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Iosif Pinelis
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Daniel Roy
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Does the truncated Hausdorff moment problem admit absolutely continuous solutions?

Let $\mu$ be a (Borel) probability measure on $[0,1]$ and define $m_j(\mu) = \int x^j\,\mu(dx)$. Let $k$ be a positive integer and consider the set $\mathcal C_{\mu,k}$ of probability measures $\nu$ on $[0,1]$ such that $m_j(\nu) = m_j(\mu)$ for $j = 1,\dotsc,k$.

We are interested in whether $\mathcal C = \mathcal C_{\mu,k}$ contains an absolutely continuous probability measure.

Some restrictions are obviously necessary: When $k=1$, evidently we must rule out $\mu \{0\} = 1$ or $\mu \{1\} = 0$, as either statement implies that $\mathcal C$ contains one point. More generally, if we define the set $$ \mathcal M_k = \{ (m_1(\nu),\dotsc,m_k(\nu) ) \} $$ of achievable moments, where $\nu$ ranges over the probability measures on $[0,1]$, then it would seem (though we have not shown) that the boundary $\partial \mathcal M_k$ corresponds to discrete distributions supported by at most $k$ points.

We are aware of some literature on truncated Hausdorff moment problems, and have looked through it, but our particular question does not seem to be addressed. When explicit representations (say, by densities that are sums of Bernstein polynomials) are used, the emphasis is usually on showing that the moments can be arbitrarily well approximated, but this does not rule out the possibility that the limit is no longer absolutely continuous.