Inequalities for standardized central moments of probability distributions
$begingroup$
It's known that the standardized central even moments of any probability distribution with a density symmetric around the mean form a non-decreasing series, the lower bound (when all are equal to 1) provided by two-point distributions such as the Bernoulli distribution.
Let's say we have two distributions (A and B), for which the first few standardized central moments are exactly the same, but the next higher order even moment of A is greater than that of B.
My question is, that given these conditions, is it possible to argue that all further standardized central even moments of distribution A are going to be greater than those of B? I have seen plenty of examples pointing towards this evidence, however, I cannot provide any proof.
Edit: true for symmetric densities, not necessarily for skewed ones.
inequality probability-distributions
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
$endgroup$
add a comment |
$begingroup$
It's known that the standardized central even moments of any probability distribution with a density symmetric around the mean form a non-decreasing series, the lower bound (when all are equal to 1) provided by two-point distributions such as the Bernoulli distribution.
Let's say we have two distributions (A and B), for which the first few standardized central moments are exactly the same, but the next higher order even moment of A is greater than that of B.
My question is, that given these conditions, is it possible to argue that all further standardized central even moments of distribution A are going to be greater than those of B? I have seen plenty of examples pointing towards this evidence, however, I cannot provide any proof.
Edit: true for symmetric densities, not necessarily for skewed ones.
inequality probability-distributions
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
$endgroup$
add a comment |
$begingroup$
It's known that the standardized central even moments of any probability distribution with a density symmetric around the mean form a non-decreasing series, the lower bound (when all are equal to 1) provided by two-point distributions such as the Bernoulli distribution.
Let's say we have two distributions (A and B), for which the first few standardized central moments are exactly the same, but the next higher order even moment of A is greater than that of B.
My question is, that given these conditions, is it possible to argue that all further standardized central even moments of distribution A are going to be greater than those of B? I have seen plenty of examples pointing towards this evidence, however, I cannot provide any proof.
Edit: true for symmetric densities, not necessarily for skewed ones.
inequality probability-distributions
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
$endgroup$
It's known that the standardized central even moments of any probability distribution with a density symmetric around the mean form a non-decreasing series, the lower bound (when all are equal to 1) provided by two-point distributions such as the Bernoulli distribution.
Let's say we have two distributions (A and B), for which the first few standardized central moments are exactly the same, but the next higher order even moment of A is greater than that of B.
My question is, that given these conditions, is it possible to argue that all further standardized central even moments of distribution A are going to be greater than those of B? I have seen plenty of examples pointing towards this evidence, however, I cannot provide any proof.
Edit: true for symmetric densities, not necessarily for skewed ones.
inequality probability-distributions
inequality probability-distributions
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
edited Dec 17 '18 at 14:08
hryghr
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
asked Dec 17 '18 at 13:54
hryghrhryghr
1114
1114
add a comment |
add a comment |
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