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Let $\phi(x)$ be a probability distribution on$[0,1]$, and consider the moment matrix $M$ where the $(i,j)^{th}$ entry is $$ M_{ij} := \int_0^1 x^{i+j}\phi(x)dx, $$ or in other words, the expectation $\mathbb E(x^{i+j})$. Is there an easy way to see that this matrix is positive semidefinite? The matrix is definitely symmetric, and $$z^T Mz = \begin{bmatrix} \sum_{i=1}^n M_{i1}z_i & \cdots & \sum_{i=1}^nM_{in}z_i \end{bmatrix} \begin{bmatrix} z_1\\\vdots\\z_n \end{bmatrix} $$ $$= \left( \sum_{i=1}^nM_{i1}z_i \right) z_1 + \cdots + \left( \sum_{i=1}^nM_{in}z_i \right) z_n $$ $$= \left( \sum_{i=1}^n\mathbb E(x^{i+1})z_i \right) z_1 + \cdots + \left( \sum_{i=1}^n\mathbb E(x^{i+n})z_i \right) z_n$$ I got stuck here aiming to prove it via the definition. Or perhaps there is an eigenvalue approach?

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Note that $M = \int_0^1 \begin{bmatrix} x \\ \vdots \\ x^n\end{bmatrix} \begin{bmatrix} x & \cdots & x^n\end{bmatrix} \phi(x) dx$, and hence $\langle v , Mv \rangle = \int_0^1 (\langle v , \begin{bmatrix} x \\ \vdots \\ x^n\end{bmatrix} \rangle )^2 \phi(x) dx \ge 0$.

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  • $\begingroup$ You want to write powers, not subscripts. $\endgroup$ – Martin Argerami Feb 11 '14 at 6:35
  • $\begingroup$ @MartinArgerami: Thanks for catching that! $\endgroup$ – copper.hat Feb 11 '14 at 6:48

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