2
$\begingroup$

In some projects I got to a conditional expectation of the form:

$$E\left[\exp\left\{\int_s^tX(r)f(r)dr\right\}\mid\mathcal F_s\right]$$

where $t\geq s$, $f$ is nice enough and $X$ is an Ito process and $\mathcal F_s$ is the information about $X$ up to time $s$. I want to compute this. Is there anyway to do so?

I tried Taylor expanding $\exp$, I tried to apply Ito formula in a few clever ways and nothing works out nicely. Is there anyway of computing this?

$\endgroup$
2
$\begingroup$

Disclaimer. I'm using $T$ instead of $t$ for the terminal time and $t_0$ instead of $s$ for the initial time. I am ignoring issues arising due to a lack of regularity and, as you say, assuming that the drift, diffusion, and $f$ are nice enough.

Since $X$ is an Ito process and hence Markov, it is sufficient to consider $$ v(t,x)\equiv\mathbb{E}\left[\exp\left\{\int_{t}^{T}X(s)f(s)ds\right\}\,\middle|\, X(t)=x\right]. $$ By the Feynman-Kac formula, it follows that \begin{align*} v_{t}+Lv+xfv & =0, & \text{on }[t_{0},T)\times\mathbb{R}\phantom{.}\\ v(T,\cdot) & =1, & \text{on }\mathbb{R}. \end{align*} where $L$ is the infinitesimal generator of $X$.

You can solve this PDE numerically to get the value of $v(t_0, X(t_0))$. Alternatively, you can use Monte Carlo simulation on the original expectation, but this is generally slower than the solving the PDE numerically due to the problem's low dimensionality.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! This helps a lot. $\endgroup$ – user658409 Aug 16 '19 at 3:03

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.