Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_t,t\geq 0$ be a Poisson process with rate parameter $\lambda$. Compute the Karhunen-Loève expansion of $X$ in interval $[0, T]$. How about the KL expansion of the centered process $X_t−\lambda t$?

The auto-correlation function of Poisson process is $R(s,t)=\lambda^2st+\lambda \min(s,t)$. By definition, KL expansion should satisfy $\int^T_0 R(s,t)\phi_n(t)dt=\lambda_n \phi_n(s)$.

I've problems figuring out how to solve the integrated equation.

For Wiener process, this link and Wikipedia article on KL expansion was useful.

This is a mirror question of this MO question.

share|cite|improve this question
up vote 5 down vote accepted

The obvious works: plugging in KL integral equation the value of $R$ and splitting the integral on $(0,T)$ into a sum of integrals on $(0,s)$ and on $(s,T)$, one gets $$ \lambda_n\phi_n(s)=\lambda^2s\int_0^Tt\phi_n(t)\mathrm dt+\lambda\int_0^st\phi_n(t)\mathrm dt+\lambda s\int_s^T\phi_n(t)\mathrm dt. $$ Differentiating this twice yields $$ \lambda_n\phi_n''(s)=-\lambda\phi_n(s), $$ from which an expression of the eigenfunctions $\phi_n$ and eigenvalues $\lambda_n$ follows.

share|cite|improve this answer
Thanks, general solution for this form of ODE is $\phi_n(t) = A Sin(\sqrt(\frac{\lambda}{\lambda_n})t)$ (The cosine term will be zero, since $\phi(0)=0$. But for finding the value for A, I have to plug-it in back to the main equation and I cannot solve this one! – Adel Ahmadyan May 5 '12 at 15:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.