Karhunen–Loève theorem expansion random variables

  • #1
cianfa72
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TL;DR Summary
About the integral involved in the definition of Karhunen–Loève theorem expansion random variable coefficients.
Hi,
in the Karhunen–Loève theorem's statement the random variables in the expansion are given by $$Z_k = \int_a^b X_te_k(t) \: dt$$
##X_t## is a zero-mean square-integrable stochastic process defined over a probability space ##(\Omega, F, P)## and indexed over a closed and bounded interval ##[a, b]##, with continuous covariance function ##K_X(s, t)##.

##e_k(t)## are elements of an orthonormal basis on ##L^2([a, b])## formed by the eigenfunctions of the linear operator ##T_{K_{X}}## with respective eigenvalues ##\lambda_k##. As explained here ##L^2([a, b])## is the Hilbert space of measurable square functions defined on the space ##([a,b], \mathcal B([a,b], P)## that have finite measure. Note that ##\mathcal B([a,b])## is the Borel ##\sigma##-algebra on the closed set ##[a,b]## and ##P## is the Lebesgue measure on it.

I'm confused about the meaning of the integral involved in the definition of random variables ##Z_k##

Which kind of integral is this ?

Ps. I asked the same question on MSE without luck.
 
Last edited:
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  • #2
It seems that integrals used in the definition of expansion coefficients are just integrals over the realizations of the stochastic process in the interval ##[a,b]##.

Every coefficient is thus a random variable defined on the same ##(\Omega, F, P)## probability space.

However the above definition requires that each function product of stochastic process's realizations (sample paths) times basis elements ##e_k(t)## has to be integrable over the closed bounded set ##[a,b]## -- see also this lecture.

In particular each of the above function products has to be a measurable function w.r.t. the Borel ##\sigma##-algebra on ##[a,b]##. Thus their integrals are actually Lebesgue integrals, I believe.
 
Last edited:

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