is uniform discrete probability density $[0,N]$. (Since we are working with continuous case and integral instead of sum this is actually a sequence of Paul Dirac's delta distribution at integer values.)
Edit I first misread as $(x_k-i)^k$, that would have given the uniform discrete distribution mentioned above. Now what instead happens is we have sum of Dirac distributions like this: $$f(x) = sum_{i=0}^N delta(x-s_i)$$
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
$endgroup$
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
add a comment |
$begingroup$
They are partial moments with respect to the minimizing reference point of an un-normalized uniformly distributed probability function.
From wikipedia:
$$mu_n^-(r) + mu_n^+(r)$$
$$r_o = min_{r}left{int_{-infty}^infty (r-x)^n f(x)dxright}$$
Where $f(x)$ is uniform discrete probability density $[0,N]$. (Since we are working with continuous case and integral instead of sum this is actually a sequence of Paul Dirac's delta distribution at integer values.)
Edit I first misread as $(x_k-i)^k$, that would have given the uniform discrete distribution mentioned above. Now what instead happens is we have sum of Dirac distributions like this: $$f(x) = sum_{i=0}^N delta(x-s_i)$$
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
$endgroup$
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
add a comment |
$begingroup$
They are partial moments with respect to the minimizing reference point of an un-normalized uniformly distributed probability function.
From wikipedia:
$$mu_n^-(r) + mu_n^+(r)$$
$$r_o = min_{r}left{int_{-infty}^infty (r-x)^n f(x)dxright}$$
Where $f(x)$ is uniform discrete probability density $[0,N]$. (Since we are working with continuous case and integral instead of sum this is actually a sequence of Paul Dirac's delta distribution at integer values.)
Edit I first misread as $(x_k-i)^k$, that would have given the uniform discrete distribution mentioned above. Now what instead happens is we have sum of Dirac distributions like this: $$f(x) = sum_{i=0}^N delta(x-s_i)$$
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
$endgroup$
They are partial moments with respect to the minimizing reference point of an un-normalized uniformly distributed probability function.
From wikipedia:
$$mu_n^-(r) + mu_n^+(r)$$
$$r_o = min_{r}left{int_{-infty}^infty (r-x)^n f(x)dxright}$$
Where $f(x)$ is uniform discrete probability density $[0,N]$. (Since we are working with continuous case and integral instead of sum this is actually a sequence of Paul Dirac's delta distribution at integer values.)
Edit I first misread as $(x_k-i)^k$, that would have given the uniform discrete distribution mentioned above. Now what instead happens is we have sum of Dirac distributions like this: $$f(x) = sum_{i=0}^N delta(x-s_i)$$
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
edited Dec 7 '18 at 9:31
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
answered Dec 6 '18 at 20:22
mathreadlermathreadler
14.9k72261
14.9k72261
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
add a comment |
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
$begingroup$
No wait, this ain't right. I will try to fix it. Ok, that should fix it, I think.
$endgroup$
– mathreadler
Dec 6 '18 at 20:37
add a comment |
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