1
$\begingroup$

This might be a simple question, but I'm having trouble with it.

Consider the Cauchy problem with final condition. \begin{equation} \begin{cases} \frac{\partial u}{\partial t}(t,x) + \mathcal{L}u(t,x) + k(t,x)u(t,x) = g(t,x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ u(T,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb} \end{equation} where \begin{equation} \mathcal{L} = \frac{1}{2}\sigma^2(t,x)\frac{\partial^2}{\partial x^2} + \mu(t,x)\frac{\partial }{\partial x}. \end{equation} I am well aware that a solution to this problem can be given in terms of the following Feynman–Kac formula: \begin{equation} u(t,x)=\mathbb{E}\left[\phi(X_T^{t,x})\exp\left\lbrace\int_t^Tk(s,X_s^{t,x})ds\right\rbrace -\int_t^Tg(s,X_s^{t,x})\exp\left\lbrace\int_t^s k(u,X_u^{t,x})du\right\rbrace ds\right] \end{equation} where $X_t$ is an Itô process that is described by: \begin{equation} dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t\,, \end{equation} with $X_0=x$.

The problem arises when I try to make the change of variables $v(t') = u(T-t)$. Now, the previous Cauchy problem with final condition, becomes a Cauchy problem with initial condition: \begin{equation} \begin{cases} -\frac{\partial v}{\partial t'}(t',x) + \mathcal{L}v(t',x) + k(T-t',x)v(t',x) = g(T-t',x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ v(0,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb2} \end{equation} with \begin{equation} \mathcal{L} = \frac{1}{2}\sigma^2(T-t',x)\frac{\partial^2}{\partial x^2} + \mu(T-t',x)\frac{\partial }{\partial x}. \end{equation}

What form does the Feynman–Kac formula take when we perform this change of variable?

$\endgroup$
4
  • $\begingroup$ In the equation for $v$, the coefficients should be time reversed as well (i.e. $k(t',x)$ should be $k(T - t',x)$). $\endgroup$ – Peter Morfe Mar 1 at 13:14
  • $\begingroup$ Thanks, @PeterMorfe, I just edited the question. $\endgroup$ – Paulo Rocha Mar 1 at 13:49
  • $\begingroup$ $\mathcal{L}$ also needs to be transformed. $\endgroup$ – Peter Morfe Mar 1 at 14:36
  • $\begingroup$ Thanks for helping improve the question. I made the necessary edits. $\endgroup$ – Paulo Rocha Mar 1 at 16:21
0
$\begingroup$

Given $t \in (0,T)$, define $\tilde{X}^{(t),x}$ to be the solution of the SDE \begin{equation*} d\tilde{X}^{(t),x}_{s} = \mu(t + s,\tilde{X}^{(t),x}_{s}) \, ds + \sigma(t + s, \tilde{X}^{(t),x}_{s}) \, d B_{s}, \quad \tilde{X}^{(t),x}_{0} = x. \end{equation*} Notice that $\tilde{X}^{(t),x}_{\cdot} = X^{t,x}_{\cdot + t}$. Hence we can write \begin{align*} u(t,x) &= \mathbb{E} \left(\phi(\tilde{X}^{(t),x}_{T - t}) \exp \left \{ \int_{0}^{T -t} k(s + t, \tilde{X}^{(t),x}_{s} \, ds \right\}\right)\\ &\quad - \mathbb{E} \left( \int_{0}^{T - t} g(s + t, \tilde{X}^{(t),x}_{s}) \exp \left\{ \int_{0}^{t} k(u + t, \tilde{X}^{(t),x}_{u}) \,du \right\} \, ds \right) \end{align*}

Now if we define $v(t',x) = u(T - t,x)$, we find \begin{align*} v(t',x) &= \mathbb{E} \left(\phi(\tilde{X}^{(T - t'),x}_{t'}) \exp \left \{ \int_{0}^{t'} k(s + T - t', \tilde{X}^{(T - t'),x}_{s} \, ds \right\}\right)\\ &\quad - \mathbb{E} \left( \int_{0}^{t'} g(s + T - t', \tilde{X}^{(T - t'),x}_{s}) \exp \left\{ \int_{0}^{T - t'} k(u + T - t', \tilde{X}^{(T - t'),x}_{u}) \,du \right\} \, ds \right) \end{align*} It's worth noticing that this formula makes a certain amount of sense when $\mu$ and $\sigma$ are temporally homogeneous, in which case $\tilde{X}^{(t)}$ is independent of $t$.

$\endgroup$

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.