# Karhunen-Loève expansion of Poisson process

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.

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## 1 Answer

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.

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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
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