What conditions on the moments make a measure a probability measure?
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For a positive Borel measure $mu$ on the real line, let $displaystyle{m_n = int_{-infty}^infty x^n dmu(x)}$, i.e. the $n$th moments of the measure. Are there any conditions on $m_n$ for when $mu$ can be made into a probability measure, i.e. $mu(mathbb{R}) < infty$? Recall that the moments of a uniform distribution on $[-1,1]$ are $1/(n+1)$ for even $n$. The measure given by density $p(x)= 1/(2|x|)$ for $x in [-1,1]$ has moments $1/n$ for even $n$. So it seems that two moment sequences with very similar asymptotic behavior can give different answers to this question.
Edit
I should specify that $m_0$ is of course not given, since one can read off the answer from the $0$th moment.
probability-theory probability-distributions moment-generating-functions moment-problem
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
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add a comment |
$begingroup$
For a positive Borel measure $mu$ on the real line, let $displaystyle{m_n = int_{-infty}^infty x^n dmu(x)}$, i.e. the $n$th moments of the measure. Are there any conditions on $m_n$ for when $mu$ can be made into a probability measure, i.e. $mu(mathbb{R}) < infty$? Recall that the moments of a uniform distribution on $[-1,1]$ are $1/(n+1)$ for even $n$. The measure given by density $p(x)= 1/(2|x|)$ for $x in [-1,1]$ has moments $1/n$ for even $n$. So it seems that two moment sequences with very similar asymptotic behavior can give different answers to this question.
Edit
I should specify that $m_0$ is of course not given, since one can read off the answer from the $0$th moment.
probability-theory probability-distributions moment-generating-functions moment-problem
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21
add a comment |
$begingroup$
For a positive Borel measure $mu$ on the real line, let $displaystyle{m_n = int_{-infty}^infty x^n dmu(x)}$, i.e. the $n$th moments of the measure. Are there any conditions on $m_n$ for when $mu$ can be made into a probability measure, i.e. $mu(mathbb{R}) < infty$? Recall that the moments of a uniform distribution on $[-1,1]$ are $1/(n+1)$ for even $n$. The measure given by density $p(x)= 1/(2|x|)$ for $x in [-1,1]$ has moments $1/n$ for even $n$. So it seems that two moment sequences with very similar asymptotic behavior can give different answers to this question.
Edit
I should specify that $m_0$ is of course not given, since one can read off the answer from the $0$th moment.
probability-theory probability-distributions moment-generating-functions moment-problem
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
$endgroup$
For a positive Borel measure $mu$ on the real line, let $displaystyle{m_n = int_{-infty}^infty x^n dmu(x)}$, i.e. the $n$th moments of the measure. Are there any conditions on $m_n$ for when $mu$ can be made into a probability measure, i.e. $mu(mathbb{R}) < infty$? Recall that the moments of a uniform distribution on $[-1,1]$ are $1/(n+1)$ for even $n$. The measure given by density $p(x)= 1/(2|x|)$ for $x in [-1,1]$ has moments $1/n$ for even $n$. So it seems that two moment sequences with very similar asymptotic behavior can give different answers to this question.
Edit
I should specify that $m_0$ is of course not given, since one can read off the answer from the $0$th moment.
probability-theory probability-distributions moment-generating-functions moment-problem
probability-theory probability-distributions moment-generating-functions moment-problem
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
edited Dec 5 '18 at 20:16
zoidberg
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
asked Dec 4 '18 at 7:03
zoidbergzoidberg
1,065113
1,065113
$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21
add a comment |
$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21
$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21
$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21
add a comment |
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$begingroup$
Not just asymptotic behavior. You can have two different distributions with exactly the same moments of all orders. A necessary condition on ${m_n}$ is $m_n^{1/n}leq m_k^{1/k}$ for $n leq k$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 7:21