Solving a backwards heat equation using stochastic calculus

Question

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 $\mathscr F_t = \mathscr F_t^W$.

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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?

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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

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*}

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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.

Answer 2

• 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.

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From Shreve's Stochastic Calculus for Finance:

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