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As has been mentioned in previous answers, the moments do not uniquely determine the distributions unless certain conditions are satisfied, such as bounded distributions. One thing you can say, is that the distribution of a random variable X is uniquely determined by the characteristic function φX(a)=E[exp(iaX)]. Letting mn=E[Xn] be the n'th moment, this can be expanded as

φX(a) = Σninanmn/n!

which is valid within its radius of convergence. So, the moments will uniquely determine the distribution as long as this has infinite radius of convergence, which is the case as long as limsupn→∞|mn/n!|1/n=0. Stirling's formula simplifies it a bit to limsupn→∞|mn|1/n/n=0. This can be proven using the dominated convergence theorem.

For example, a distribution is bounded by K if |mn|≤Kn, which satisfies this condition.

On the other hand, given any unbounded set S, it is not hard possible to construct two different distinct distributions supported in S the positive integers and with the same moments. (I'll add To do thisin , you need to find a moment)sequence of real numbers cn satisfying Σncnnr=0 for all r (and converging absolutely). This doesn't involve anything more than solving some linear equations to solve this for any finite set of powers r. Then, by keeping adding more terms to extend to all positive integers r, you get the infinite sequence cn. The two distributions can then be obtained by taking the positive and negative parts of cn.

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As has been mentioned in previous answers, the moments do not uniquely determine the distributions unless certain conditions are satisfied, such as bounded distributions. One thing you can say, is that the distribution of a random variable X is uniquely determined by the characteristic function φX(a)=E[exp(iaX)]. Letting mn=E[Xn] be the n'th moment, this can be expanded as

φX(a) = Σninanmn/n!

which is valid within its radius of convergence. So, the moments will uniquely determine the distribution as long as this has infinite radius of convergence, which is the case as long as limsupn→∞|mn/n!|1/n=0. Stirling's formula simplifies it a bit to limsupn→∞|mn|1/n/n=0. This can be proven using the dominated convergence theorem.

For example, a distribution is bounded by K if |mn|≤Kn, which satisfies this condition.

On the other hand, given any unbounded set S, it is not hard to construct two different distributions supported in S and with the same moments. (I'll add this in a moment).

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As has been mentioned in previous answers, the moments do not uniquely determine the distributions unless certain conditions are satisfied, such as bounded distributions. One thing you can say, is that the distribution of a random variable X is uniquely determined by the characteristic function φX(a)=E[exp(iaX)]. Letting mn=E[Xn] be the n'th moment, this can be expanded as

φX(a) = Σninanmn/n!

which is valid within its radius of convergence. So, the moments will uniquely determine the distribution as long as this has infinite radius of convergence, which is the case as long as limsupn→∞|mn/n!|1/n=0. For example, a distribution is bounded by K if |mn|≤Kn, which satisfies this condition.

On the other hand, given any unbounded set S, it is not hard to construct two different distributions supported in S and with the same moments. (I'll add this in a moment).