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.
Best Answer
No, it is not true that the relation between lower order standardized central moments of two different symmetric distributions determines the relation for those of higher order. It can be disproven by counterexamples. For instance, take the symmetric triangular distribution and the exponential power distribution with $\beta=4$.
Both have second standardized central moments equal to $1$ (as all distributions, since we normalize with the respective powers of the variance). Due to the symmetry, both have odd moments equal to $0$. From order 4 to 12, the triangular distribution has higher moments (e.g. kurtosis: $2.4$ to $2.1884$), but from 14 to 20, the moments of the exponential fourth power distribution exceed those of the triangular.