Conditional Expectations and the Markov Property in Feynman–Kac
Summary
The document examines a backward heat equation with a quadratic terminal condition and asks why the Markov property is relevant when evaluating a conditional expectation. For a process driven by Brownian motion with constant volatility, the terminal value is the current state plus a scaled Brownian increment. The increment is independent of the information available at the current time, so its conditional moments give a solution that depends on the current state and elapsed time.
The replies distinguish calculating this expectation from proving the Feynman–Kac result. The calculation can use independence of future Brownian increments directly. The Markov property explains why conditioning on the current state can replace conditioning on the full information set, and it supports the broader argument that the conditional expectation defines a function of time and state satisfying the PDE. The discussion is conceptual rather than a full proof, and the source excerpts leave the general theorem’s assumptions unstated.
Key ideas
- Future Brownian increments are independent of the information available at the current time.
- For this constant-volatility process, the conditional second moment yields the quadratic solution plus a variance term.
- The Markov property connects conditioning on the full information set with conditioning on the current state.
- The direct expectation calculation does not itself require proving the Markov property.
- A general Feynman–Kac proof uses the process’s Markov structure to establish the PDE relationship.
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Full text
# Solving a backwards heat equation using stochastic calculus
# Solving a backwards heat equation using stochastic calculus
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
## Answer by Gordon (score 1, accepted)
https://quant.stackexchange.com/a/22271
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.
## Answer by BCLC (score 0)
https://quant.stackexchange.com/a/80256
- 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:Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.