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Consider a random scalar variable $X$ with arbitrary measure. I'm after a basis of polynomial functions $\{p_k\}_{k=0}^\infty$ which are orthonormal with respect to $X$ in the sense that

\begin{equation} \mathbb{E}_X [p_k(X)p_{k'}(X)] = \delta_{kk'}. \end{equation}

When discussing orthogonal polynomial bases, the measure of integration is usually assumed. For example, if $X \sim \mathcal{N}(0,1)$, then $\{p_k\}_{k=0}^\infty$ are the Hermite polynomials. However, it seems there ought to exist generic expressions for such orthogonal polynomials, with the coefficients given in terms of moments of $X$. For example, applying the typical Gram-Schmidt procedure, one can quickly find that

\begin{align} p_0(X) &= 1 \\ p_1(X) &= \frac{X - \mathbb{E}[X]}{\sqrt{\text{Var}[x]}} \\ p_2(X) &= \ \ \ ... \end{align}

Are there known expressions for the rest of this polynomial basis (or even just the next few elements)? In light of the expression for $p_1$, perhaps centered moments or cumulants are involved.

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  • $\begingroup$ For arbitrary measure, I see no reason to think this is any simpler than the general Gram-Schmidt procedure en.wikipedia.org/wiki/Gram–Schmidt_process $\endgroup$
    – Gerald Edgar
    Aug 17, 2022 at 0:20
  • $\begingroup$ Well there's a 3-term relation so it's not an entirely generic Gram-Schmidt situation. But I still wouldn't expect a general formula that's much more useful than generic Gram-Schmidt. $\endgroup$
    – Noam D. Elkies
    Aug 17, 2022 at 0:31
  • $\begingroup$ I see! Do you know of anywhere the next few terms might be written out? $\endgroup$
    – dotdashdashdash
    Aug 17, 2022 at 0:43
  • $\begingroup$ Actually you don't even need a measure. One can develop the theory of orthogonal polynomials starting from a scalar product on polynomials of the form L(pq) for a linear form L on polynomials. Check e.g. Akhiezer's book quoted below. $\endgroup$
    – Pietro Majer
    Aug 17, 2022 at 6:20

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For example, you can write orthogonal polynomials as determinants

$p_n(x) = c_n \, \det \begin{bmatrix} m_0 & m_1 & m_2 &\cdots & m_n \\ m_1 & m_2 & m_3 &\cdots & m_{n+1} \\ \vdots&\vdots&\vdots&\ddots& \vdots \\ m_{n-1} &m_n& m_{n+1} &\cdots &m_{2n-1}\\ 1 & x & x^2 & \cdots & x^n \end{bmatrix}$,

where $c_n$ is some constant for normalization and $m_k$ is the k-th moment.

A good book concerning orthogonal polynomials is Akhiezer, The Classical Moment problem

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  • $\begingroup$ amazing! this is what I was looking for. is there any easy way to see why these polynomials are orthogonal (e.g. using some rule for the product of two determinants)? $\endgroup$
    – dotdashdashdash
    Aug 25, 2022 at 22:53
  • $\begingroup$ ah, nvm, I see it - you can see $p_n$ is orthogonal to any monomial in $\{1, ..., x^{n-1}\}$ by just taking the expectation and noting that the bottom row is now a bunch of moments equal to one of the above rows, so the determinant's zero, and that's enough. sweet trick. thanks! $\endgroup$
    – dotdashdashdash
    Aug 26, 2022 at 0:22

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