I was taught the collection of a distribution's moments uniquely defined the distribution. Recently, I have been studying Pareto distributions, which have infinite means for shape parameters less than 1. It was also my understanding that if a moment of rank $\rho$ does not converge, all moments of rank greater than $\rho$ will also not converge. As such, for a Pareto distribution with shape $\alpha$ less than 1, its moments are:
$moment = \left\{
\begin{array}{lr}
\infty & : order \ge 1\\
1 & : order = 0
\end{array}
\right.\\$
The moments are identical for multiple different Pareto distributions, violating my initial statement, a contradiction.
Could anyone help point out where my lack of understanding lies?
Edit: fixed some notation
Best Answer
Getting the right distribution from moments is a tricky business. First off, the sequence of moments has to be valid. You can see a technical condition involving Hankel matrices in the link. Basically, it says you can't just write down an arbitrary sequence of numbers and claim they are moments of some distribution. In your case, the moments actually come from a distribution, so you don't have to worry about this.
In the case that all moments exist (are finite), then there are two subcases. If your distribution is supported on a finite interval $[a,b]$ then there is a unique measure that corresponds to them. If the interval is infinite, then the measure is not necessarily unique. You need an extra condition, something like Carleman's condition to guarantee uniqueness.
Finally, there's your case, which is a truncated moment problem: you only have existence of the first $k$ moments. This problem has even less uniqueness. This is because you can cook up plenty of distributions, such as power laws that look like $d/(1+a(x-b)^c)$ which can sometimes match all your moments. In your example, take $1/(1+|x-1|^3),$ which has mean 1 but no second moment.