Feynman Kac and discounted expectations

Definition Let $X$ be an $n$-dimensional diffusion

under a stated probability measure, where $W$ is a $d$-dimensional Brownian motion under that measure, $b$ takes values in $\mathbb{R}^n$, $\Sigma$ is an $n \times d$ loading matrix and $a = \Sigma\Sigma^{\top}$. Its generator acts on a function $\psi$ with continuous second derivatives by

Given a discount rate $c(t, x)$ and a running payment rate $f(t, x)$, the backward PDE for a function $u(t, x)$ is

on $[0, T) \times \mathbb{R}^n$. It is called backward because, with $a$ positive semidefinite, it is solved from data at the final time $T$ toward earlier times, and in the remaining time $\tau = T - t$ it becomes a forward parabolic equation. For continuous coefficients and a function $\psi$ with continuous second derivatives and compact support, Itô’s formula shows that

as $h \downarrow 0$, so the generator is the expected rate of change of $\psi(X)$. A reverse route explains the shape of the PDE. If $u(t, x)$ is an expected discounted cash flow, conditioning on $X_{t+h}$ with the Markov property and expanding the first short step with Itô’s formula gives $\partial_t u + \mathcal{L}_t u - c\,u + f = 0$ in the limit $h \to 0$, provided $u$ is smooth enough for the expansion. That proviso is a separate regularity result, so this route explains the equation rather than proving it.

Definition A terminal value problem consists of the backward PDE and a terminal condition,

the first for $0 \le t < T$ and $x \in \mathbb{R}^n$ and the second for every $x$, to be solved in a stated solution class. In pricing, $g$ is the payoff at maturity, $f$ the rate of a payment received continuously before maturity and $c$ the short rate as a function of the state. The data are given at the final time only. On the whole space no spatial boundary condition is imposed, and its role is taken by the growth condition of the solution class, without which the problem can have more than one solution. For a positive state such as a GBM price, $\mathbb{R}^n$ is replaced by $(0, \infty)$. The problem is linear, so the solutions for several payoffs and payment streams add.

Definition A classical solution of the terminal value problem is a function $u$ that is continuous on $[0, T] \times \mathbb{R}^n$, has continuous partial derivatives $\partial_t u$, $\partial_i u$ and $\partial_{ij} u$ on $[0, T) \times \mathbb{R}^n$, satisfies the PDE at every point of $[0, T) \times \mathbb{R}^n$ and equals $g$ at $t = T$. In short,

No derivative is required at $t = T$, so a payoff with a kink is allowed, but the solution must be continuous up to maturity. This is the regularity that lets Itô’s formula be applied to $u(s, X_s)$ for $s < T$. Whether a classical solution exists for given data is a separate question. Existence theorems need conditions such as a uniformly positive definite matrix $a$ and sufficiently regular coefficients, and degenerate or nonsmooth problems are handled with viscosity solutions. Both lie beyond this lesson.

Definition Suppose the coefficients $b$ and $\Sigma$ are continuous, globally Lipschitz in $x$ uniformly in $t$, and of linear growth, $\lvert b(t, x) \rvert + \lvert \Sigma(t, x) \rvert \le A(1 + \lvert x \rvert)$. Then for every starting point $(t, x)$ the equation has a unique strong solution $X^{t,x}$ on $[t, T]$, and for every $m \ge 1$ there is a constant $C_m$ with

The controlled-growth class consists of the classical solutions with $\lvert u(t, x) \rvert \le A(1 + \lvert x \rvert^m)$ for some constants $A$ and $m$ and all $(t, x) \in [0, T] \times \mathbb{R}^n$, for problems with a bounded continuous discount rate $c$ and a continuous running payment with $\lvert f(t, x) \rvert \le A(1 + \lvert x \rvert^m)$. In this class the moment bound makes the discounted solution along the path, together with the accumulated running payments, dominated by an integrable random variable, which is the step the Feynman-Kac proof needs. Without a growth restriction uniqueness fails even for the heat equation. Tychonoff constructed a solution of $\partial_\tau v = \partial_{xx} v$ that vanishes at $\tau = 0$, is not identically zero and is not bounded by $M e^{a x^2}$ for any constants $M$ and $a$. For a positive state such as a GBM price on $(0, \infty)$, a natural bound is $A(1 + x^m + x^{-m})$, and the needed moments come from the finite power moments of the running maximum and minimum of the price.

Definition For $0 \le t \le s \le T$ the pathwise discount factor is

The discounted conditional expectation of a terminal payoff $g(X_T)$ and a running payment at the rate $f(s, X_s)$ is

computed under the measure in which $X$ has the stated dynamics, whenever the expectation is finite. For a Markov diffusion it is a function of $t$ and $X_t$ only. When $c$ depends on the state, $D_{t,T}$ is random and must stay inside the expectation. By Jensen’s inequality its conditional mean is at least the exponential of minus the conditional mean of $\int_t^T c(v, X_v)\,dv$, with equality only when that integral is known at time $t$. With a constant rate $c = r$ the factor leaves the expectation and the expression is $e^{-r(T - t)}\,\mathbb{E}[g(X_T) \mid \mathcal{F}_t]$ plus the discounted running payments. With the money-market account $B_t = \exp\bigl(\int_0^t c(v, X_v)\,dv\bigr)$ one has $D_{t,s} = B_t/B_s$, so under a pricing measure $\mathbb{Q}$ the expression is the arbitrage-free price of the cash flows with the money-market numeraire. The measure is part of the data. For a dividend-free stock, replication produced the generator with drift $r x$, which is the generator under $\mathbb{Q}$. Putting the physical drift $\mu x$ into the generator gives a different PDE and a different expectation, not the replication price.

Definition Assume that $b$ and $\Sigma$ satisfy the Lipschitz and linear growth conditions of the controlled-growth card under a stated measure, that $c$ is continuous and bounded, and that $f$ and $g$ are continuous with polynomial growth. Let $u$ be a classical solution of the terminal value problem in the controlled-growth class. Then for every $(t, x) \in [0, T] \times \mathbb{R}^n$,

where $\mathbb{E}_{t,x}$ is the expectation under that measure for the solution started at $X_t = x$. In particular the problem has at most one solution in the class. For a solution started at an earlier time, $u(t, X_t)$ equals the discounted conditional expectation given $\mathcal{F}_t$. To prove the formula, let

for $t \le s < T$. The factor $D_{t,s}$ has finite variation with $dD_{t,s} = -c(s, X_s) D_{t,s}\,ds$, so the product rule has no covariation term, and Itô’s formula with the PDE gives

with the functions evaluated at $(s, X_s)$. So $M$ is a local martingale on $[t, T)$. Stopping at the first time after $t$ that $\lvert X_s \rvert$ reaches $n$, or at $T - 1/n$ if that comes first, keeps the integrand bounded, so the stopped integral is a true martingale and $u(t, x)$ equals the expectation of the stopped value of $M$. If $\lvert c \rvert \le \bar c$, the growth bounds dominate every stopped value by

which is integrable by the moment bound. As $n \to \infty$ the stopping times tend to $T$, the continuity of $u$ up to $T$ carries the stopped values to $M_T$, and dominated convergence gives the formula. For uniqueness, two solutions in the class with the same $g$ and $f$ differ by a solution with zero data, whose representation is zero. A local martingale alone would give the stopped identity but not the limit, since without domination or uniform integrability the mean of $M_T$ can differ from $M_t$.