A moment problem

Question

What is the p.d.f. for the moments $m_n=1/n$ ?

(They are obtained from $\int_0^1 x^n/x\ dx $, but clearly $1/x$ is not a p.d.f. on $[0,1]$)

Answer 1

Suppose $\mu$ is a Borel measure on $\mathbb{R}$ such that there exist $a \in \mathbb{R}$ and $N \in \mathbb{N}_0$ satisfying

$$ \int_{\mathbb{R}} x^n \, \mu(\mathrm{d}x) = \frac{a}{n} \tag{*} $$

for all $n \geq N$. Then it is easy to see that $a \geq 0$ and that $\mu$ is supported on $[-1, 1]$. Moreover, we know that $\nu(\mathrm{d}x) = x^N \mu(\mathrm{d}x)$ satisfies:

$$ \int_{\mathbb{R}} x^n \, \nu(\mathrm{d}x) = \int_{\mathbb{R}} x^{n+N} \, \mu(\mathrm{d}x) = \frac{a}{N+n}, \qquad \forall n \geq 0. $$

Since the Hausdorff moment problem has a unique solution, the above condition determines $\nu$ as $\nu(\mathrm{d}x) = ax^{N-1} \, \mathrm{d}x$. Therefore, we conclude

$$ \mu(\mathrm{d}x) = \frac{a}{x}\mathbf{1}_{(0,1)}(x) \, \mathrm{d}x. $$

Of course, this cannot be a probability measure.