1
$\begingroup$

I have the following PDE: $$ u_t + r x u_x + \frac{\sigma^2 x^2}{2} u_{xx} + h(t,x) u_y - ru =0 $$ $$ u(x,T,y) = y $$

I wanted to check whether the following representation is correct (I used Feynman-Kac theorem):

$$ u(x,t,y) = E[ y e^{-r(T-t)} | x(T) = x] $$

Thanks!

$\endgroup$
1
$\begingroup$

What I get is $$ u(x,t,y) = \mathbf{E} \left[ y(T)e^{-t(T-t)} \right. \left| X(t) = x, Y(t) = y \right], $$ where the processes $X$, and $Y$ follow $$ dX(t) = rX(t) dt + \sigma X(t) dW(t),\text{ and } dY(t) = h( t,X(t))dt. $$

$\endgroup$
1
  • $\begingroup$ In this case, can we write: $$ u(x,t,y) = \mathbf{E} [\int_{0}^{T} {h(t,X(t)) dt} \hspace{0.1cm} e^{-r(T-t)} | X(t) = x], $$ $\endgroup$ – kagami Sep 27 '14 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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