Discount Factors and Potential Terms

Discounting Along a Diffusion

A discount factor accumulated over time is determined by the rate encountered along the path. If the rate is constant, the factor over an interval of length $T-t$ is $e^{-c(T-t)}$. If the rate depends on the state of a diffusion, the corresponding factor is $e^{-\int_t^T c(X_s)\,ds}$. In the backward equation, this pathwise weighting appears as the potential term $-c(x)u(t,x)$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. All stochastic integrals below are Itô integrals. Their integrands are progressively measurable and belong to $\mathcal{L}^2$ on the stated finite time interval: that is, $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Expectations in this tutorial are under $\mathbb{P}$; no change to a risk-neutral measure $\mathbb{Q}$ is being made.

For a starting time $t$ and state $x$, let $X_s=X_s^{t,x}$, $t\le s\le T$, be the strong solution of $dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s$, with $X_t=x$. Assume $\mu$ and $\sigma$ are continuous in time, globally Lipschitz in the state uniformly over $[0,T]$, and have linear growth: for some constant $C$, $|\mu(s,x)|+|\sigma(s,x)|\le C(1+|x|)$. These conditions give a nonexplosive strong solution and finite second moments on finite time intervals. The drift is $\mu$; the diffusion coefficient is $\sigma$. Its generator on twice differentiable functions is $\mathcal{L}_s f(x)=\mu(s,x)f'(x)+\tfrac12\sigma(s,x)^2f''(x)$.

Definition: For a bounded, continuous rate $c(x)\ge0$, the discount factor accumulated by the diffusion from $t$ to $s$ is $D_{t,s}=\exp\!\left(-\int_t^s c(X_r)\,dr\right)$. The potential term associated with this rate is the zeroth-order term $-c(x)u(t,x)$ in the backward equation. The rate is evaluated at the current state inside the time integral; in general, $D_{t,s}$ is random.

The Discounted Feynman–Kac Equation

Theorem: Let $c$ be bounded, continuous, and nonnegative, and let $g$ be bounded and continuous. Suppose $u$ is a bounded classical solution, with bounded $u_x$, of $$ u_t+\mathcal{L}_t u-c(x)u=0,\qquad u(T,x)=g(x). $$ Then, for every $t\in[0,T]$ and $x\in\mathbb{R}$, $$ u(t,x)=\mathbb{E}\!\left[\left.e^{-\int_t^T c(X_s)\,ds}g(X_T)\right|X_t=x\right]. $$ In this formula the expectation means that the diffusion is started at $x$ at time $t$. Conversely, any two bounded classical solutions with bounded spatial derivative and the same terminal data agree.

Proof. Fix $t,x$, and abbreviate $D_s=D_{t,s}$. Since $c$ is bounded and continuous and $X$ has continuous paths, $D_s$ has finite variation and satisfies $dD_s=-c(X_s)D_s\,ds$, with $D_t=1$. Apply Itô’s formula to $u(s,X_s)$. The function is $u(s,y)$; its relevant derivatives are $u_t(s,y)$, $u_x(s,y)$, and $u_{xx}(s,y)$. Using $dB_s\,dB_s=ds$, $dB_s\,ds=0$, and $ds\,ds=0$, the formula is

du(s,X_s)=\left(u_t+\mu(s,X_s)u_x+\tfrac12\sigma(s,X_s)^2u_{xx}\right)(s,X_s)\,ds +\sigma(s,X_s)u_x(s,X_s)\,dB_s.

The product rule for $D_su(s,X_s)$ has no quadratic-covariation term involving $D$, because $D$ has finite variation. Thus

d\big(D_su(s,X_s)\big) =D_s\big(u_t+\mathcal{L}_s u-c(X_s)u\big)(s,X_s)\,ds +D_s\sigma(s,X_s)u_x(s,X_s)\,dB_s.

The drift is zero by the PDE. The remaining stochastic integral has integrand $H_s=D_s\sigma(s,X_s)u_x(s,X_s)$. It is progressively measurable. Since $c\ge0$, $0<D_s\le1$; since $u_x$ is bounded, $|H_s|^2\le \|u_x\|_\infty^2|\sigma(s,X_s)|^2$. The linear growth condition and the finite second moments of $X$ imply $\mathbb{E}\int_t^T|H_s|^2\,ds<\infty$. The Itô integral is therefore a true square-integrable martingale, not merely a local martingale: its conditional expectation of every future increment given the current filtration is zero. Integrating from $t$ to $T$ and taking conditional expectation given $\mathcal{F}_t$ gives $u(t,x)=\mathbb{E}[D_Tu(T,X_T)\mid\mathcal{F}_t]$. Substituting $u(T,X_T)=g(X_T)$ gives the representation. If $u_1,u_2$ are two such solutions with the same terminal data, apply the representation to their difference, whose terminal value is zero; the difference is zero at every $(t,x)$. This proves uniqueness in the stated class. $\square$

The theorem is a verification result: it proves the representation for a classical solution under the stated assumptions. Classical existence and regularity for the corresponding parabolic problem are supplied by the standard classical Feynman–Kac result from the earlier material on terminal payoffs. The calculation here identifies exactly how discounting modifies that equation.

Worked Examples

Worked Example: Constant Discount Rate

Let $X$ be Brownian motion, so $dX_s=dB_s$, and let the terminal payoff be $g(x)=1$. Take the constant rate $c=\tfrac34$ and horizon $T=2$. Compute the value $u(0,x)$ and check its equation.

The rate does not depend on the path, so the factor is $e^{-\int_0^2(3/4)\,ds}=e^{-3/2}$. Since the payoff is one, its expectation is also one, and hence $u(t,x)=e^{-(3/4)(2-t)}$. In particular, $u(0,x)=e^{-3/2}$, independent of $x$.

For the PDE check, $u_t=\tfrac34u$, $u_x=0$, and $u_{xx}=0$. Brownian motion has $\mathcal{L}u=\tfrac12u_{xx}=0$, so $u_t+\mathcal{L}u-\tfrac34u=\tfrac34u-0-\tfrac34u=0$. At $t=2$, $u(2,x)=e^0=1=g(x)$.

Check: at $t=0$, the factor computed directly is $e^{-(3/4)(2-0)}=e^{-3/2}$, matching the PDE solution. The rate is subtracted in the equation, even though the time derivative of this particular solution is positive.

Worked Example: A State-Dependent Rate Along a Deterministic Diffusion

Consider the degenerate diffusion $dX_s=1\,ds+0\,dB_s$ on the state space $x\ge0$, started at $X_t=x\ge0$. Let $c(x)=x$ and $g(x)=x$. Find the discounted value and verify the backward equation.

Write $\tau=T-t$. The path is $X_s=x+(s-t)$, so the accumulated rate is

\int_t^T c(X_s)\,ds=\int_t^T(x+s-t)\,ds=x\tau+\tfrac12\tau^2.

The terminal state is $X_T=x+\tau$. There is no randomness in this example, so the expectation is the value of the discounted payoff on this path:

u(t,x)=e^{-x\tau-\tau^2/2}(x+\tau).

To verify the PDE, set $F=e^{-x\tau-\tau^2/2}$. Then $F_\tau=-(x+\tau)F$ and $F_x=-\tau F$. Since $\tau=T-t$, differentiating gives $u_t=F((x+\tau)^2-1)$. Differentiating in $x$ gives $u_x=F(1-\tau(x+\tau))$ and $u_{xx}=F(\tau^2(x+\tau)-2\tau)$. Here $\mu=1$, $\sigma=0$, so $\mathcal{L}u=u_x$. Substitution yields

u_t+\mathcal{L}u-xu =F\big((x+\tau)^2-1+1-\tau(x+\tau)-x(x+\tau)\big)=0,

because $(x+\tau)^2-\tau(x+\tau)-x(x+\tau)=0$. At $t=T$, $\tau=0$, so $u(T,x)=x=g(x)$. This also checks the rate’s state dependence: the exponent contains $x\tau+\tfrac12\tau^2$, not the constant-rate expression $x\tau$.

Worked Example: Bounds for a Brownian State-Dependent Rate

Let $X_s=B_s$, started at $X_0=0$, let $T=2$, let $g(x)=1$, and set $c(x)=1+\tfrac12\sin^2x$. Bound the discounted value without trying to evaluate the full expectation.

For every state, $0\le\sin^2x\le1$, so $1\le c(x)\le\tfrac32$. Integrating over the interval of length $2$, every continuous sample path satisfies $2\le\int_0^2c(B_s)\,ds\le3$. Exponentiation reverses the order because $y\mapsto e^{-y}$ is decreasing. Therefore

e^{-3}\le e^{-\int_0^2c(B_s)\,ds}\le e^{-2}.

Taking expectations preserves these inequalities, giving $e^{-3}\le u(0,0)\le e^{-2}$. The associated backward equation is $u_t+\tfrac12u_{xx}-(1+\tfrac12\sin^2x)u=0$, with $u(2,x)=1$. In contrast with the constant-rate example, the factor cannot be pulled outside the expectation: it depends on the Brownian path through $B_s$.

Check: replacing $c$ by its smallest possible value $1$ gives the upper bound $e^{-(1)(2)}=e^{-2}$; replacing it by its largest possible value $\tfrac32$ gives the lower bound $e^{-(3/2)(2)}=e^{-3}$. The order of the bounds is therefore consistent with stronger discounting producing a smaller value.

Why the Potential Has a Minus Sign

The sign follows from differentiating the discount factor: $dD_s=-c(X_s)D_s\,ds$. When the product $D_su(s,X_s)$ is differentiated, this contributes $-c(X_s)u(s,X_s)\,ds$. For the product to have zero drift, the PDE must therefore contain $-cu$. This is the same cancellation used in the proof, and it does not depend on whether the rate is constant or state-dependent.

A common mistake is to write $u_t+\mathcal{L}u+cu=0$. For a constant positive rate $c$ and terminal payoff $g=1$, this incorrect sign would give $u_t+cu=0$, with $u(T)=1$, hence $u(t)=e^{c(T-t)}$. For $c=\tfrac34$, $T=2$, and $t=0$, that answer is $e^{3/2}>1$, whereas the defining discount factor is $e^{-3/2}<1$. The correction is to use $u_t+\mathcal{L}u-cu=0$, which gives $e^{-3/2}$ as required.

For $c\ge0$, the factor lies between zero and one, so it reduces the weight on later payoffs. If a bounded rate is allowed to be negative, the same PDE calculation remains valid, but the exponential can exceed one; it then represents growth weighting rather than discounting. The hypotheses and the interpretation should not be conflated.

Check Your Understanding

Use the representation and the product calculation to test your understanding of discount factors and potential terms.

For a constant rate $c$ and terminal payoff $g=1$, derive $u(t,x)$ and substitute it into the backward equation.
Explain why a state-dependent rate generally cannot be moved outside the expectation in the Feynman–Kac representation.
For the deterministic diffusion $dX_s=1\,ds$, starting from $x\ge0$, and $c(x)=x$, compute the accumulated rate over an interval of length $\tau$.
In the discounted Itô product, identify the term that forces the potential to have a negative sign.
State why the stochastic integral in the proof is a true martingale under the hypotheses, rather than only a local martingale.

Discount Factors and Potential Terms

What You'll Learn