4

For a positive Borel measure $\mu$ on the real line, let $\displaystyle{m_n = \int_{-\infty}^\infty x^n d\mu(x)}$, 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.

zoidberg
  • 1,315

1 Answers1

1

If there is no condition on $m_0$ we can say nothing about the total masse of the measure solution. In fact if we assume that $\mu$ is a solution for the moment problem $$ m_k=\int_{\mathbb{R}}t^k d\mu(t) \qquad \textrm{for all }\; k\ge 1. $$ then $\mu_\alpha=\mu+\alpha \delta_0$ will be also solution for the same problem for every $\alpha \ge0$, with total mass $\mu_\alpha(\mathbb{R})=\mu(\mathbb{R})+\alpha.$

Hamza
  • 3,831