Given the PDE

$$\frac{\partial F}{\partial t} + \frac{1}{2}\sigma^2 \frac{\partial^2 F}{\partial x^2} = 0$$

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

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

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 such, 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


1 Answer 1


Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma dW_t, \end{align*} where $W$ is a standard Brownian motion. Since, for $0 \leq t \leq T$, \begin{align*} X_T = X_t + \sigma (W_T-W_t), \end{align*} based on Feynman–Kac formula, the solution is given by \begin{align*} F(t, x) &= E\left(X_T^2 \mid X_t = x\right)\\ &=E\Big(\big[x + \sigma (W_T-W_t)\big]^2 \Big)\\ &=x^2 + (T-t)\sigma^2. \end{align*}

Copied from comment:

Here, the Markov property is not explicitly employed. However, only with the Markov property, we can convert the conditional expectation w.r.t. $\mathscr F_t$ as the conditional expectation w.r.t. $X_t$, and can express the expectation as a function of $X_t$, which can then lead the solution by the Feyman-Kac formula. See the proof in Section 6.4 of the book Stochastic Calculus for Finance II by Shreve.

  • $\begingroup$ So the Markov property is unnecessary? That's my concern $\endgroup$
    – BCLC
    Commented Dec 13, 2015 at 13:12
  • $\begingroup$ The Markov property is already used - the $\mathscr{F}_t$ is replaced by $X_t$. $\endgroup$
    – Gordon
    Commented Dec 13, 2015 at 14:24
  • $\begingroup$ What do you mean Gordon? Which part has the replacement? $\endgroup$
    – BCLC
    Commented Dec 13, 2015 at 14:28
  • $\begingroup$ This is a big topic. The book by Karatzas and Shreve has a good explanation. $\endgroup$
    – Gordon
    Commented Dec 13, 2015 at 14:49
  • 1
    $\begingroup$ Thanks for providing more background information. I also need more systematic learning. $\endgroup$
    – Gordon
    Commented Dec 13, 2015 at 15:44

Your Answer

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

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