Is a sample of size $N$ uniquely described by $N$ sample moments?
$begingroup$
I have the intuition that a sample of size $n$ should be able to be described uniquely by just N moments. But I don't know if that's true.
The idea is that, first, it is obvious that a sample of $n = 1$ can be uniquely describe by just the sample mean. For $n = 2$, knowing the first two (uncentered) moments is also enough because (if I did my maths correctly)
$$
X_1 = 2overline{X} - X_2
$$
and
$$
M_2 = 1/2X_2^2 - 2M_1X_2+2M_1^2
$$
From which I can solve for $X_2$ and then $X_1$ by knowing $M_1$ and $M_2$.
It seems intuitive that this must hold on for $n > 2 $. The generalization for a sample of size $n$ and $m$ moments would be
begin{matrix}
x_1 + x_2 + ... + x_n = nM_1\
x_1^2 + x_2^2 + ... + x_n^2 = nM_2 \
vdots \
x_1^m + x_2^m + ... + x_n^m = nM_m
end{matrix}
It these were linear equations, they would only have an unique solution for $n = m$ if the system described a full rank matrix. But they aren't, so... no dice.
From an information theory standpoint it seems to me (again, intuitively) that you shouldn't need more than $N$ numbers to represent any other $N$ numbers.
Note that I'm not trying to estimate the population distribution by the sample moments. I'm talking about describing a specific vector of $n$ values.
Also note that I assume that the order of samples does not matter so that $(x_1, x_2)$ and $(x_2, x_2)$ would be treated as the same solution.
Any hints on this problem?
statistics
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
$endgroup$
add a comment |
$begingroup$
I have the intuition that a sample of size $n$ should be able to be described uniquely by just N moments. But I don't know if that's true.
The idea is that, first, it is obvious that a sample of $n = 1$ can be uniquely describe by just the sample mean. For $n = 2$, knowing the first two (uncentered) moments is also enough because (if I did my maths correctly)
$$
X_1 = 2overline{X} - X_2
$$
and
$$
M_2 = 1/2X_2^2 - 2M_1X_2+2M_1^2
$$
From which I can solve for $X_2$ and then $X_1$ by knowing $M_1$ and $M_2$.
It seems intuitive that this must hold on for $n > 2 $. The generalization for a sample of size $n$ and $m$ moments would be
begin{matrix}
x_1 + x_2 + ... + x_n = nM_1\
x_1^2 + x_2^2 + ... + x_n^2 = nM_2 \
vdots \
x_1^m + x_2^m + ... + x_n^m = nM_m
end{matrix}
It these were linear equations, they would only have an unique solution for $n = m$ if the system described a full rank matrix. But they aren't, so... no dice.
From an information theory standpoint it seems to me (again, intuitively) that you shouldn't need more than $N$ numbers to represent any other $N$ numbers.
Note that I'm not trying to estimate the population distribution by the sample moments. I'm talking about describing a specific vector of $n$ values.
Also note that I assume that the order of samples does not matter so that $(x_1, x_2)$ and $(x_2, x_2)$ would be treated as the same solution.
Any hints on this problem?
statistics
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
$endgroup$
1
$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25
add a comment |
$begingroup$
I have the intuition that a sample of size $n$ should be able to be described uniquely by just N moments. But I don't know if that's true.
The idea is that, first, it is obvious that a sample of $n = 1$ can be uniquely describe by just the sample mean. For $n = 2$, knowing the first two (uncentered) moments is also enough because (if I did my maths correctly)
$$
X_1 = 2overline{X} - X_2
$$
and
$$
M_2 = 1/2X_2^2 - 2M_1X_2+2M_1^2
$$
From which I can solve for $X_2$ and then $X_1$ by knowing $M_1$ and $M_2$.
It seems intuitive that this must hold on for $n > 2 $. The generalization for a sample of size $n$ and $m$ moments would be
begin{matrix}
x_1 + x_2 + ... + x_n = nM_1\
x_1^2 + x_2^2 + ... + x_n^2 = nM_2 \
vdots \
x_1^m + x_2^m + ... + x_n^m = nM_m
end{matrix}
It these were linear equations, they would only have an unique solution for $n = m$ if the system described a full rank matrix. But they aren't, so... no dice.
From an information theory standpoint it seems to me (again, intuitively) that you shouldn't need more than $N$ numbers to represent any other $N$ numbers.
Note that I'm not trying to estimate the population distribution by the sample moments. I'm talking about describing a specific vector of $n$ values.
Also note that I assume that the order of samples does not matter so that $(x_1, x_2)$ and $(x_2, x_2)$ would be treated as the same solution.
Any hints on this problem?
statistics
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
$endgroup$
I have the intuition that a sample of size $n$ should be able to be described uniquely by just N moments. But I don't know if that's true.
The idea is that, first, it is obvious that a sample of $n = 1$ can be uniquely describe by just the sample mean. For $n = 2$, knowing the first two (uncentered) moments is also enough because (if I did my maths correctly)
$$
X_1 = 2overline{X} - X_2
$$
and
$$
M_2 = 1/2X_2^2 - 2M_1X_2+2M_1^2
$$
From which I can solve for $X_2$ and then $X_1$ by knowing $M_1$ and $M_2$.
It seems intuitive that this must hold on for $n > 2 $. The generalization for a sample of size $n$ and $m$ moments would be
begin{matrix}
x_1 + x_2 + ... + x_n = nM_1\
x_1^2 + x_2^2 + ... + x_n^2 = nM_2 \
vdots \
x_1^m + x_2^m + ... + x_n^m = nM_m
end{matrix}
It these were linear equations, they would only have an unique solution for $n = m$ if the system described a full rank matrix. But they aren't, so... no dice.
From an information theory standpoint it seems to me (again, intuitively) that you shouldn't need more than $N$ numbers to represent any other $N$ numbers.
Note that I'm not trying to estimate the population distribution by the sample moments. I'm talking about describing a specific vector of $n$ values.
Also note that I assume that the order of samples does not matter so that $(x_1, x_2)$ and $(x_2, x_2)$ would be treated as the same solution.
Any hints on this problem?
statistics
statistics
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
edited Dec 10 '18 at 14:28
Elio Campitelli
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
asked Dec 10 '18 at 3:29
Elio CampitelliElio Campitelli
1235
1235
1
$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25
add a comment |
1
$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25
1
1
$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25
add a comment |
1 Answer
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$begingroup$
We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n.
From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).
(Summing up a discussion with Elio via twiiter)
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
$endgroup$
add a comment |
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$begingroup$
We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n.
From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).
(Summing up a discussion with Elio via twiiter)
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
$endgroup$
add a comment |
$begingroup$
We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n.
From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).
(Summing up a discussion with Elio via twiiter)
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
$endgroup$
add a comment |
$begingroup$
We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n.
From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).
(Summing up a discussion with Elio via twiiter)
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
$endgroup$
We can think of x_1,x_2, ...x_n as being the roots of a polynomial p of degree n.
From the Newton identities we can compute from the given moments M_1, M_2, ...,M_n the elementary symmetric polynomials in x_1,x_2,... x_n and hence the coefficients of the polynomial p. But the set of roots (without order) of p is determined by p. Hence, the values of x_1,x_2,...x_n are determined by M_1,M_2, ...,M_n except for a permutation of the values. This is the best that you can hope as the moments are invariant under such a permutation (as Elio example shows).
(Summing up a discussion with Elio via twiiter)
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
answered Dec 10 '18 at 21:13
Pablo De NapoliPablo De Napoli
1795
1795
add a comment |
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$begingroup$
It's not true that you can determine the values of each of $n$ variables with the first $n$ moments. You can certainly determine the set of values in the sample but you can't know what, for instance, $x_1$ happens to be. Consider a sample size of two with first and second raw moments 1/2 and 1. There are two solutions: $left(x_1to frac{1}{2} left(1-sqrt{3}right),x_2to frac{1}{2} left(sqrt{3}+1right)right)$ is one solution and $left(x_1to frac{1}{2} left(sqrt{3}+1right),x_2to frac{1}{2} left(1-sqrt{3}right)right)$ is another.
$endgroup$
– JimB
Dec 10 '18 at 6:26
$begingroup$
Fair point. I was implicitly assuming that the order didn't matter. (Although I intuit that if we defined each $x_i$ as a tuple $(X^a_i; X^b_i)$, then we could have an analogous situation but with the combined moments; covariance and the like).
$endgroup$
– Elio Campitelli
Dec 10 '18 at 14:25