Subset variance order preserving function
$begingroup$
Given a finite set with real numbers. X = {x1, x2, x3}. There can be a unique order defined for all the subsets using Variance operator.
e.g. X = {1, 2, 4}.
$$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
Var({1, 4}) > Var({1, 2, 4}) > Var({2, 4}) > Var({4}) = Var({2}) = Var({1}) = Var({}).
Is there any nontrivial function f(x) such that f(X) = {f(x1), f(x2), f(x3),...} preserve the same order?
f(x) = x and f(x) = -x would work.
order-theory transformation variance
asked Jan 8 at 4:39
yupbankyupbank
12
$endgroup$
add a comment |
$begingroup$
Given a finite set with real numbers. X = {x1, x2, x3}. There can be a unique order defined for all the subsets using Variance operator.
e.g. X = {1, 2, 4}.
$$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
Var({1, 4}) > Var({1, 2, 4}) > Var({2, 4}) > Var({4}) = Var({2}) = Var({1}) = Var({}).
Is there any nontrivial function f(x) such that f(X) = {f(x1), f(x2), f(x3),...} preserve the same order?
f(x) = x and f(x) = -x would work.
order-theory transformation variance
asked Jan 8 at 4:39
yupbankyupbank
12
$endgroup$
1
$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37
add a comment |
$begingroup$
Given a finite set with real numbers. X = {x1, x2, x3}. There can be a unique order defined for all the subsets using Variance operator.
e.g. X = {1, 2, 4}.
$$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
Var({1, 4}) > Var({1, 2, 4}) > Var({2, 4}) > Var({4}) = Var({2}) = Var({1}) = Var({}).
Is there any nontrivial function f(x) such that f(X) = {f(x1), f(x2), f(x3),...} preserve the same order?
f(x) = x and f(x) = -x would work.
order-theory transformation variance
asked Jan 8 at 4:39
yupbankyupbank
12
$endgroup$
Given a finite set with real numbers. X = {x1, x2, x3}. There can be a unique order defined for all the subsets using Variance operator.
e.g. X = {1, 2, 4}.
$$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
Var({1, 4}) > Var({1, 2, 4}) > Var({2, 4}) > Var({4}) = Var({2}) = Var({1}) = Var({}).
Is there any nontrivial function f(x) such that f(X) = {f(x1), f(x2), f(x3),...} preserve the same order?
f(x) = x and f(x) = -x would work.
order-theory transformation variance
order-theory transformation variance
asked Jan 8 at 4:39
yupbankyupbank
12
asked Jan 8 at 4:39
yupbankyupbank
12
edited Jan 8 at 16:42
yupbank
asked Jan 8 at 4:39
yupbankyupbank
12
asked Jan 8 at 4:39
yupbankyupbank
12
asked Jan 8 at 4:39
yupbankyupbank
12
12
1
$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37
add a comment |
1
$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37
1
1
$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37
add a comment |
1 Answer
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$begingroup$
The singleton and empty sets are not ordered by the variance.
The variance creates a preorder.
f(x) = ax, a /= 0 and f(x) = a + x
preserve the variance preorder.
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
$endgroup$
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
add a comment |
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$begingroup$
The singleton and empty sets are not ordered by the variance.
The variance creates a preorder.
f(x) = ax, a /= 0 and f(x) = a + x
preserve the variance preorder.
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
$endgroup$
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
add a comment |
$begingroup$
The singleton and empty sets are not ordered by the variance.
The variance creates a preorder.
f(x) = ax, a /= 0 and f(x) = a + x
preserve the variance preorder.
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
$endgroup$
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
add a comment |
$begingroup$
The singleton and empty sets are not ordered by the variance.
The variance creates a preorder.
f(x) = ax, a /= 0 and f(x) = a + x
preserve the variance preorder.
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
$endgroup$
The singleton and empty sets are not ordered by the variance.
The variance creates a preorder.
f(x) = ax, a /= 0 and f(x) = a + x
preserve the variance preorder.
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
answered Jan 8 at 21:08
William ElliotWilliam Elliot
9,1962820
9,1962820
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
add a comment |
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
$begingroup$
yeah, any variance invariant transformation would do. what if f(x) itself is a discrete set mapping, is there anyway to prove the mapping is preserving order or not?
$endgroup$
– yupbank
Jan 8 at 21:33
add a comment |
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$begingroup$
There are no duplicates nor order in sets. Define the variance operator of a set.
$endgroup$
– William Elliot
Jan 8 at 5:51
$begingroup$
the variance operator of a set is just calculate variance of all the elements inside of the set , consider, the set is finite and elements is defined in R. $$ {displaystyle operatorname {Var} (X)={frac {1}{n}}sum _{i=1}^{n}(x_{i}-mu )^{2},} $$ and $${displaystyle mu ={frac {1}{n}}sum _{i=1}^{n}x_{i}.}$$
$endgroup$
– yupbank
Jan 8 at 16:37