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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

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    $\begingroup$ 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, 2015 at 3:15
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    $\begingroup$ @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, 2015 at 3:36
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    $\begingroup$ I cannot answer you at the moment, but I will try to find some reference later if time permits. $\endgroup$
    – Henry.L
    Sep 8, 2015 at 3:43
  • $\begingroup$ @Henry.L Thank you. Any guesses as of the moment? $\endgroup$
    – BCLC
    Sep 8, 2015 at 4:09
  • $\begingroup$ maybe you can provide the actual text of the question $\endgroup$
    – seanv507
    Dec 9, 2015 at 19:33

2 Answers 2

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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:

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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.

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  • $\begingroup$ 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, 2015 at 18:56
  • $\begingroup$ Alex R., edited question $\endgroup$
    – BCLC
    Dec 9, 2015 at 22:14

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