Feynman-Kac Representation for Terminal Payoffs

From a Backward Equation to a Terminal Payoff

A backward equation describes how a value at time $t$ depends on the distribution of a diffusion at a later terminal time $T$. The Feynman–Kac representation makes that connection explicit: solve the diffusion forward from the current state, evaluate a terminal payoff, and take its expected value after applying a discount along the path. The representation is useful both for interpreting PDE solutions and for computing values when the diffusion’s distribution is known.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to the filtration. For clarity, the statements below use a one-dimensional diffusion. Assume its coefficients are continuous in time and globally Lipschitz in the state variable, uniformly over $t\in[0,T]$, and satisfy a linear growth bound: for some $C<\infty$, $|\mu(t,x)|+|\sigma(t,x)|\le C(1+|x|)$. These conditions ensure a unique nonexplosive strong solution to the SDE for each initial time and state. A strong solution is constructed on the given space using the given Brownian motion; a weak solution allows the probability space and Brownian motion to be part of the construction. Pathwise uniqueness means solutions driven by the same Brownian motion agree almost surely, whereas uniqueness in law means their distributions agree.

Definition: Given an initial state $X_t=x$, the diffusion is $dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s$ for $s\in[t,T]$. Its generator acts on twice differentiable $f$ by $\mathcal{L}_s f(x)=\mu(s,x)f'(x)+\tfrac12\sigma(s,x)^2f''(x)$. A bounded function $g$ is the terminal payoff, and $c(s,x)$ is the discount rate, or potential. The accumulated discount from $t$ to $T$ is $\exp(-\int_t^T c(s,X_s)\,ds)$.

Every Ito integral below is with respect to $B$. Its integrand is assumed adapted and progressively measurable, with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, unless localization is explicitly stated. The symbol $dB_s\,dB_s=ds$ records Brownian quadratic variation; $dB_s\,ds=0$ and $ds\,ds=0$. These are Ito multiplication rules, not ordinary chain-rule rules. A Stratonovich integral, written $\int H_s\circ dB_s$, uses a different convention and is not the integral in the representation here.

The Bounded-Solution Representation

Theorem (Feynman–Kac representation): Suppose $c$ is bounded and continuous, $g$ is bounded and continuous, and $X$ satisfies the coefficient conditions above. Let $u$ be bounded, continuous on $[0,T]\times\mathbb{R}$, and of class $C^{1,2}$ on $[0,T)\times\mathbb{R}$. If $$ \partial_t u(t,x)+\mathcal{L}_t u(t,x)-c(t,x)u(t,x)=0,\qquad u(T,x)=g(x), $$ then, for $0\le t<T$, $$ u(t,x)=\mathbb{E}\left[\exp\left(-\int_t^T c(s,X_s)\,ds\right)g(X_T)\,\middle|\,X_t=x\right]. $$ The conditional notation means expectation for the diffusion started at state $x$ at time $t$. This is the standard classical Feynman–Kac theorem; its full proof is given in the later treatment of that theorem.

The formula specifies the measure: the expectation is under $\mathbb{P}$ for the diffusion as defined. In a pricing model, if the diffusion and PDE are specified under a risk-neutral measure $\mathbb{Q}$, the expectation must instead be taken under $\mathbb{Q}$, written $\mathbb{E}^{\mathbb{Q}}$. One cannot switch between $\mathbb{P}$ and $\mathbb{Q}$ without changing the model. In particular, $\mu$ is the drift and $\sigma$ the diffusion coefficient in the measure being used.

The theorem says that the bounded solution is represented by terminal data, weighted by the discount accumulated along the realized path. If $c$ is positive, paths are discounted; if $c$ is negative, the weight can grow. Boundedness of $c$ ensures the weight is finite and bounded on a fixed finite time interval. The theorem is about a bounded solution with the stated regularity and terminal condition; it does not by itself assert a representation for arbitrary generalized PDE solutions.

Two Consequences of the Representation

Proposition: If $|g(x)|\le M$ for all $x$ and $|c(t,x)|\le C$, then the represented function satisfies $|u(t,x)|\le M e^{C(T-t)}$. If $c\ge0$, the sharper bound $|u(t,x)|\le M$ holds.

Proof. For every path and every $s\in[t,T]$, the bound $|c(s,X_s)|\le C$ gives $-C(T-t)\le-\int_t^T c(s,X_s)\,ds\le C(T-t)$. Exponentiating yields a positive discount factor at most $e^{C(T-t)}$. Since $|g(X_T)|\le M$, the absolute value of their product is at most $M e^{C(T-t)}$. Taking expectation preserves this bound. If $c\ge0$, the discount factor is at most $1$, giving $|u(t,x)|\le M$. $\square$

Lemma: If $Z$ is standard normal, then for real $a,b$, $\mathbb{E}[\cos(a+bZ)]=\cos(a)e^{-b^2/2}$.

Proof. Define $\varphi(b)=\mathbb{E}[e^{ibZ}]$. Differentiation under the expectation is valid because $\mathbb{E}|Z|<\infty$. The standard normal density $\rho(z)=(2\pi)^{-1/2}e^{-z^2/2}$ satisfies $\rho'(z)=-z\rho(z)$. Integration by parts, with the boundary term zero because $\rho(z)$ decays to zero, gives $\mathbb{E}[Ze^{ibZ}]=ib\mathbb{E}[e^{ibZ}]$. Thus $\varphi'(b)=i\mathbb{E}[Ze^{ibZ}]=-b\varphi(b)$, and $\varphi(0)=1$. Solving this scalar differential equation gives $\varphi(b)=e^{-b^2/2}$. Taking the real part of $e^{ia}\varphi(b)$ proves the stated identity. $\square$

Worked Representations

Worked Example: Brownian Motion with Drift and Discount

Let $dX_s=ds+2\,dB_s$, let $T=1$, and take terminal payoff $g(x)=\cos x$ and constant discount rate $c=\tfrac12$. Find $u(t,x)$, and check the PDE and terminal value.

Over a time interval of length $h=T-t$, $X_T=x+h+2(B_T-B_t)$. The Brownian increment has the same distribution as $\sqrt{h}Z$, so the random part has the same distribution as $2\sqrt{h}Z$. The normal-cosine lemma gives $$ \mathbb{E}[\cos(X_T)\mid X_t=x] =\cos(x+h)e^{-(2\sqrt{h})^2/2} =\cos(x+h)e^{-2h}. $$ The discount factor is $e^{-h/2}$. Therefore $$ u(t,x)=e^{-5h/2}\cos(x+h),\qquad h=1-t. $$

To check the equation, write $\lambda=\tfrac52$ and $\theta=x+h$. Then $u_t=\lambda e^{-\lambda h}\cos\theta+e^{-\lambda h}\sin\theta$, $u_x=-e^{-\lambda h}\sin\theta$, and $u_{xx}=-e^{-\lambda h}\cos\theta=-u$. Here $\mathcal{L}u=u_x+2u_{xx}=-e^{-\lambda h}\sin\theta-2u$, while $-cu=-\tfrac12u$. Since $\lambda=\tfrac52$, the terms in $\partial_tu+\mathcal{L}u-cu$ cancel. At $t=1$, $h=0$, so $u(1,x)=\cos x=g(x)$.

Check: At $t=0,x=0$, the formula gives $u(0,0)=e^{-5/2}\cos(1)$. Directly, $X_1=1+2Z$; the payoff expectation is $e^{-2}\cos(1)$, and the discount is $e^{-1/2}$. Their product is exactly $e^{-5/2}\cos(1)$.

Worked Example: Ornstein–Uhlenbeck Diffusion

Let $dX_s=-X_s\,ds+dB_s$, with terminal time $T=\ln 2$, initial state $X_0=0$, terminal payoff $g(x)=\cos x$, and constant discount rate $c=\tfrac14$. Compute the represented value at time $0$.

The explicit solution at time $h$ is $X_h=e^{-h}X_0+\int_0^h e^{-(h-s)}\,dB_s$. The integrand is deterministic and progressively measurable, and its squared integral is $\int_0^h e^{-2(h-s)}\,ds<\infty$, so this Ito integral is in $\mathcal{L}^2$. At $X_0=0$, the solution is centered normal with variance $$ v(h)=\int_0^h e^{-2(h-s)}\,ds=\int_0^h e^{-2r}\,dr=\frac{1-e^{-2h}}{2}. $$ At $h=\ln2$, $e^{-2h}=e^{-\ln4}=\tfrac14$, hence $v(h)=\tfrac{1-\frac14}{2}=\tfrac38$. The normal-cosine lemma gives $\mathbb{E}[\cos(X_h)]=e^{-v(h)/2}=e^{-3/16}$. The discount is $e^{-h/4}=e^{-(\ln2)/4}=2^{-1/4}$. Thus the represented value is $2^{-1/4}e^{-3/16}$.

Check: The state distribution used above has mean $0$, so there is no phase shift in the cosine expectation. The variance calculation gives $\tfrac38$, and the exponent in the cosine expectation is exactly $-\tfrac12\cdot\tfrac38=-\tfrac3{16}$. Multiplying by the independently computed discount gives the stated value.

Worked Example: A Bounded Payoff of Geometric Brownian Motion

Suppose $S_0=1$ and $dS_s=S_s\,ds+\sqrt{2}S_s\,dB_s$ on $[0,1]$. Take $g(s)=\cos(\log s)$ for $s>0$ and constant discount rate $c=\tfrac12$. Find the represented value at time $0$.

First compute the logarithm carefully. For $f(s)=\log s$, $f'(s)=1/s$, $f''(s)=-1/s^2$. Ito’s formula, including its second-derivative correction, gives $$ d\log S_s=\frac{1}{S_s}dS_s+\frac12\left(-\frac{1}{S_s^2}\right)(dS_s)^2 =\left(1-\frac12\cdot2\right)ds+\sqrt{2}\,dB_s =\sqrt{2}\,dB_s. $$ Here $(dS_s)^2=2S_s^2\,ds$, from $dB_s\,dB_s=ds$, $dB_s\,ds=0$, and $ds\,ds=0$. Therefore $\log S_1=\sqrt{2}B_1$, which is normal with mean $0$ and variance $2$. The normal-cosine lemma gives $\mathbb{E}[\cos(\log S_1)]=e^{-2/2}=e^{-1}$. The discount over one time unit is $e^{-1/2}$, so the value is $e^{-3/2}$.

Check: The payoff is bounded in absolute value by $1$, and the positive stock process keeps $\log S_1$ defined. The computed expectation $e^{-1}$ lies between $-1$ and $1$; multiplying by the discount $e^{-1/2}$ gives $e^{-3/2}$, consistent with the bound.

Common Pitfall: Forgetting the Measure or the Discount

A common error is to report only $\mathbb{E}[g(X_T)\mid X_t=x]$ when the PDE contains the term $-cu$. That omits the pathwise discount. For the first example, this wrong calculation gives $e^{-2}\cos(1)$, while the PDE includes $c=\tfrac12$ and the correct value is $e^{-1/2}e^{-2}\cos(1)=e^{-5/2}\cos(1)$. The missing factor changes the answer even though the diffusion distribution and terminal payoff are unchanged.

Another error is to apply the representation under a convenient measure without checking which measure defines the SDE and PDE. An expectation under $\mathbb{P}$ is not automatically an expectation under the risk-neutral measure $\mathbb{Q}$. Likewise, omitting the Ito correction when transforming a diffusion can change its drift, as the logarithm calculation above illustrates. The drift and diffusion coefficient, the measure, the terminal payoff, and the discount must all match the PDE being represented.

Check Your Understanding

Use the representation and hypotheses above to answer the following questions.

For constant $c\ge0$ and $|g|\le M$, explain why the represented value has absolute value at most $M$.
For $dX_s=\mu\,ds+\sigma\,dB_s$ and payoff $g(x)=\cos(kx)$, write the conditional terminal expectation over an interval of length $h$.
In the Ornstein–Uhlenbeck example, recompute the variance at $h=\ln 2$ from its defining integral.
For the geometric Brownian motion example, identify the Ito second-derivative correction in $d\log S_s$ and explain why it cancels the drift.
What changes in the representation if the diffusion and PDE are specified under $\mathbb{Q}$, rather than $\mathbb{P}$?