6
$\begingroup$

Given the PDE

$$\frac{\partial G}{\partial t} + 0.5\sigma^2 \frac{\partial^2 G}{\partial x^2} = 0$$

with condition $G(T,x) = x^2$, one can use the Feynman-Kac formula to arrive at

$$G(t,x) = E[X_T^2 | X_t = x] = E[ (X_t \pm \sigma(W_T - W_t))^2 |X_t = x] = x^2 + (T-t)\sigma^2$$

where $W_t$ is standard Brownian motion and $X_t$ is the stochastic process satisfying either:

$$dX_t = \pm \sigma dW_t$$

where the $X_t$'s and $W_t$'s are in the filtered probability space $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,t]}, \mathbb P)$ where $\mathscr F_t = \mathscr F_t^W$.


I am supposed to evaluate

$$E[ (X_t \pm \sigma(W_T - W_t))^2 |X_t]$$

and then later plug in $X_t = x$.

Apparently, in evaluating

$$E[ (X_t \pm \sigma(W_T - W_t))^2 |X_t]$$

I am to use the Markov property to say that

$$E[ (X_t \pm \sigma(W_T - W_t))^2 |X_t] = E[ (X_t \pm \sigma(W_T - W_t))^2 | \mathscr{F_t}]$$

Why exactly do we need to use the Markov property?

I know that $W_T - W_t$ is independent of $\mathscr{F_t}$. I think that $\because X_t \in m \mathscr F_t$, $W_T - W_t$ is independent also of $X_t$.

If I am wrong, why?

If I am right, why is the Markov property needed?


The problem seems to be taken from Bjork's Arbitrage Theory in Continuous Time. I got the problem from my class notes. Neither Bjork nor Wikipedia seems to use the Markov property

enter image description here

enter image description here

enter image description here

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Because Markov property is stronger than mere moment independency, and I want to point out that this is a SDE, in a better terminology. $\endgroup$ – Henry.L Sep 8 '15 at 3:15
  • 2
    $\begingroup$ @Henry.L It is not an SDE, I think. Feynman-Kac solves PDEs using stochastic processes. Anyway, $W_T - W_t$ IS independent of $X_t$? What's the purpose of using a stronger property here? $\endgroup$ – BCLC Sep 8 '15 at 3:36
  • 1
    $\begingroup$ I cannot answer you at the moment, but I will try to find some reference later if time permits. $\endgroup$ – Henry.L Sep 8 '15 at 3:43
  • $\begingroup$ @Henry.L Thank you. Any guesses as of the moment? $\endgroup$ – BCLC Sep 8 '15 at 4:09
  • $\begingroup$ maybe you can provide the actual text of the question $\endgroup$ – seanv507 Dec 9 '15 at 19:33
1
$\begingroup$

Perhaps what's meant here is that since $\mathcal{F}_t$ is a filtration, then for $s\leq t$, $\mathcal{F}_s\subseteq \mathcal{F}_t$. In other words the filtration contains all information up to time $t$ so that you really are invoking the Markov property since $X_t$ really just specifies $X_t$ at time $t$ only.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What? I mean why do we need to use Markov property to change to $\mathscr F_t$ ?Why is $X_t$ not sufficient? I mean why does the conclusion not hold if we don't condition on $\mathscr F_t$ but instead just $X_t$? $\endgroup$ – BCLC Dec 9 '15 at 18:56
  • $\begingroup$ Alex R., edited question $\endgroup$ – BCLC Dec 9 '15 at 22:14
1
$\begingroup$

In applying the Feynman-Kac formula, there is no need to use the Markov property.

In proving the Feynman-Kac formula, the Markov property is needed.

Showing that $G(t,x)$ indeed satisfies the PDE requires showing that $G(t,X_t)$ is a martingale which relies on $X_t$ having the Markov property, which it has because it is a solution of an SDE.

Or something like that.


From Shreve's Stochastic Calculus for Finance:


enter image description here


enter image description here


enter image description here


enter image description here


enter image description here


enter image description here

| cite | improve this answer | |
$\endgroup$

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.