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Feynman Kac

This page has two goals:

  1. practice solving PDEs using the probabilistic representation
  2. make explicit where the forward equation sits inside Feynman–Kac

Application Heat

Solve

\[ V_t + \frac12\sigma^2 V_{xx}=0,\qquad V(T,x)=x^2. \]

Take the process \(dX_s=\sigma\,dW_s\). By Feynman–Kac (no discount, no running payoff),

\[ V(t,x)=\mathbb E_{t,x}[X_T^2]. \]

Since \(X_T=x+\sigma(W_T-W_t)\),

\[ V(t,x)=x^2+\sigma^2(T-t). \]

Exercise: verify directly that this satisfies the PDE and terminal condition.

Application heat

Solve

\[ V_t + \mu V_x + \frac12\sigma^2 V_{xx}=0,\qquad V(T,x)=x^2. \]

Take \(dX_s=\mu\,ds+\sigma\,dW_s\). Then \(X_T=x+\mu(T-t)+\sigma(W_T-W_t)\), so

\[ V(t,x)=\mathbb E[X_T^2]=\big(x+\mu(T-t)\big)^2+\sigma^2(T-t). \]

Where forward

Let \(p(x,t;y,T)\) be the transition density of \(X\). Then (in the simplest case)

\[ u(x,t)=\mathbb E[f(X_T)\mid X_t=x]=\int f(y)\,p(x,t;y,T)\,dy. \]
  • For fixed \((x,t)\), as a function of \((y,T)\), the density \(p\) satisfies the Kolmogorov forward equation.
  • For fixed \((y,T)\), as a function of \((x,t)\), the same \(p\) satisfies the Kolmogorov backward equation.

So even when you “solve the backward PDE for \(u\),” the representation

\[ u(x,t)=\int f(y)\,p(x,t;y,T)\,dy \]

quietly contains the forward evolution through \(p\).

Summary

  • Backward PDE: differential law for \(u\)
  • Forward PDE: differential law for \(p\)
  • Feynman–Kac: the bridge \(u = \int f\,p\)

Exercises

Exercise 1. Solve the PDE \(V_t + \frac{1}{2}\sigma^2 V_{xx} = 0\) with terminal condition \(V(T, x) = x^3\) using the Feynman-Kac representation. (Hint: if \(X_T = x + \sigma(W_T - W_t)\), compute \(\mathbb{E}[X_T^3]\) using the moments of the normal distribution.)

Solution to Exercise 1

With \(dX_s = \sigma\,dW_s\) and \(X_t = x\), the terminal value is \(X_T = x + \sigma(W_T - W_t)\). Let \(Z = W_T - W_t \sim N(0, T - t)\).

Using the Feynman-Kac representation with \(r = 0\), \(f = 0\):

\[ V(t, x) = \mathbb{E}[X_T^3 \mid X_t = x] = \mathbb{E}[(x + \sigma Z)^3] \]

Expanding the cube:

\[ (x + \sigma Z)^3 = x^3 + 3x^2(\sigma Z) + 3x(\sigma Z)^2 + (\sigma Z)^3 \]

Taking expectations and using \(\mathbb{E}[Z] = 0\), \(\mathbb{E}[Z^2] = T - t\), \(\mathbb{E}[Z^3] = 0\) (odd moments of a Gaussian vanish):

\[ V(t, x) = x^3 + 3x\sigma^2(T - t) \]

Verification: We check that \(V_t + \frac{1}{2}\sigma^2 V_{xx} = 0\). We have \(V_t = -3x\sigma^2\), \(V_x = 3x^2 + 3\sigma^2(T-t)\), and \(V_{xx} = 6x\). Therefore:

\[ V_t + \frac{1}{2}\sigma^2 V_{xx} = -3x\sigma^2 + \frac{1}{2}\sigma^2(6x) = -3x\sigma^2 + 3x\sigma^2 = 0 \;\checkmark \]

Terminal condition: \(V(T, x) = x^3 + 0 = x^3\). \(\checkmark\)

Exercise 2. Solve \(V_t + \mu V_x + \frac{1}{2}\sigma^2 V_{xx} - rV = 0\) with \(V(T, x) = x\) using the discounted Feynman-Kac formula. Verify your answer satisfies the PDE by direct substitution.

Solution to Exercise 2

The PDE is \(V_t + \mu V_x + \frac{1}{2}\sigma^2 V_{xx} - rV = 0\) with \(V(T, x) = x\).

By the discounted Feynman-Kac formula with \(dX_s = \mu\,ds + \sigma\,dW_s\):

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

Since \(X_T = x + \mu(T - t) + \sigma(W_T - W_t)\), we have \(\mathbb{E}[X_T \mid X_t = x] = x + \mu(T - t)\), so:

\[ V(t, x) = e^{-r(T-t)}(x + \mu(T - t)) \]

Verification by substitution: Let \(\tau = T - t\). Then \(V = e^{-r\tau}(x + \mu\tau)\).

\[ V_t = re^{-r\tau}(x + \mu\tau) - e^{-r\tau}\mu = rV - \mu e^{-r\tau} \]
\[ V_x = e^{-r\tau}, \quad V_{xx} = 0 \]

Substituting:

\[ V_t + \mu V_x + \frac{1}{2}\sigma^2 V_{xx} - rV = rV - \mu e^{-r\tau} + \mu e^{-r\tau} + 0 - rV = 0 \;\checkmark \]

Terminal condition: \(V(T, x) = e^0(x + 0) = x\). \(\checkmark\)

Exercise 3. Consider the PDE \(V_t + \frac{1}{2}\sigma^2 V_{xx} = 0\) with \(V(T, x) = e^{ax}\) for constant \(a\). Use Feynman-Kac to compute \(V(t, x) = \mathbb{E}[e^{aX_T} | X_t = x]\) where \(dX_s = \sigma\,dW_s\). Verify that your answer satisfies the PDE.

Solution to Exercise 3

With \(dX_s = \sigma\,dW_s\) and \(g(x) = e^{ax}\), the Feynman-Kac formula gives:

\[ V(t, x) = \mathbb{E}[e^{aX_T} \mid X_t = x] \]

Since \(X_T = x + \sigma(W_T - W_t)\) and \(W_T - W_t \sim N(0, T - t)\):

\[ V(t, x) = \mathbb{E}\!\left[e^{a(x + \sigma Z)}\right] = e^{ax}\,\mathbb{E}\!\left[e^{a\sigma Z}\right] \]

where \(Z \sim N(0, T - t)\). Using the moment generating function \(\mathbb{E}[e^{\lambda Z}] = e^{\lambda^2(T-t)/2}\) with \(\lambda = a\sigma\):

\[ V(t, x) = e^{ax}\,e^{a^2\sigma^2(T-t)/2} = \exp\!\left(ax + \frac{1}{2}a^2\sigma^2(T - t)\right) \]

Verification: \(V_t = -\frac{1}{2}a^2\sigma^2 V\), \(V_x = aV\), \(V_{xx} = a^2 V\). Therefore:

\[ V_t + \frac{1}{2}\sigma^2 V_{xx} = -\frac{1}{2}a^2\sigma^2 V + \frac{1}{2}\sigma^2 a^2 V = 0 \;\checkmark \]

Terminal condition: \(V(T, x) = e^{ax + 0} = e^{ax}\). \(\checkmark\)

Exercise 4. Explain the relationship between the backward and forward Kolmogorov equations in the context of the Feynman-Kac formula. If \(u(x,t) = \int f(y)\,p(x,t;y,T)\,dy\), identify which equation governs \(u\) as a function of \((x,t)\) and which governs \(p\) as a function of \((y,T)\).

Solution to Exercise 4

The Feynman-Kac formula involves two functions of the transition density \(p(x, t; y, T)\):

  • As a function of \((x, t)\) (initial condition), \(p\) satisfies the Kolmogorov backward equation: \(\partial_t p + \mathcal{L}_x p = 0\), where \(\mathcal{L}_x\) acts on the \(x\) variable.
  • As a function of \((y, T)\) (terminal state and time), \(p\) satisfies the Kolmogorov forward (Fokker-Planck) equation: \(\partial_T p = \mathcal{L}_y^* p\), where \(\mathcal{L}_y^*\) is the adjoint of \(\mathcal{L}\) acting on \(y\).

The function \(u(x, t) = \int f(y)\,p(x, t; y, T)\,dy\) is governed by the backward equation as a function of \((x, t)\): since \(\mathcal{L}_x\) acts on \(p\) and \(f(y)\) does not depend on \(x\) or \(t\), the integral inherits the backward PDE \(\partial_t u + \mathcal{L}_x u = 0\).

The density \(p(x, t; y, T)\) appearing inside the integral satisfies the forward equation as a function of \((y, T)\). So even though we "solve the backward PDE for \(u\)," the probabilistic representation \(u = \int f\,p\,dy\) implicitly encodes the forward evolution of probability densities through \(p\).

Exercise 5. Solve \(V_t + \mu V_x + \frac{1}{2}\sigma^2 V_{xx} = 0\) with \(V(T, x) = \max(x, 0)\) using Feynman-Kac. Write \(V(t, x) = \mathbb{E}[\max(X_T, 0) | X_t = x]\) and express the answer in terms of the standard normal CDF \(\Phi\).

Solution to Exercise 5

With \(dX_s = \mu\,ds + \sigma\,dW_s\), \(X_t = x\), and \(g(x) = \max(x, 0)\):

\[ V(t, x) = \mathbb{E}[\max(X_T, 0) \mid X_t = x] \]

We have \(X_T = x + \mu(T - t) + \sigma\sqrt{T - t}\,Z\) where \(Z \sim N(0,1)\). Let \(m = x + \mu(T - t)\) and \(s = \sigma\sqrt{T - t}\). Then \(X_T = m + sZ\) and:

\[ V = \mathbb{E}[\max(m + sZ, 0)] = \int_{-m/s}^{\infty}(m + sz)\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz \]

Splitting the integral:

\[ V = m\,\Phi\!\left(\frac{m}{s}\right) + s\,\phi\!\left(\frac{m}{s}\right) \]

where \(\Phi\) is the standard normal CDF and \(\phi\) is the standard normal PDF. Substituting back:

\[ V(t, x) = \bigl(x + \mu(T - t)\bigr)\,\Phi\!\left(\frac{x + \mu(T - t)}{\sigma\sqrt{T - t}}\right) + \sigma\sqrt{T - t}\;\phi\!\left(\frac{x + \mu(T - t)}{\sigma\sqrt{T - t}}\right) \]

Exercise 6. Consider \(V_t + \frac{1}{2}\sigma^2 V_{xx} + f(x) = 0\) with \(V(T, x) = 0\) where \(f(x) = 1\). Use the Feynman-Kac formula with running payoff to show that \(V(t, x) = T - t\). Verify by substitution into the PDE.

Solution to Exercise 6

The PDE is \(V_t + \frac{1}{2}\sigma^2 V_{xx} + 1 = 0\) with \(V(T, x) = 0\). This has \(r = 0\), \(g(x) = 0\) (terminal), and running payoff \(f(x) = 1\).

By the Feynman-Kac formula with running payoff:

\[ V(t, x) = \mathbb{E}\!\left[\int_t^T 1\,ds \,\Big|\, X_t = x\right] = T - t \]

The expectation is simply the length of the interval \([t, T]\), since \(f = 1\) is constant and there is no discounting.

Verification: \(V(t, x) = T - t\), so \(V_t = -1\), \(V_{xx} = 0\). Then:

\[ V_t + \frac{1}{2}\sigma^2 V_{xx} + 1 = -1 + 0 + 1 = 0 \;\checkmark \]

Terminal condition: \(V(T, x) = T - T = 0\). \(\checkmark\)

Exercise 7. Explain why the Feynman-Kac formula provides two equivalent computational methods: solving the PDE (finite differences) and computing the expectation (Monte Carlo). For a one-dimensional problem on a fine grid, which method is typically more efficient? For a five-dimensional problem, which method is preferred and why?

Solution to Exercise 7

The Feynman-Kac formula establishes the equivalence \(u(t,x) = \mathbb{E}[e^{-\int r\,ds}g(X_T) \mid X_t = x]\), which is simultaneously a PDE solution and an expected value. This gives two computational methods:

  • PDE approach (finite differences): Discretize the spatial domain and march backward from the terminal condition. Cost scales as \(O(N^d)\) where \(N\) is the number of grid points per dimension and \(d\) is the number of spatial dimensions.
  • Monte Carlo approach: Simulate paths of \(X_t\) and average the discounted payoff. Cost scales as \(O(M \cdot n)\) where \(M\) is the number of paths and \(n\) is the number of time steps, with convergence rate \(O(1/\sqrt{M})\) independent of dimension.

One-dimensional problem: Finite differences are typically more efficient. With \(N\) grid points, the PDE solver has cost \(O(N)\) per time step (tridiagonal system), giving all option values across the entire grid simultaneously. Monte Carlo requires many paths to achieve comparable accuracy.

Five-dimensional problem: Monte Carlo is strongly preferred. The PDE finite difference grid requires \(O(N^5)\) points, which becomes computationally intractable even for moderate \(N\) (e.g., \(N = 100\) gives \(10^{10}\) points). Monte Carlo, by contrast, has convergence rate \(O(1/\sqrt{M})\) regardless of dimension, making it the only feasible approach. This exponential growth of PDE grid complexity with dimension is known as the curse of dimensionality.