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Consider the unit interval $I$ with a continuous probability measure $\mu$, and consider a smooth random variable $f:I\to \mathbb{R}$. We can define its cumulative distribution function and probability density function in the regular way. There is a number of known way to obtain the latter two functions using histograms, kernel smoothing and so on. Both methods rely on the statistically motivated assumption that we randomly sample $f$.

Assume now that we have sampled $f$ on the Gauss-Legendre quadrature point $\lbrace \alpha_{i}^N \rbrace_{i=1}^N$, which are the roots of orthogonal polynomials w.r.t. $\mu$, $p_n (\alpha)$. We can now obtain an $N$-th order approximation of $f$ using $p_n$, $$ f \approx \sum\limits_{n=0}^{N-1} p_n (\alpha) \hat{f}(n) \, .$$

My Question: Is there an elegant way of approximating the CDF and/or PDF of $f$ only with the spectral coefficients $\hat{f} (n)$ and the value at tha quadrature point $f(\alpha _j ^N )$?

Note that generally speaking, $f$ is not the same as its PDF.

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Gauss quadrature approximations of probability distributions undergird some quite new numerical methods for approximating the solution of SDEs of McKean type; see Section 2 of https://arxiv.org/abs/1608.06741 and references therein. In such SDEs, the coefficients depend on both the current state of the process and the distribution of the solution.

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    $\begingroup$ Nawaf, In section 2 they approximate $\phi$ with GL quadrature, but they don't construct its pdf or cdf, do they? $\endgroup$
    – Amir Sagiv
    Commented Aug 29, 2016 at 16:23
  • $\begingroup$ I would arg that what they do is precisely what you want, since they show how to approximate $\mu(\phi)$ with GL quadrature, not $\phi$ itself. Also their case is a bit more involved since they have to propagate the weights and quadrature points in their GL approximation under the dynamics of the SDE of McKean type. They propose a clever procedure to do this based on the Euler-Maruyama method. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Aug 29, 2016 at 17:52
  • $\begingroup$ PS: that is why I recommended this paper to you in the first place, since it seemed to be an "elegant" approach by necessity. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Aug 29, 2016 at 18:00

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