Discounted Feynman-Kac Formula¶

The discounted Feynman-Kac formula extends the basic Feynman-Kac result by incorporating a killing or discounting term $r(t, x)$. This extension is essential for finance, where the time value of money requires discounting future cash flows. The discounting rate may be constant (as in Black-Scholes) or state-dependent (as in stochastic interest rate models), leading to a rich interplay between the PDE and probabilistic representations.

Motivation: Time Value of Money¶

In finance, a dollar received in the future is worth less than a dollar today. If the risk-free rate is $r$, then receiving $\$1$ at time $T$ is worth $e^{-r(T-t)}$ today at time $t$.

For a derivative with payoff $g(X_T)$ at maturity $T$, the present value at time $t$ is:

\[ V(t, x) = \mathbb{E}\!\left[e^{-r(T-t)}\,g(X_T) \,\Big|\, X_t = x\right] \]

When $r$ depends on the state $X_t$ (e.g., stochastic interest rates), the discount factor becomes path-dependent:

\[ V(t, x) = \mathbb{E}\!\left[e^{-\int_t^T r(s, X_s)\,ds}\,g(X_T) \,\Big|\, X_t = x\right] \]

The discounted Feynman-Kac formula tells us which PDE this expectation satisfies.

The Discounted Feynman-Kac Formula¶

Statement¶

This is the $f = 0$ special case of the general theorem. Recall (see § The Classical Feynman–Kac Formula): for $dX_s = \mu\,ds + \sigma\,dW_s$ with $X_t = x$, the function

\[ u(t, x) = \mathbb{E}\!\left[e^{-\int_t^T r(s, X_s)\,ds}\,g(X_T) \,\Big|\, X_t = x\right] \]

satisfies the discounted FK PDE

\[ \partial_t u + \mathcal{L}u - r\,u = 0, \quad u(T, \cdot) = g, \]

where $\mathcal{L} = \mu\,\partial_x + \tfrac12\sigma^2\,\partial_{xx}$.

The -ru Term: Three Equivalent Interpretations¶

The discounting appears as the zeroth-order term $-r\,u$ in the PDE — a "sink" that continuously removes value at rate $r$. This page is the canonical reference for the three interpretations:

Interpretation	Field	Description
Discounting	Finance	Future cash flows are discounted at rate $r$
Killing	Probability	The process is "killed" at exponential rate $r$
Absorption	Physics	Particles are absorbed at rate $r$

Constant Discounting¶

Recall (see § Special Cases — Case 2): when $r$ is constant, $e^{-r(T-t)}$ factors out and $u(t,x) = e^{-r(T-t)}\,\mathbb{E}[g(X_T) \mid X_t = x]$, equivalent to $u = e^{-r(T-t)}v$ where $v$ solves the undiscounted backward equation $\partial_t v + \mathcal{L}v = 0$.

State-Dependent Discounting¶

When $r = r(t, X_t)$ depends on the stochastic process, the discount factor $e^{-\int_t^T r(s, X_s)\,ds}$ is path-dependent and cannot be factored out of the expectation. Two canonical settings are worth distinguishing.

Stochastic interest rates. When $X_t = r_t$ is the short rate itself, the killing rate and the state variable coincide, and the FK expectation $u(t,r) = \mathbb{E}[e^{-\int_t^T r_s\,ds} \mid r_t = r]$ becomes the zero-coupon bond price. Recall (see § FK Formula — Application 2: Bond Pricing) for the Vasicek model, the corresponding PDE, and the affine solution $P(t,r) = A(t,T)e^{-B(t,T)r}$.

State-dependent killing. In credit risk, the default intensity $\lambda(X_t)$ depends on a credit-quality process $X_t$ that is not the killing rate. The survival probability

\[ Q(t, x) = \mathbb{E}\!\left[e^{-\int_t^T \lambda(X_s)\,ds} \,\Big|\, X_t = x\right] \]

satisfies $\partial_t Q + \mathcal{L}Q - \lambda\,Q = 0$ with $Q(T, x) = 1$. Here $\lambda$ and $X$ play distinct roles: $X$ drives the dynamics, $\lambda(X)$ controls the killing rate. This is the structural distinction missing from the bond case, where the two collapse.

The Killing Interpretation¶

The discounting term has a clean probabilistic interpretation via killing.

Killed Diffusion¶

Define a killed process $(X_t, \zeta)$ where $\zeta$ is an independent exponential killing time with rate $r(t, X_t)$. Formally:

\[ \mathbb{P}(\zeta > T \mid \text{path of } X) = e^{-\int_t^T r(s, X_s)\,ds} \]

Then:

\[ u(t, x) = \mathbb{E}[g(X_T)\,\mathbf{1}_{\{\zeta > T\}} \mid X_t = x] \]

Interpretation: The process lives until a random killing time $\zeta$. We only collect the payoff $g(X_T)$ if the process survives to maturity.

Why -ru in the PDE?¶

The $-ru$ term represents the rate of "leakage" from the system. At each instant $ds$, a fraction $r\,ds$ of the remaining probability mass is killed. The PDE tracks the expected value conditional on survival, and the $-ru$ term accounts for the continual erosion of this value.

\[ \frac{\partial u}{\partial t} = \underbrace{\mathcal{L}u}_{\text{diffusion}} - \underbrace{r\,u}_{\text{killing}} \]

Derivation of the -ru Term¶

From the Undiscounted Backward Equation¶

Start with the process $Y_s = (X_s, R_s)$ where $R_s = \int_t^s r(\tau, X_\tau)\,d\tau$ is the accumulated discount. Define:

\[ v(t, x, 0) = \mathbb{E}[e^{-R_T}\,g(X_T) \mid X_t = x, R_t = 0] \]

The function $v(t, x, \rho)$ satisfies the backward equation for the extended process:

\[ \partial_t v + \mu\,\partial_x v + \frac{1}{2}\sigma^2\,\partial_{xx}v + r\,\partial_\rho v = 0 \]

Since $v$ depends on $\rho$ only through $e^{-\rho}\,w(t, x)$ (by the multiplicative structure of the discount factor):

\[ v(t, x, \rho) = e^{-\rho}\,u(t, x) \]

Substituting: $\partial_\rho v = -e^{-\rho}\,u$, so:

\[ e^{-\rho}\!\left[\partial_t u + \mathcal{L}u - r\,u\right] = 0 \]

This gives $\partial_t u + \mathcal{L}u - r\,u = 0$. $\square$

Extension: Running Payoff¶

Adding a running payoff $f$ extends the formula to $u = \mathbb{E}[e^{-\int r\,ds}g(X_T) + \int e^{-\int r\,d\tau}f\,ds]$, satisfying $\partial_t u + \mathcal{L}u - ru + f = 0$. See Running Payoff Feynman–Kac for details.

Financial Applications¶

The discounted Feynman–Kac PDE underlies several pricing models. Full derivations:

Black–Scholes option pricing: see Feynman–Kac Option Pricing.
Bond pricing (Vasicek, CIR): see Feynman–Kac Formula § Bond Pricing.
Dividend-paying stock: replace the drift $rS$ by $(r-q)S$ while keeping the discount rate $r$.

Comparison: With and Without Discounting¶

Aspect	Undiscounted ($r = 0$)	Discounted ($r > 0$)
PDE	$\partial_t u + \mathcal{L}u = 0$	$\partial_t u + \mathcal{L}u - ru = 0$
Solution	$\mathbb{E}[g(X_T)]$	$\mathbb{E}[e^{-\int r\,ds}\,g(X_T)]$
Process	$u(s, X_s)$ is a martingale	$e^{-\int_t^s r\,d\tau}\,u(s, X_s)$ is a martingale
Max principle	$\max u$ on parabolic boundary	Still holds (with $r \geq 0$)
Steady state	$\mathcal{L}u = 0$	$\mathcal{L}u - ru = 0$
Financial name	Undiscounted expectation	Risk-neutral pricing

Summary¶

\[ \boxed{ \partial_t u + \mathcal{L}u - r\,u = 0 \quad \Longleftrightarrow \quad u(t,x) = \mathbb{E}\!\left[e^{-\int_t^T r(s,X_s)\,ds}\,g(X_T) \,\Big|\, X_t = x\right] } \]

Component	PDE Role	Probabilistic Role	Financial Role
$\mathcal{L}u$	Diffusion operator	Generator of $X_t$	Drift + volatility of underlying
$-r\,u$	Zeroth-order sink	Killing rate	Time value of money
$g(x)$	Terminal condition	Terminal payoff	Derivative payoff at maturity
$r(t,x)$	PDE coefficient	Killing intensity	Short rate / discount rate

The discounted Feynman-Kac formula is the mathematical foundation of risk-neutral pricing. The $-r\,u$ term in the PDE encodes the time value of money, transforming expected future payoffs into present values. This single extension of the Kolmogorov backward equation underlies virtually all of derivatives pricing theory.

Exercises¶

Exercise 1. For a European call option under $\mathbb{Q}$ with constant $r = 0.05$, $\sigma = 0.30$, and $S_0 = 100$, write the Black-Scholes PDE with the $-rV$ discounting term and identify each component: generator $\mathcal{L}$, discount rate, and terminal condition for strike $K = 100$.

Solution to Exercise 1

The Black-Scholes PDE with the given parameters is:

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

Substituting $r = 0.05$ and $\sigma = 0.30$:

\[ \frac{\partial V}{\partial t} + 0.05\,S\frac{\partial V}{\partial S} + \frac{1}{2}(0.09)\,S^2\frac{\partial^2 V}{\partial S^2} - 0.05\,V = 0 \]

Components:

Generator $\mathcal{L}$: Under $\mathbb{Q}$, the stock follows $dS_t = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}$, so $\mu(S) = rS = 0.05S$ and $\sigma_{\text{diff}}(S) = \sigma S = 0.30S$. The generator is $\mathcal{L}V = 0.05S\,\partial_S V + \frac{1}{2}(0.09)S^2\,\partial_{SS}V$.
Discount rate: The constant risk-free rate $r = 0.05$, appearing in the $-rV = -0.05V$ term.
Terminal condition: For a call with strike $K = 100$, we have $V(T, S) = (S - 100)^+$.

Exercise 2. Show that if $r$ is constant, the discounted Feynman-Kac expectation $u(t,x) = \mathbb{E}[e^{-r(T-t)}g(X_T) | X_t = x]$ can be written as $e^{-r(T-t)}v(t,x)$ where $v$ solves the undiscounted backward equation $\partial_t v + \mathcal{L}v = 0$. Verify by substitution that $u$ then satisfies $\partial_t u + \mathcal{L}u - ru = 0$.

Solution to Exercise 2

Let $v(t, x) = \mathbb{E}[g(X_T) \mid X_t = x]$, which solves the undiscounted backward equation $\partial_t v + \mathcal{L}v = 0$ with $v(T, x) = g(x)$.

Define $u(t, x) = e^{-r(T-t)}v(t, x)$. Since $r$ is constant, the discount factor $e^{-r(T-t)}$ does not depend on the path of $X$, so:

\[ u(t, x) = e^{-r(T-t)}\mathbb{E}[g(X_T) \mid X_t = x] = \mathbb{E}[e^{-r(T-t)}g(X_T) \mid X_t = x] \]

which is the discounted Feynman-Kac expectation.

Verification by substitution. Compute each derivative of $u = e^{-r(T-t)}v$:

\[ \partial_t u = re^{-r(T-t)}v + e^{-r(T-t)}\partial_t v = ru + e^{-r(T-t)}\partial_t v \]

\[ \partial_x u = e^{-r(T-t)}\partial_x v, \quad \partial_{xx}u = e^{-r(T-t)}\partial_{xx}v \]

Therefore:

\[ \mathcal{L}u = e^{-r(T-t)}\mathcal{L}v \]

Substituting into $\partial_t u + \mathcal{L}u - ru$:

\[ \partial_t u + \mathcal{L}u - ru = ru + e^{-r(T-t)}\partial_t v + e^{-r(T-t)}\mathcal{L}v - ru = e^{-r(T-t)}(\partial_t v + \mathcal{L}v) = 0 \]

since $v$ solves $\partial_t v + \mathcal{L}v = 0$. $\checkmark$

Exercise 3. In the Vasicek model with $\kappa = 0.5$, $\theta = 0.04$, $\sigma_r = 0.01$, and $r_0 = 0.03$, the zero-coupon bond price satisfies $\partial_t P + \kappa(\theta - r)\partial_r P + \frac{1}{2}\sigma_r^2 \partial_{rr}P - rP = 0$ with $P(T,r) = 1$. Explain why $r$ appears both in the drift coefficient $\kappa(\theta - r)$ and as the discount rate $-rP$. What makes this PDE different from the standard Black-Scholes PDE?

Solution to Exercise 3

In the Vasicek model, the short rate $r_t$ is the state variable itself. This creates a dual role for $r$:

In the drift $\kappa(\theta - r)$: The short rate is mean-reverting toward $\theta$, and the current value $r$ determines the drift of the process. This is the standard role of the state variable in the generator $\mathcal{L}$.
In the discount term $-rP$: The same variable $r$ serves as the instantaneous discounting rate. This arises because the bond price is $P(t, r) = \mathbb{E}[e^{-\int_t^T r_s\,ds} \mid r_t = r]$, and the Feynman-Kac formula produces the $-r\,P$ term.

Difference from Black-Scholes: In the Black-Scholes PDE, the discount rate $r$ is a fixed constant, independent of the state variable $S$. The state variable $S$ appears in the drift ($rS$) and the diffusion ($\sigma S$), but $r$ in the $-rV$ term is simply a parameter. In the Vasicek PDE, the discount rate is the state variable: the coefficient of $-P$ is $r$ itself, which varies stochastically. This makes the problem inherently different because the discounting is path-dependent, and the discount factor $e^{-\int_t^T r_s\,ds}$ cannot be factored out of the expectation.

Exercise 4. Consider the discounted Feynman-Kac formula with running payoff: $u(t,x) = \mathbb{E}[\int_t^T e^{-\int_t^s r\,d\tau}f(s, X_s)\,ds | X_t = x]$. This satisfies $\partial_t u + \mathcal{L}u - ru + f = 0$ with $u(T,x) = 0$. Give a financial example where a running payoff arises, and explain why the terminal condition is zero.

Solution to Exercise 4

Financial example: A coupon bond that pays continuous coupons at rate $c$ per unit time. The holder receives a stream of payments $f(s, X_s) = c$ over $[t, T]$, each discounted to present value. With no terminal payoff ($g = 0$, or equivalently with a face value $g = 1$ at maturity for a standard bond), the value is:

\[ u(t, x) = \mathbb{E}\!\left[\int_t^T e^{-\int_t^s r(\tau, X_\tau)\,d\tau}\,c\,ds \,\Big|\, X_t = x\right] \]

Other examples include: a stock paying continuous dividends at rate $qS$, where $f(s, S_s) = qS_s$; or a floating-rate note with running interest payments.

Why the terminal condition is zero: The formula $u(t,x) = \mathbb{E}[\int_t^T e^{-\int_t^s r\,d\tau}f(s, X_s)\,ds \mid X_t = x]$ represents only the running payoff component. At maturity $T$, there are no remaining future running payments to collect (the integral $\int_T^T \cdots\,ds = 0$), so $u(T, x) = 0$. If there were also a terminal payoff $g(X_T)$, it would be added separately, and the terminal condition would be $u(T, x) = g(x)$.

Exercise 5. In the killing interpretation, a process $X_t$ is killed at an exponential rate $r(t,X_t)$. Write the survival probability $\mathbb{P}(\zeta > T | \text{path of } X) = e^{-\int_t^T r(s,X_s)\,ds}$. If $r$ is constant at $r = 0.05$ and $T - t = 10$, compute the survival probability. Explain the financial interpretation in terms of discounting.

Solution to Exercise 5

With constant killing rate $r = 0.05$ and time horizon $T - t = 10$:

\[ \mathbb{P}(\zeta > T \mid \text{path of } X) = e^{-\int_t^T r\,ds} = e^{-0.05 \times 10} = e^{-0.5} \approx 0.6065 \]

So approximately $60.65\%$ of the probability mass survives to time $T$.

Financial interpretation: The survival probability $e^{-r(T-t)}$ is exactly the discount factor used to compute present values. A payoff of $\$1$ received at time $T$ is worth $e^{-0.5} \approx \$0.6065$ today. In the killing interpretation, we imagine the process being "killed" at random with intensity $r = 0.05$ per year. After 10 years, the probability that the process has not been killed is $e^{-0.5}$. The expected payoff, conditional on survival, times the survival probability gives the discounted expected value -- this is exactly how risk-neutral pricing works.

Exercise 6. For a dividend-paying stock with continuous yield $q$ under $\mathbb{Q}$: $dS_t = (r - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}$. Write the PDE for an option price $V(t,S)$ including the $-rV$ discounting term. Identify how $q$ affects the drift but not the discount rate.

Solution to Exercise 6

Under $\mathbb{Q}$, the dividend-paying stock follows $dS_t = (r - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}$, so the drift is $\mu(S) = (r - q)S$ and the diffusion coefficient is $\sigma_{\text{diff}}(S) = \sigma S$.

Applying the discounted Feynman-Kac formula with discount rate $r$, the PDE for $V(t, S)$ is:

\[ \frac{\partial V}{\partial t} + (r - q)S\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

How $q$ affects the PDE: The dividend yield $q$ enters only through the drift term $(r - q)S\,\partial_S V$. Compared to the non-dividend case (where the drift coefficient is $rS$), the effective risk-neutral growth rate is reduced from $r$ to $r - q$. Intuitively, dividends "leak" value from the stock at rate $q$, so the capital appreciation under $\mathbb{Q}$ is slower.

The discount rate is unchanged: The $-rV$ term remains at rate $r$, not $r - q$. This is because discounting reflects the time value of money (the risk-free rate), which is independent of whether the stock pays dividends. The dividend yield affects the dynamics of the underlying but not the rate at which future payoffs are discounted.

Exercise 7. Derive the $-ru$ term in the discounted Feynman-Kac PDE by considering the extended process $Y_s = (X_s, R_s)$ where $R_s = \int_t^s r(\tau, X_\tau)\,d\tau$. Show that $v(t, x, \rho) = e^{-\rho}u(t,x)$ satisfies $\partial_\rho v = -e^{-\rho}u$, and use this to derive $\partial_t u + \mathcal{L}u - ru = 0$ from the backward equation for the extended process.

Solution to Exercise 7

Consider the extended process $Y_s = (X_s, R_s)$ where $dX_s = \mu(s, X_s)\,ds + \sigma(s, X_s)\,dW_s$ and $dR_s = r(s, X_s)\,ds$ with $R_t = 0$. The function:

\[ v(t, x, \rho) = \mathbb{E}\!\left[e^{-R_T}\,g(X_T) \mid X_t = x, R_t = \rho\right] \]

satisfies the backward equation for the joint process $(X_s, R_s)$:

\[ \partial_t v + \mu\,\partial_x v + \frac{1}{2}\sigma^2\,\partial_{xx}v + r\,\partial_\rho v = 0 \]

The term $r\,\partial_\rho v$ arises because $R_s$ has drift $r(s, X_s)$ and zero diffusion.

Now, by the multiplicative structure of $e^{-R_T} = e^{-\rho}\,e^{-\int_t^T r\,ds}$ (since $R_T = \rho + \int_t^T r(s, X_s)\,ds$), we can write:

\[ v(t, x, \rho) = e^{-\rho}\,u(t, x) \]

where $u(t, x) = \mathbb{E}[e^{-\int_t^T r\,ds}\,g(X_T) \mid X_t = x]$.

Computing the partial derivative: $\partial_\rho v = -e^{-\rho}\,u$.

Substituting into the backward equation:

\[ e^{-\rho}\,\partial_t u + e^{-\rho}\,\mu\,\partial_x u + \frac{1}{2}e^{-\rho}\,\sigma^2\,\partial_{xx}u + r\,(-e^{-\rho}\,u) = 0 \]

Dividing through by $e^{-\rho} > 0$:

\[ \partial_t u + \mu\,\partial_x u + \frac{1}{2}\sigma^2\,\partial_{xx}u - r\,u = 0 \]

which is $\partial_t u + \mathcal{L}u - r\,u = 0$. $\square$

Aspect	Undiscounted (\(r = 0\))	Discounted (\(r > 0\))
PDE	\(\partial_t u + \mathcal{L}u = 0\)	\(\partial_t u + \mathcal{L}u - ru = 0\)
Solution	\(\mathbb{E}[g(X_T)]\)	\(\mathbb{E}[e^{-\int r\,ds}\,g(X_T)]\)
Process	\(u(s, X_s)\) is a martingale	\(e^{-\int_t^s r\,d\tau}\,u(s, X_s)\) is a martingale
Max principle	\(\max u\) on parabolic boundary	Still holds (with \(r \geq 0\))
Steady state	\(\mathcal{L}u = 0\)	\(\mathcal{L}u - ru = 0\)
Financial name	Undiscounted expectation	Risk-neutral pricing

Component	PDE Role	Probabilistic Role	Financial Role
\(\mathcal{L}u\)	Diffusion operator	Generator of \(X_t\)	Drift + volatility of underlying
\(-r\,u\)	Zeroth-order sink	Killing rate	Time value of money
\(g(x)\)	Terminal condition	Terminal payoff	Derivative payoff at maturity
\(r(t,x)\)	PDE coefficient	Killing intensity	Short rate / discount rate