How the Feynman–Kac Equation Connects Stochastic Processes and Pricing

# Is there an intuitive explanation for the Feynman-Kac-Theorem?


# Is there an intuitive explanation for the Feynman-Kac-Theorem?












The Feynman-Kac theorem states that for an Ito-process of the form $$dX_t = \mu(t, X_t)dt + \sigma(t, X_t)dW_t$$ there is a measurable function $g$ such that $$g_t(t,x) + g_x(t, x) \mu(t,x) + \frac{1}{2} g_{xx}(t,x)\sigma(t,x)^2 = 0$$ with an appropriate boundary condition $h$: $g(T,x) = h(x)$. We also know that $g(t,x)$ is of the form $$g(t,x)=\mathbb{E}\left[h(X_T) \big| X_t=x\right].$$

This means that I can price an option with payoff function $h(x)$ at $T$ by solving the differential equation without regard to the stochastic process.

Is there an intuitive explanation how it is possible to model the stochastic behaviour of the Ito-process by a differential equation, even though the differential equation does not have a stochastic component?

## Answer by user25064 (score 19, accepted)

https://quant.stackexchange.com/a/10367

Martingales + Markovian

Here is the motivation. Conditional expectations are martingales by the tower property of conditional expectations (an easy exercise to show). Suppose $r=0$, by the risk neutral pricing theorem $E^\star\left[h(X_T)\bigg|\mathscr{F}_t,\,X_t=x\right]$ is the price of any derivative security with $X$ as the underlying asset and payoff function $h$ assuming for the moment that the underlying security and the derivative itself pay no intermediate cashflows. In a Markovian setting, it must be the case that the price of the derivative is a measurable function of the current asset price and the time to maturity only, say a function $g(t, x)$. Then, by Ito's lemma $d(g(t, x))=\ldots$. Because $g$ is a (shifted) martingale, the drift term must be equal to zero. The boundary condition comes from no arbitrage, see this by noticing what is $g(T, x)$ from the definition given at first (remember measurability when taking conditional expectation).

## Answer by Probilitator (score 9)

https://quant.stackexchange.com/a/10360

The Feynman-Kac theorem primarily makes sense in a pricing context. If you know that some function solves the Feynman-Kac equation you can represent it's soluation as an Expectation with respect to the process. (confer this document)

On the other hand a pricing function solves the FK-PDE. Thus often one would try solving the PDE to get a closed form pricing formula. (confer this document starting with page 22)

You wouldn't use the Feynman-Kac to simulate a stochastic process. On the other hand you can use a stochastic process in order to find a solution to the FK-PDE (see here)

Edit 26.02.2014: I found a document that tries to explain the connection between the transition density and the FK-PD ( see here starting with page 5)

Also there is a connection between the FK-Formula and the Sturm-Liouville equations that can be used for the decomposition of Brownian paths. (see this paper)

## Answer by Ross D (score 5)

https://quant.stackexchange.com/a/10370

The way I think of it is that the PDE describes the flow of a time dependent probability distribution. The stochastic process describes individual realisations (random walks with a drift), but if you ran a large number of them you'd build up a distribution.

The PDE says how that distribution changes in time (first term) due to deterministic drift (the second term) and diffusion (the third term, which is the link between 'lots of random walkers' and the spreading probability distribution which describes how far they've got, on average). Usually the probability distribution starts off as a delta function due to the known initial condition.

## Answer by davidhigh (score 4)

https://quant.stackexchange.com/a/37855

Let's approach this answer in two steps.

First, I find it quite intuitive, that for a given stochastic PDE there exists a deterministic PDE that evolves the density to a later time. This equation is the forward Kolmogorov or Fokker-Plank equation. Why is it intuitive? One also knows the future distribution of a Brownian motion (by definition), why should this change for a more complex stochastic term?

Second, once you got the forward equation, it's a matter of mathematics to also derive a time-reversed version of it. This is the Feynman-Kac equation, and it propagates a distribution backwards in time.