The SDE–PDE Bridge

This chapter explores the profound connection between stochastic differential equations (SDEs) and partial differential equations (PDEs). The infinitesimal generator serves as the bridge, transforming questions about random processes into deterministic equations.

The Central Theme

Expected values of diffusion processes satisfy partial differential equations.

This single insight underlies option pricing, filtering theory, quantum mechanics, and much of modern applied mathematics.

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Why Do PDEs Arise from SDEs?

Consider a diffusion process \(X_t\) and ask: "What is the expected value of some quantity \(g(X_T)\), given that the process starts at \(X_0 = x\)?"

\[u(x) = \mathbb{E}[g(X_T) \mid X_0 = x]\]

This is a function of the starting point \(x\). The remarkable fact is that this function satisfies a PDE—even though the underlying process is random.

The Intuition

Think of \(u(x)\) as a "value function" that tells you the expected outcome starting from position \(x\). As you vary \(x\):

• Nearby points should have similar values (continuity)
• The local dynamics of the SDE determine how values propagate (the generator)
• Time evolution follows from the Markov property

These constraints force \(u\) to satisfy a specific PDE.

A Simple Example

For Brownian motion \(X_t = W_t\) with \(g(x) = x^2\):

\[u(t, x) = \mathbb{E}[(x + W_t)^2] = x^2 + t\]

This function satisfies the heat equation:

\[\frac{\partial u}{\partial t} = \frac{1}{2}\frac{\partial^2 u}{\partial x^2}\]

The connection is not coincidental—it's structural.

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The Generator: Foundation of Everything

For the diffusion \(dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t\), the infinitesimal generator is:

\[\boxed{\mathcal{L} = \mu(x)\frac{\partial}{\partial x} + \frac{\sigma^2(x)}{2}\frac{\partial^2}{\partial x^2}}\]

The generator captures:

• First-order term \(\mu(x)\partial_x\): the deterministic drift
• Second-order term \(\frac{\sigma^2(x)}{2}\partial_{xx}\): the random fluctuations

It uniquely characterizes the process in law (via the martingale problem).

What the Generator Tells Us

For any smooth function \(f\):

\[(\mathcal{L}f)(x) = \lim_{t \downarrow 0} \frac{\mathbb{E}_x[f(X_t)] - f(x)}{t}\]

This is the instantaneous expected rate of change of \(f(X_t)\) when starting from \(x\).

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The Hierarchy of Descriptions

The SDE–PDE connection unfolds through a hierarchy:

┌─────────────────────────────────────┐ │ INFINITESIMAL GENERATOR │ │ (local dynamics, t → 0 limit) │ └─────────────────┬───────────────────┘ │ Markov property extends to finite time ▼ ┌─────────────────────────────────────┐ │ KOLMOGOROV BACKWARD EQUATION │ │ ∂ₜu = Lu (expected values) │ └─────────────────┬───────────────────┘ │ integration by parts (duality) ▼ ┌─────────────────────────────────────┐ │ KOLMOGOROV FORWARD EQUATION │ │ ∂ₜp = L*p (probability density) │ └─────────────────┬───────────────────┘ │ add discounting/killing ▼ ┌─────────────────────────────────────┐ │ FEYNMAN–KAC FORMULA │ │ ∂ₜu + Lu - ru = 0 (pricing) │ └─────────────────────────────────────┘

Why Does the Generator Lead to the Backward Equation?

The arrow labeled "Markov property extends to finite time" deserves explanation — it is the logical core of the entire hierarchy.

Setup. Define:

\[u(x_0, t_0) = \mathbb{E}[g(X_t) \mid X_{t_0} = x_0]\]

This is a function of the starting point \((x_0, t_0)\). We want to find what PDE it satisfies.

The generator is purely infinitesimal. By definition, \(\mathcal{L}_{x_0}\) captures only what happens as \(h \to 0\), with the time argument frozen:

\[(\mathcal{L}_{x_0} f)(x_0) = \lim_{h \downarrow 0} \frac{\mathbb{E}_{x_0}[f(X_{t_0+h})] - f(x_0)}{h}\]

This is a local rule — it tells you the instantaneous rate of change in the spatial variable \(x_0\), but says nothing directly about finite time horizons.

Step 1: Apply the Markov property (tower property).

Take a small step \(h > 0\) forward from \(t_0\). By the Markov property, once you know where the process is at \(t_0 + h\), the earlier history is irrelevant:

\[u(x_0, t_0) = \mathbb{E}\!\left[\mathbb{E}[g(X_t) \mid X_{t_0+h}] \;\Big|\; X_{t_0} = x_0\right] = \mathbb{E}_{x_0}\!\left[u(X_{t_0+h},\, t_0+h)\right]\]

Step 2: Rewrite as a difference and take \(h \to 0\).

Since \(u(x_0, t_0) = \mathbb{E}_{x_0}[u(X_{t_0+h}, t_0+h)]\), we can write:

\[\frac{\partial u}{\partial t_0} = \lim_{h \to 0} \frac{\mathbb{E}_{x_0}[u(X_{t_0+h},\, t_0+h)] - u(x_0, t_0)}{h}\]

Step 3: Split the increment into spatial and temporal parts.

Add and subtract \(\mathbb{E}_{x_0}[u(X_{t_0+h}, t_0)]\) to separate the two effects:

\[\frac{\partial u}{\partial t_0} = \lim_{h \to 0} \frac{\mathbb{E}_{x_0}[u(X_{t_0+h},\, t_0)] - u(x_0, t_0)}{h} + \lim_{h \to 0} \frac{\mathbb{E}_{x_0}[u(X_{t_0+h},\, t_0+h) - u(X_{t_0+h},\, t_0)]}{h}\]

The first term is exactly the generator applied to \(u(\cdot, t_0)\) as a function of \(x_0\):

\[\lim_{h \to 0} \frac{\mathbb{E}_{x_0}[u(X_{t_0+h},\, t_0)] - u(x_0, t_0)}{h} = \mathcal{L}_{x_0}\, u(x_0, t_0)\]

The second term: expanding \(u(X_{t_0+h}, t_0+h) - u(X_{t_0+h}, t_0) = \partial_{t_0} u \cdot h + O(h^2)\), so after dividing by \(h\) and taking the limit this contributes \(\mathbb{E}_{x_0}[\partial_{t_0} u(X_{t_0+h}, t_0)] \to \partial_{t_0} u(x_0, t_0)\).

Why the second term cannot be ignored

The difference \(u(X_{t_0+h}, t_0+h) - u(X_{t_0+h}, t_0)\) is \(O(h)\) — it is of leading order in \(h\) and does not vanish in the limit. This is why you cannot simply substitute \(u(X_{t_0+h}, t_0)\) for \(u(X_{t_0+h}, t_0+h)\) and read off the generator directly. The two-step split is necessary.

Combining both terms and rearranging:

\[\frac{\partial u}{\partial t_0} - \partial_{t_0} u(x_0, t_0) = \mathcal{L}_{x_0}\, u(x_0, t_0)\]

which gives the backward equation with the correct sign convention (since \(t_0\) runs backward toward the terminal condition at \(t\)):

\[\boxed{-\partial_{t_0} u = \mathcal{L}_{x_0} u, \qquad u(x_0, t) = g(x_0)}\]

Why Markov is essential

Without the Markov property, \(\mathbb{E}[g(X_t) \mid X_{t_0+h}]\) would depend on the full history before \(t_0 + h\), not just the current state \(X_{t_0+h}\). The tower step in Step 1 would collapse, and you could not reduce a finite-time expectation to a single application of \(\mathcal{L}_{x_0}\).

In summary:

Step	Role
Generator \(\mathcal{L}_{x_0}\)	Encodes local dynamics in the spatial variable \(x_0\) at each instant
Markov property	Enables tower property: \(u(x_0,t_0) = \mathbb{E}_{x_0}[u(X_{t_0+h}, t_0+h)]\)
Two-step split	Separates spatial increment (→ generator) from temporal increment (→ \(\partial_{t_0} u\))
Result	\(-\partial_{t_0} u = \mathcal{L}_{x_0} u\) holds for all \(t_0 < t\)

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Each level answers a different question:

Level	Object	Equation	Question Answered
Generator	\(\mathcal{L}\)	\((\mathcal{L}f)(x) = \lim_{t\downarrow 0}\frac{\mathbb{E}_x[f(X_t)] - f(x)}{t}\)	What happens infinitesimally?
Backward	\(u(t,x) = \mathbb{E}_x[g(X_t)]\)	\(\partial_t u = \mathcal{L}u\)	Expected value starting from \(x\)?
Forward	\(p(x,t)\) density	\(\partial_t p = \mathcal{L}^* p\)	Where does probability mass go?
Feynman–Kac	\(u(t,x) = \mathbb{E}_x[e^{-\int r \, ds} g(X_T)]\)	\(\partial_t u + \mathcal{L}u - ru = 0\)	Discounted expected value?

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The Three Fundamental PDEs

1. Kolmogorov Backward Equation

\[\frac{\partial u}{\partial t} = \mathcal{L} u, \quad u(0, x) = g(x)\]

Solution: \(u(t, x) = \mathbb{E}[g(X_t) \mid X_0 = x]\)

Interpretation: How expected values evolve forward in time.

Applications:

• Transition probabilities
• Heat equation (for Brownian motion)
• Expected hitting times

See Kolmogorov Backward Equation.

2. Kolmogorov Forward Equation (Fokker–Planck)

\[\frac{\partial p}{\partial t} = \mathcal{L}^* p, \quad p(x, 0) = \delta(x - x_0)\]

Solution: \(p(x, t)\) is the probability density of \(X_t\) given \(X_0 = x_0\).

Interpretation: How probability density spreads over time.

Applications:

• Evolution of distributions
• Stationary distributions (\(\mathcal{L}^* p_\infty = 0\))
• Score functions in diffusion models

See Kolmogorov Forward Equation.

3. Feynman–Kac Formula

\[\frac{\partial u}{\partial t} + \mathcal{L} u - r(x)u = 0, \quad u(T, x) = g(x)\]

Solution: \(u(t, x) = \mathbb{E}\left[e^{-\int_t^T r(X_s) ds} g(X_T) \mid X_t = x\right]\)

Interpretation: Discounted expected values (present value of future payoff).

Applications:

• Option pricing (Black–Scholes PDE)
• Bond pricing with stochastic rates
• Optimal stopping problems

See Feynman–Kac Formula.

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Forward vs Backward: The Duality

The backward and forward equations are adjoints of each other, connected by integration by parts.

Aspect	Backward	Forward
Equation	\(\partial_t u = \mathcal{L} u\)	\(\partial_t p = \mathcal{L}^* p\)
Operator	\(\mathcal{L}\) (generator)	\(\mathcal{L}^*\) (adjoint)
Acts on	Test functions / payoffs	Probability densities
Initial/terminal	Initial condition \(g\)	Initial point mass \(\delta\)
Question	"Expected value of \(g\)?"	"Where is the mass?"
Time direction	Information flows backward	Density flows forward

The duality relation:

\[\int_{\mathbb{R}} f(x) (\mathcal{L}g)(x) \, dx = \int_{\mathbb{R}} (\mathcal{L}^* f)(x) g(x) \, dx\]

This connects the two perspectives:

\[\mathbb{E}[g(X_T)] = \int g(x) p(x, T) \, dx\]

The backward equation gives \(\mathbb{E}[g(X_T)]\) directly; the forward equation gives the density \(p\) which you integrate against \(g\).

See Forward–Backward Duality.

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The Transition Density: Where Both Meet

The transition density \(p(x, t \mid x_0, t_0)\) satisfies both equations simultaneously:

• Backward in \((x_0, t_0)\): \(\partial_{t_0} p + \mathcal{L}_{x_0} p = 0\)
• Forward in \((x, t)\): \(\partial_t p = \mathcal{L}_x^* p\)

This is a deep fact: the same object satisfies two different PDEs in different variables.

Physical interpretation:

• Fix the destination \((x, t)\), vary the origin \((x_0, t_0)\) → backward equation
• Fix the origin \((x_0, t_0)\), vary the destination \((x, t)\) → forward equation

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Applications Overview

Finance: Option Pricing

The Black–Scholes PDE:

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

is exactly the Feynman–Kac equation for geometric Brownian motion with constant discounting.

Key insight: Pricing = solving a PDE = computing an expectation under the risk-neutral measure.

Physics: Heat Conduction and Diffusion

The heat equation:

\[\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial x^2}\]

is the Fokker–Planck equation for Brownian motion. Temperature evolution = probability density evolution.

Machine Learning: Diffusion Models

Score-based generative models use:

• Forward process: Add noise via SDE, density evolves by Fokker–Planck
• Reverse process: Remove noise using learned score \(\nabla \log p\)
• Training: Match the score function

The entire framework rests on the SDE–PDE connection.

Filtering: Kalman and Beyond

The Zakai equation for nonlinear filtering:

\[d\rho_t = \mathcal{L}^* \rho_t \, dt + \rho_t h(x) \, dY_t\]

combines Fokker–Planck evolution with measurement updates.

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Chapter Roadmap

This chapter develops the SDE–PDE connection systematically:

Heat Equation (Section 3.2)

The simplest case: Brownian motion and the heat equation.

• Heat Equation Overview
• Fundamental Solution
• Scaling and Invariance
• Maximum Principle
• Heat Equation and Brownian Motion

Kolmogorov Equations (Section 3.3)

The general theory for diffusion processes.

• Kolmogorov Forward (Fokker–Planck)
• Kolmogorov Backward
• Forward–Backward Duality

Feynman–Kac Formula (Section 3.4)

Discounting, killing, and the connection to finance.

• Feynman–Kac Formula
• Running Payoff Extension
• Applications

Pricing PDEs (Section 3.5)

Applications to derivative pricing.

• Discounting and Killing
• Greeks from PDE
• Boundary Conditions

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Key Formulas at a Glance

Object	Formula
Generator	\(\mathcal{L} = \mu(x)\partial_x + \frac{\sigma^2(x)}{2}\partial_{xx}\)
Adjoint	\(\mathcal{L}^* f = -\partial_x[\mu f] + \frac{1}{2}\partial_{xx}[\sigma^2 f]\)
Backward PDE	\(\partial_t u = \mathcal{L} u\)
Forward PDE	\(\partial_t p = \mathcal{L}^* p\)
Feynman–Kac	\(\partial_t u + \mathcal{L} u - ru = 0\)
Probabilistic solution	\(u(t,x) = \mathbb{E}_x[e^{-\int r \, ds} g(X_T)]\)

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Historical Perspective

The SDE–PDE connection has deep historical roots:

• 1900s: Bachelier's thesis on option pricing (before Black–Scholes!)
• 1905: Einstein's theory of Brownian motion
• 1931: Kolmogorov's rigorous foundation of diffusion theory
• 1947: Feynman's path integral formulation of quantum mechanics
• 1948: Kac's rigorous proof connecting path integrals to PDEs
• 1973: Black–Scholes formula, applying Feynman–Kac to finance
• 2020s: Diffusion models in generative AI, using the forward/backward connection

The mathematics developed for physics became the foundation of quantitative finance, and now powers modern machine learning.

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See Also

• Infinitesimal Generator — the operator that starts it all
• Dynkin's Formula — integral form of the backward equation
• Generator and Martingales — martingale characterization
• Invariant Measures — long-time behavior from Fokker–Planck

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Exercises

Exercise 1. Consider the Ornstein-Uhlenbeck process \(dX_t = -\theta X_t\,dt + \sigma\,dW_t\) with \(\theta > 0\). Write down the infinitesimal generator \(\mathcal{L}\) and verify that the function \(u(t, x) = x\,e^{-\theta t}\) satisfies \(\partial_t u = \mathcal{L} u\) (the Kolmogorov backward equation with terminal condition \(u(0,x) = x\)).

Solution to Exercise 1

The Ornstein-Uhlenbeck process \(dX_t = -\theta X_t\,dt + \sigma\,dW_t\) has drift \(\mu(x) = -\theta x\) and diffusion coefficient \(\sigma(x) = \sigma\). The infinitesimal generator is:

\[ \mathcal{L} = -\theta x\frac{\partial}{\partial x} + \frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2} \]

Now verify that \(u(t, x) = xe^{-\theta t}\) satisfies \(\partial_t u = \mathcal{L} u\).

Left-hand side:

\[ \frac{\partial u}{\partial t} = -\theta x e^{-\theta t} \]

Right-hand side: Compute each term of \(\mathcal{L} u\):

\[ \frac{\partial u}{\partial x} = e^{-\theta t}, \qquad \frac{\partial^2 u}{\partial x^2} = 0 \]

Therefore:

\[ \mathcal{L} u = -\theta x \cdot e^{-\theta t} + \frac{\sigma^2}{2} \cdot 0 = -\theta x e^{-\theta t} \]

Since \(\partial_t u = -\theta x e^{-\theta t} = \mathcal{L} u\), the function \(u(t,x) = xe^{-\theta t}\) satisfies the Kolmogorov backward equation. This is consistent with the probabilistic interpretation: \(\mathbb{E}[X_t \mid X_0 = x] = xe^{-\theta t}\) for the OU process, which is the conditional mean decaying exponentially toward zero.

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Exercise 2. For a general one-dimensional diffusion \(dX_t = \mu(x)\,dt + \sigma(x)\,dW_t\), the adjoint operator \(\mathcal{L}^*\) acting on a density \(p\) is

\[ \mathcal{L}^* p = -\frac{\partial}{\partial x}[\mu(x)\,p] + \frac{1}{2}\frac{\partial^2}{\partial x^2}[\sigma^2(x)\,p] \]

Verify the duality relation \(\int f(x)\,(\mathcal{L}g)(x)\,dx = \int (\mathcal{L}^* f)(x)\,g(x)\,dx\) by integration by parts, assuming \(f\) and \(g\) vanish at \(\pm\infty\).

Solution to Exercise 2

We need to verify \(\int f\,(\mathcal{L}g)\,dx = \int (\mathcal{L}^* f)\,g\,dx\) where:

\[ \mathcal{L}g = \mu g' + \frac{1}{2}\sigma^2 g'', \qquad \mathcal{L}^* f = -(\mu f)' + \frac{1}{2}(\sigma^2 f)'' \]

Consider the left-hand side, integrating term by term.

First term (drift):

\[ \int_{-\infty}^{\infty} f(x)\,\mu(x)\,g'(x)\,dx \]

Integrate by parts (boundary terms vanish since \(f, g \to 0\) at \(\pm\infty\)):

\[ = -\int_{-\infty}^{\infty} \frac{\partial}{\partial x}[\mu(x)\,f(x)]\,g(x)\,dx = -\int_{-\infty}^{\infty} (\mu f)'(x)\,g(x)\,dx \]

Second term (diffusion):

\[ \int_{-\infty}^{\infty} f(x)\,\frac{1}{2}\sigma^2(x)\,g''(x)\,dx \]

Integrate by parts once:

\[ = -\int_{-\infty}^{\infty} \frac{\partial}{\partial x}\!\left[\frac{1}{2}\sigma^2(x)\,f(x)\right] g'(x)\,dx \]

Integrate by parts again:

\[ = \int_{-\infty}^{\infty} \frac{\partial^2}{\partial x^2}\!\left[\frac{1}{2}\sigma^2(x)\,f(x)\right] g(x)\,dx \]

Combining both terms:

\[ \int f\,(\mathcal{L}g)\,dx = \int \left\{-(\mu f)' + \frac{1}{2}(\sigma^2 f)''\right\} g\,dx = \int (\mathcal{L}^* f)\,g\,dx \]

This confirms the duality relation. \(\square\)

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Exercise 3. For Brownian motion (\(\mu = 0\), \(\sigma = 1\)), the transition density is \(p(x, t \mid x_0, 0) = \frac{1}{\sqrt{2\pi t}} e^{-(x - x_0)^2/(2t)}\). Verify directly that this density satisfies both the forward equation \(\partial_t p = \frac{1}{2}\partial_{xx} p\) in \((x, t)\) and the backward equation \(\partial_t p + \frac{1}{2}\partial_{x_0 x_0} p = 0\) in \((x_0, t)\) (with \(t\) treated as the same variable in both cases).

Solution to Exercise 3

The Brownian motion transition density is \(p(x, t \mid x_0, 0) = \frac{1}{\sqrt{2\pi t}}\exp\!\left(-\frac{(x-x_0)^2}{2t}\right)\).

Forward equation: \(\partial_t p = \frac{1}{2}\partial_{xx} p\) with derivatives taken in \((x, t)\), holding \(x_0\) fixed.

Compute \(\partial_t p\). Let \(\phi = -\frac{(x-x_0)^2}{2t}\), so \(p = (2\pi t)^{-1/2}e^\phi\).

\[ \partial_t p = p\left(-\frac{1}{2t} + \frac{(x-x_0)^2}{2t^2}\right) \]

Compute \(\partial_{xx} p\):

\[ \partial_x p = p\cdot\frac{-(x-x_0)}{t} \]

\[ \partial_{xx} p = p\left(\frac{(x-x_0)^2}{t^2} - \frac{1}{t}\right) \]

Therefore \(\frac{1}{2}\partial_{xx}p = p\left(\frac{(x-x_0)^2}{2t^2} - \frac{1}{2t}\right) = \partial_t p\). The forward equation is verified.

Backward equation: \(\partial_t p + \frac{1}{2}\partial_{x_0 x_0} p = 0\) with derivatives in \((x_0, t)\), holding \(x\) fixed. (Note: here \(t\) plays the role of the time variable in both equations.)

\[ \partial_{x_0} p = p\cdot\frac{(x - x_0)}{t}, \qquad \partial_{x_0 x_0} p = p\left(\frac{(x-x_0)^2}{t^2} - \frac{1}{t}\right) \]

Note that \(\partial_{x_0 x_0}p = \partial_{xx}p\) since \((x - x_0)^2\) is symmetric and only the sign of the first derivative changes. Therefore:

\[ \partial_t p + \frac{1}{2}\partial_{x_0 x_0}p = p\left(-\frac{1}{2t} + \frac{(x-x_0)^2}{2t^2}\right) + \frac{1}{2}p\left(\frac{(x-x_0)^2}{t^2} - \frac{1}{t}\right) \]

Wait -- this does not give zero. The issue is the sign convention. For Brownian motion with \(\mu = 0\), the backward equation in the standard form is \(\partial_t p = \frac{1}{2}\partial_{x_0 x_0}p\) (or equivalently \(-\partial_{t_0}p = \frac{1}{2}\partial_{x_0 x_0}p\) when differentiating in the initial time \(t_0\)). Since \(p\) depends on \(t - t_0\) and we set \(t_0 = 0\), we have \(\partial_{t_0}p = -\partial_t p\). The backward equation in \((x_0, t_0)\) becomes:

\[ -\partial_t p = -\frac{1}{2}\partial_{x_0 x_0}p \quad \Longleftrightarrow \quad \partial_t p = \frac{1}{2}\partial_{x_0 x_0}p \]

Since \(\partial_{x_0 x_0}p = \partial_{xx}p\) (by the symmetry \((x - x_0)^2\)), this is the same computation as the forward equation, and it holds. \(\square\)

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Exercise 4. Let \(X_t\) follow geometric Brownian motion \(dX_t = rX_t\,dt + \sigma X_t\,dW_t\). Define the value function \(u(t, x) = \mathbb{E}[e^{-r(T-t)} g(X_T) \mid X_t = x]\) for a European payoff \(g\). Show that \(u\) satisfies the Black-Scholes PDE

\[ \frac{\partial u}{\partial t} + rx\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2 x^2 \frac{\partial^2 u}{\partial x^2} - ru = 0 \]

by identifying the generator, applying Feynman-Kac, and matching the discounting term.

Solution to Exercise 4

For geometric Brownian motion under the risk-neutral measure, \(dX_t = rX_t\,dt + \sigma X_t\,dW_t\), the generator is:

\[ \mathcal{L} = rx\frac{\partial}{\partial x} + \frac{1}{2}\sigma^2 x^2\frac{\partial^2}{\partial x^2} \]

The value function is \(u(t, x) = \mathbb{E}\!\left[e^{-r(T-t)}g(X_T) \mid X_t = x\right]\).

By the Feynman-Kac theorem, the function \(u(t, x) = \mathbb{E}_x\!\left[e^{-\int_t^T r\,ds}\,g(X_T)\right]\) satisfies:

\[ \frac{\partial u}{\partial t} + \mathcal{L}u - r\,u = 0, \qquad u(T, x) = g(x) \]

Substituting the generator:

\[ \frac{\partial u}{\partial t} + rx\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2 x^2\frac{\partial^2 u}{\partial x^2} - ru = 0 \]

This is exactly the Black-Scholes PDE with terminal condition \(u(T, x) = g(x)\).

The key steps are:

1. Identify the generator: From \(dX_t = rX_t\,dt + \sigma X_t\,dW_t\), we read off \(\mu(x) = rx\) and \(\sigma(x) = \sigma x\), giving \(\mathcal{L} = rx\partial_x + \frac{1}{2}\sigma^2 x^2\partial_{xx}\)
2. Apply Feynman-Kac: The discounted expectation \(u = \mathbb{E}[e^{-r(T-t)}g(X_T)]\) satisfies \(\partial_t u + \mathcal{L}u - ru = 0\)
3. Match the discounting term: The constant discount rate \(r\) produces the \(-ru\) term in the PDE, which converts the generator equation into the pricing PDE \(\square\)

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Exercise 5. Consider the diffusion \(dX_t = \kappa(\theta - X_t)\,dt + \xi\sqrt{X_t}\,dW_t\) (the CIR process). Write down the generator \(\mathcal{L}\) and the Kolmogorov forward equation for the transition density \(p(x, t)\). Then find the stationary density \(p_\infty(x)\) by solving \(\mathcal{L}^* p_\infty = 0\) and identify it as a Gamma distribution.

Solution to Exercise 5

The CIR process \(dX_t = \kappa(\theta - X_t)\,dt + \xi\sqrt{X_t}\,dW_t\) has \(\mu(x) = \kappa(\theta - x)\) and \(\sigma(x) = \xi\sqrt{x}\).

Generator:

\[ \mathcal{L} = \kappa(\theta - x)\frac{\partial}{\partial x} + \frac{\xi^2 x}{2}\frac{\partial^2}{\partial x^2} \]

Kolmogorov forward equation: \(\partial_t p = \mathcal{L}^* p\) where:

\[ \mathcal{L}^* p = -\frac{\partial}{\partial x}[\kappa(\theta - x)\,p] + \frac{1}{2}\frac{\partial^2}{\partial x^2}[\xi^2 x\,p] \]

Stationary density: Set \(\mathcal{L}^* p_\infty = 0\):

\[ -\frac{d}{dx}[\kappa(\theta - x)\,p_\infty] + \frac{1}{2}\frac{d^2}{dx^2}[\xi^2 x\,p_\infty] = 0 \]

Integrate once:

\[ -\kappa(\theta - x)p_\infty + \frac{1}{2}\frac{d}{dx}[\xi^2 x\,p_\infty] = C_1 \]

With the boundary condition that \(p_\infty \to 0\) as \(x \to \infty\) and the flux vanishes, we set \(C_1 = 0\):

\[ \frac{1}{2}\xi^2\frac{d}{dx}[x\,p_\infty] = \kappa(\theta - x)p_\infty \]

Expand the left side: \(\frac{1}{2}\xi^2(p_\infty + xp_\infty') = \kappa(\theta - x)p_\infty\). Rearranging:

\[ xp_\infty' = \left(\frac{2\kappa\theta}{\xi^2} - 1\right)p_\infty - \frac{2\kappa}{\xi^2}x\,p_\infty \]

Let \(\alpha = \frac{2\kappa\theta}{\xi^2}\) and \(\beta = \frac{2\kappa}{\xi^2}\). Then:

\[ xp_\infty' = (\alpha - 1)p_\infty - \beta x\,p_\infty \]

This ODE has the solution \(p_\infty(x) = C\,x^{\alpha - 1}e^{-\beta x}\) for \(x > 0\), which is the Gamma distribution \(\text{Gamma}(\alpha, \beta)\) with shape \(\alpha = \frac{2\kappa\theta}{\xi^2}\) and rate \(\beta = \frac{2\kappa}{\xi^2}\).

The normalizing constant is \(C = \frac{\beta^\alpha}{\Gamma(\alpha)}\), giving:

\[ p_\infty(x) = \frac{\beta^\alpha}{\Gamma(\alpha)}\,x^{\alpha-1}\,e^{-\beta x}, \quad x > 0 \]

The Feller condition \(2\kappa\theta > \xi^2\) (i.e., \(\alpha > 1\)) ensures that \(p_\infty(0) = 0\), so the process stays strictly positive.

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Exercise 6. The Feynman-Kac formula states that \(u(t,x) = \mathbb{E}_x[e^{-\int_t^T r(X_s)\,ds}\,g(X_T)]\) solves \(\partial_t u + \mathcal{L}u - r(x)u = 0\) with \(u(T,x) = g(x)\). For the special case \(r(x) = r\) (constant) and \(dX_t = \mu\,dt + \sigma\,dW_t\) (arithmetic Brownian motion), verify that \(u(t, x) = e^{-r(T-t)}\,\mathbb{E}_x[g(X_T)]\) satisfies the PDE by direct substitution, using the fact that \(X_T \mid X_t = x\) is \(N(x + \mu(T-t),\, \sigma^2(T-t))\).

Solution to Exercise 6

For constant \(r\) and arithmetic Brownian motion \(dX_t = \mu\,dt + \sigma\,dW_t\), we have \(X_T \mid X_t = x \sim N(x + \mu(T-t),\, \sigma^2(T-t))\).

The Feynman-Kac solution is:

\[ u(t, x) = e^{-r(T-t)}\,\mathbb{E}_x[g(X_T)] = e^{-r(T-t)} \int_{-\infty}^{\infty} g(y)\,\frac{1}{\sigma\sqrt{2\pi(T-t)}}\exp\!\left(-\frac{(y - x - \mu(T-t))^2}{2\sigma^2(T-t)}\right)dy \]

Let \(\tau = T - t\), so \(u(t,x) = e^{-r\tau}\,F(x, \tau)\) where \(F(x, \tau) = \mathbb{E}_x[g(X_T)]\).

Compute \(\partial_t u\): Since \(\tau = T - t\), \(\partial_t = -\partial_\tau\):

\[ \partial_t u = re^{-r\tau}F - e^{-r\tau}\partial_\tau F \]

Compute \(\partial_x u\): The dependence on \(x\) enters only through the Gaussian mean \(x + \mu\tau\):

\[ \partial_x F = \int g(y)\,\frac{y - x - \mu\tau}{\sigma^2\tau}\,\phi\,dy, \quad \partial_{xx}F = \int g(y)\left(\frac{(y-x-\mu\tau)^2}{\sigma^4\tau^2} - \frac{1}{\sigma^2\tau}\right)\phi\,dy \]

where \(\phi\) is the Gaussian density. These are standard results for Gaussian convolutions.

Now \(F\) satisfies the backward (heat-type) equation \(\partial_\tau F = \mu\partial_x F + \frac{1}{2}\sigma^2\partial_{xx}F\) (the Kolmogorov backward equation for ABM). Therefore \(\partial_\tau F = \mathcal{L}F\).

Substituting into the PDE \(\partial_t u + \mathcal{L}u - ru = 0\):

\[ \partial_t u + \mathcal{L}u - ru = (re^{-r\tau}F - e^{-r\tau}\partial_\tau F) + e^{-r\tau}\mathcal{L}F - re^{-r\tau}F \]

\[ = e^{-r\tau}(-\partial_\tau F + \mathcal{L}F) = 0 \]

since \(\partial_\tau F = \mathcal{L}F\). The PDE is satisfied. At \(t = T\) (\(\tau = 0\)), the Gaussian collapses to a point mass at \(x\), giving \(u(T, x) = g(x)\). \(\square\)

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Exercise 7. The transition density \(p(x, t \mid x_0, t_0)\) satisfies the backward equation in \((x_0, t_0)\) and the forward equation in \((x, t)\) simultaneously. Explain why these two PDEs do not contradict each other, despite involving different differential operators (\(\mathcal{L}_{x_0}\) vs. \(\mathcal{L}_x^*\)). In your answer, clarify the role of which variables are held fixed and which are varied in each equation.

Solution to Exercise 7

The transition density \(p(x, t \mid x_0, t_0)\) is a function of four variables: \((x, t, x_0, t_0)\). The two PDEs operate on different pairs of these variables:

Backward equation: Fix the destination \((x, t)\) and differentiate in the origin \((x_0, t_0)\):

\[ \frac{\partial p}{\partial t_0} + \mathcal{L}_{x_0}\,p = 0 \]

Here \(\mathcal{L}_{x_0} = \mu(x_0)\partial_{x_0} + \frac{1}{2}\sigma^2(x_0)\partial_{x_0 x_0}\) acts on \(p\) as a function of \(x_0\), with \((x, t)\) treated as fixed parameters. This equation asks: "How does the transition probability change as we vary the starting point?"

Forward equation: Fix the origin \((x_0, t_0)\) and differentiate in the destination \((x, t)\):

\[ \frac{\partial p}{\partial t} = \mathcal{L}_x^*\,p \]

Here \(\mathcal{L}_x^* = -\partial_x[\mu(x)\,\cdot\,] + \frac{1}{2}\partial_{xx}[\sigma^2(x)\,\cdot\,]\) acts on \(p\) as a function of \(x\), with \((x_0, t_0)\) treated as fixed parameters. This equation asks: "How does the probability density spread over time?"

Why there is no contradiction: The two equations involve differentiation with respect to different variables. The backward equation constrains \(p\) as a function of \((x_0, t_0)\) for each fixed \((x, t)\), while the forward equation constrains \(p\) as a function of \((x, t)\) for each fixed \((x_0, t_0)\). These are independent sets of constraints on the same function, and a sufficiently smooth function of four variables can satisfy both simultaneously.

An analogy: a function \(f(x, y) = e^{x+y}\) satisfies \(\partial_x f = f\) (an equation in \(x\)) and \(\partial_y f = f\) (an equation in \(y\)) simultaneously without contradiction, because the two equations operate on different variables.

The operators \(\mathcal{L}_{x_0}\) and \(\mathcal{L}_x^*\) are different (\(\mathcal{L}\) vs. its adjoint) precisely because they act on different arguments and encode different physical questions. The generator \(\mathcal{L}_{x_0}\) describes how expected values depend on starting position; the adjoint \(\mathcal{L}_x^*\) describes how probability mass flows through space. These are dual perspectives on the same Markov process, and duality is what guarantees that a single transition density can satisfy both equations.