4
$\begingroup$

Is there a way to solve analytically the Fredholm integral equation of the second kind $$ \int_0^{100} K(s, t) f(s) ds = \lambda f(t) $$ where the kernel has the piecewise 'linear' form \begin{align} K_1(s, t) &= \min \left(s, t \right) \left(a_i+b_i \left(s - i\right)\right) \left(c_{ij} + d_{ij} \left(t- i - j \right) \right), \\ i &= \lfloor s \rfloor = \text{integer part of } s , \\ j &= \lfloor t \rfloor. \end{align} That is the value of the coefficients $a, b, c ,d$ depends only on the integer part of $s$ and $t$, and they are such that the kernel is continuous.

If instead, I use a kernel which is piecewise 'geometric' $$ K_2(s, t) = \min \left(s, t \right) a_i b_i ^ {\left(s - i\right)} c_{ij} d_{ij} ^ {\left(t- i - j \right)} , $$ is an analytic solution easier to find?

I'm trying to find the Karhunen-Loeve expansion of a stopped Brownian motion $$ Z_t = B_t I_t, $$ where $B$ is a standard Brownian motion and $I$ is a process which is one until a jump occurs, after which $I$ is zero.
When I calculate the covariance of $Z_t$, I fall on $K_1$ or $K_2$ depending on how I 'smooth' the discrete distribution of $I$.

I tried to mimic the derivation of the K-L expansion for Brownian motion.
But it leads nowhere, because $\partial_t K(s,t)$ is not constant.
What other methods can I use to solve the Fredholm equation?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.