I am interested in eigenfunction expansions of random functions. We know that the autocorrelation of brownian motion, $\{ B_t \}_{t \geq 0}$, is given by $$ E[B_t B_s] = \min\{s,t \}, $$ which can be expanded as $\sum_{k=0}^{\infty} \lambda_k \phi_k(s) \phi_k(t)$ where the $\lambda_k = \pi^2 (k - 1/2)^2$ are eigenvalues corresponding to $\phi_k (t) = \sqrt{2} \sin(\pi(k - 1/2)t)$. From this one may show that $$B_t = \sum_{k = 0}^{\infty} \xi_k \lambda_k^{-1/2} \phi_k(t)$$ where $\xi_k$ are independent standard normal variables, and $t$ is between $0$ and $1$.

For functions on $[0,1]$, many alternatives to Fourier expansions have been put forward, often with higher order of convergence. A key example is the Lanczos representation of $f(x)$, given by $$ f(x) = \sum_{k=0}^{p-1} [f^{(k)}(1) - f^{(k)}(0)]{B_{k+1}(x)\over k!} + \int_0^1 f(t) dt + 2\sum_{k=1}^{\infty} [c_k(\cos(2\pi kx) + s_k\sin(2\pi kx)] $$ with $B_k(x)$ equal to the $k^{th}$ Bernoulli polynomial and $c_k + is_k = \left({-1 \over 2k\pi i}\right)^p\int_0^1 f^{(p)}(t) (e^{2k\pi i t} - 1) dt$.

The convergence here is of order $\lambda_k^p$, but the higher convergence rate is gained at the cost of calculating the relevant Bernoulli polynomials up to order $p+1$. An obvious question is whether a similar expansion is possible for the expansion of a random function, such as the Brownian motion $B_t$. Any insight into obstacles or inroads in this direction would be highly appreciated.

  • $\begingroup$ *is required to reach the possible sample paths $\endgroup$ – Titus Nov 7 '14 at 7:27
  • $\begingroup$ 10 vs 100 terms. Terms => steps to find result. 'Randomness' - avoiding repeating same values. $\endgroup$ – user103028 Nov 14 '14 at 6:09

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