One way is to appeal to the Feynman-Kac formula, which states in this case, that under certain regularity conditions¹¹ The standard assumption is that $\psi$ is twice continuously-differentiable and $\mathbb{E}^{\mathbb{Q}}[\lvert\psi(X(T))\rvert]<\infty$. However, for the particular PDE we consider, these assumptions can be relaxed. on the function $\psi\colon(0,\infty)\to\mathbb{R}$, the function $f\colon[0,T]\times(0,\infty)\to\mathbb{R}$ defined by

satisfies the partial differential equation (1) with the terminal condition

(The stochastic process $X(t)$ is to satisfy equation (5).)

Now, equations (6) and (2) look quite similar, but with an important difference: the conditional expectation on equation (6) is with respect to the event $\{X(t)=x\}$, while equation (2) is conditioning on the filtration $\mathcal{F}_{t}$. We claim that these are actually the same thing — that is, for fixed $t$, the random variable $\mathbb{E}^{\mathbb{Q}}[V(T)\mid\mathcal{F}_{t}]$ can be written as a function of $X(t)$ — provided that the terminal condition is itself a function of $X(T)$:

We will demonstrate this claim later. Assuming that the claim holds, we may thus set $V(t)=f(t,X(t))$, and the Black-Scholes partial differential equation follows from the Feynman-Kac formula as explained earlier.

Formally, any $\mathcal{F}_{t}$-measurable stochastic process $X(t)$ that ensures $\mathbb{E}[\psi(X(T))\mid\mathcal{F}_{t}]$ is always a measurable function of $X(t)$, for any Borel-measurable $\psi$, is called a Markov process, or is said to have the Markov property. Thus we want to show that the stock process $X(t)$ has the Markov property. For this, we will need the following explicit solution for it at time $T$ in terms of an initial condition at time $t$:

Then

Since $X(t)$ is $\mathcal{F}_{t}$-measurable, it may be treated as a constant $x$ while taking conditional expectations with respect to $\mathcal{F}_{t}$. And $\widetilde{W}(T)-\widetilde{W}(t)$ is a random variable independent of $\mathcal{F}_{t}$ (and hence of $X(t)$). Therefore the expression that appears on the right-hand side above is a function of $x=X(t)$, and would be unchanged if the conditioning is changed from $\mathcal{F}_{t}$ to $X(t)$.

There is also a more direct method of deriving equation (1) using Itō’s formula for Itō processes.

Set $V(t)=f(t,X(t))$ as before. Make the assumption that $f$ is twice continuously differentiable, to be checked later; then we can apply Itō’s formula to expand $dV(t)$: (partial derivatives evaluated at $x=X(t)$)

By theorems on the uniqueness of solutions to stochastic differential equations, the coefficients of the correponding $dt$ and $dX(t)$ terms in equations (4) and (7) must be equal almost surely. Equating the coefficients, we find:

Equation (9) is essentially equation (1), except that $x$ has been replaced by $X(t)$ in some places. However, because $X(t)$ is a random variable that takes on all values on $(0,\infty)$, and $f$ and its derivatives are assumed to be continuous, equation (9) must hold for arbitrary values $x$ substituted for $X(t)$. Hence we obtain equation (1).

To verify the initial assumption that $f$ is twice continuously differentiable, we write formula (2) in a more explicit form (using the Markov property for $X(t)$ as before):

where $p(x,y,\tau)$ is the transition density of the stock process from $x$ to $y$ over a time interval of length $\tau$. In fact, from the solution for $X(t)$, the density $p(x,\cdot,\tau)$ is the density for the log-normal random variable whose logarithm has mean $\log x+(r-\tfrac{1}{2}\sigma^{2})\tau$ and variance $\sigma^{2}\tau$. By differentiation under the integral sign, we see that $f(t,x)$ must be twice continously differentiable for $0\leq t<T$ and $0<x<\infty$.