Skip to content

Kolmogorov Backward Equation

The Kolmogorov backward equation describes how expected values of functions of a diffusion process depend on the initial condition. It is the PDE satisfied by \(u(t,x) = \mathbb{E}_x[g(X_t)]\), with the generator acting on the initial point.

Related Content

Setting

Time-Homogeneous Case

Consider the diffusion:

\[dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t\]

with generator:

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

Time-Inhomogeneous Case

For time-dependent coefficients:

\[dX_t = \mu(X_t, t)\,dt + \sigma(X_t, t)\,dW_t\]

the generator becomes time-dependent:

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

The Two Forms of the Backward Equation

The backward equation appears in two equivalent forms depending on the time convention. Understanding both is essential because different fields prefer different formulations.

Form 1: Initial Value Problem (Forward in Time)

Question: Given initial condition \(g\), how does \(\mathbb{E}_x[g(X_t)]\) evolve as \(t\) increases?

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

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

Time runs forward: \(t = 0 \to t = T\).

Interpretation: Start with a "snapshot" \(g\) of some quantity at \(t=0\). As time progresses, the expectation of this quantity (evaluated at where the process ends up) evolves according to \(\mathcal{L}\).

Form 2: Terminal Value Problem (Backward in Time)

Question: Given terminal payoff \(g\) at time \(T\), what is the expected value at earlier time \(t\)?

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

Here \(v(t, x) = \mathbb{E}[g(X_T) \mid X_t = x]\).

Time runs backward: \(t = T \to t = 0\).

Interpretation: We know the payoff at maturity \(T\) and want to find its present value at earlier times.

Relationship Between the Forms

Aspect Form 1 (IVP) Form 2 (TVP)
PDE \(\partial_t u = \mathcal{L}u\) \(\partial_t v + \mathcal{L}v = 0\)
Condition \(u(0,x) = g(x)\) \(v(T,x) = g(x)\)
Interpretation Evolve from initial data Propagate from terminal data
Time direction Forward Backward

They are the same equation: If \(u\) solves Form 1, then \(v(t,x) = u(T-t, x)\) solves Form 2.

Which Form to Use?

  • Form 1: Natural for transition densities, heat equation perspective, physics
  • Form 2: Natural for finance (option pricing), optimal control, dynamic programming

Finance typically uses Form 2 because we know the terminal payoff and want present value.

The Sign Convention Mystery

The sign difference (\(\partial_t u = \mathcal{L}u\) vs \(\partial_t v + \mathcal{L}v = 0\)) often confuses newcomers. The key insight:

  • In Form 1, increasing \(t\) means more time for evolution → expectation changes
  • In Form 2, increasing \(t\) means less time until maturity → opposite effect

The time-reversal \(v(t,x) = u(T-t, x)\) relates them: \(\partial_t v = -\partial_t u |_{T-t}\).

Transition Density Form

Let \(p(t; x, y)\) be the transition density:

\[\mathbb{P}(X_t \in dy \mid X_0 = x) = p(t; x, y)\,dy\]

Notation

We write \(p(t; x, y)\) with semicolon to emphasize:

  • \((t, x)\): backward variables (time elapsed, starting point)
  • \(y\): terminal state (where the process ends up)

Backward equation (acting on initial point \(x\)):

\[\boxed{\frac{\partial p}{\partial t}(t; x, y) = \mathcal{L}_x p(t; x, y)}\]

where \(\mathcal{L}_x\) differentiates with respect to \(x\):

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

Initial condition: \(p(0; x, y) = \delta(x - y)\).

This is a remarkable fact: the transition density satisfies two different PDEs:

Equation Variables Operator
Backward \((t, x)\) with \(y\) fixed \(\mathcal{L}_x\)
Forward \((t, y)\) with \(x\) fixed \(\mathcal{L}_y^*\)

Expected Value Form

For any (nice) function \(g\):

\[u(t, x) = \mathbb{E}[g(X_t) \mid X_0 = x] = \int g(y)\, p(t; x, y)\,dy\]

Theorem: \(u\) satisfies the backward equation:

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

This is the most common form encountered in applications.

Derivation

Method 1: From the Generator Definition

By definition of the infinitesimal generator:

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

This shows \(\partial_t u = \mathcal{L}u\) at \(t = 0\).

By the Markov property, the same argument applies at any time \(t\):

\[ \frac{\partial u}{\partial t}(t, x) = \lim_{h \downarrow 0} \frac{u(t+h, x) - u(t, x)}{h} = \lim_{h \downarrow 0} \frac{\mathbb{E}_x[\mathbb{E}_{X_t}[g(X_h)]] - \mathbb{E}_x[g(X_t)]}{h} \]
\[ = \mathbb{E}_x\left[\lim_{h \downarrow 0} \frac{\mathbb{E}_{X_t}[g(X_h)] - g(X_t)}{h}\right] = \mathbb{E}_x[(\mathcal{L}g)(X_t)] \]

For the function \(u(t, \cdot)\), this gives \(\frac{\partial u}{\partial t}(t, x) = (\mathcal{L}_x u)(t, x)\).

Method 2: Verification via Itô's Lemma

This is Verification, Not Derivation

This method assumes \(u\) solves the PDE and verifies consistency. It does not derive the PDE from first principles.

Suppose \(u\) solves \(\partial_t u = \mathcal{L}u\). Define \(v(t,x) = u(T-t, x)\), so \(\partial_t v + \mathcal{L}v = 0\).

Apply Itô's lemma to \(v(t, X_t)\):

\[dv(t, X_t) = \left(\frac{\partial v}{\partial t} + \mathcal{L}v\right)dt + \frac{\partial v}{\partial x}\sigma(X_t)\,dW_t = 0 \cdot dt + \frac{\partial v}{\partial x}\sigma(X_t)\,dW_t\]

So \(v(t, X_t)\) is a local martingale.

Under integrability conditions, taking expectations:

\[v(0, x) = \mathbb{E}_x[v(T, X_T)] = \mathbb{E}_x[u(0, X_T)] = \mathbb{E}_x[g(X_T)]\]

This confirms \(v(0,x) = u(T, x) = \mathbb{E}_x[g(X_T)]\). ✓

The Martingale Insight

The verification reveals a deep connection: \(v(t, X_t)\) is a martingale precisely because \(v\) solves the PDE. This is the essence of the Feynman–Kac formula and explains why PDEs arise naturally in stochastic analysis.

Method 3: Chapman–Kolmogorov Approach

Derivation via Chapman–Kolmogorov

The transition density satisfies the Chapman–Kolmogorov equation:

\[p(t+h; x, y) = \int p(h; x, z) p(t; z, y) \, dz\]

Differentiating with respect to \(t\) at \(t=0\):

\[\frac{\partial p}{\partial t}(h; x, y)\Big|_{h=0^+} = \lim_{h \downarrow 0} \frac{p(h; x, y) - \delta(x-y)}{h}\]

Using the local expansion of the transition density (moment conditions):

  • \(\int (z-x) p(h; x, z) dz = \mu(x) h + o(h)\)
  • \(\int (z-x)^2 p(h; x, z) dz = \sigma^2(x) h + o(h)\)

A Taylor expansion argument yields the backward equation.

Connection to Forward Equation (Fokker–Planck)

The backward and forward equations are adjoints of each other.

Equation Acts on Variable PDE
Backward Initial point \(x\) \(\partial_t p = \mathcal{L}_x p\)
Forward Terminal point \(y\) \(\partial_t p = \mathcal{L}_y^* p\)

where the adjoint generator is:

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

Duality: For test functions \(f, g\):

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

(plus boundary terms)

Physical Interpretation

  • Backward: "From which starting points do we reach a given target?" (Value function perspective)
  • Forward: "Where does the probability mass flow over time?" (Distribution perspective)

See Forward–Backward Duality for the detailed adjoint relationship.

Examples

Example 1: Brownian Motion

SDE: \(dX_t = dW_t\) (so \(\mu = 0\), \(\sigma = 1\))

Backward equation:

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

This is the heat equation.

Solution with \(u(0,x) = g(x)\):

\[u(t, x) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} e^{-(y-x)^2/2t} g(y)\,dy\]

The kernel \(\frac{1}{\sqrt{2\pi t}} e^{-(y-x)^2/2t}\) is the heat kernel = transition density of Brownian motion.

Explicit example: For \(g(x) = x^2\):

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

Verification: \(\partial_t u = 1\), \(\partial_{xx} u = 2\), so \(\partial_t u = \frac{1}{2}\partial_{xx} u\). ✓

Example 2: Brownian Motion with Drift

SDE: \(dX_t = \mu\,dt + \sigma\,dW_t\)

Backward equation:

\[\frac{\partial u}{\partial t} = \mu\frac{\partial u}{\partial x} + \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}\]

This is the advection-diffusion equation.

Solution with \(u(0,x) = g(x)\):

\[u(t, x) = \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi t}} \exp\left(-\frac{(y - x - \mu t)^2}{2\sigma^2 t}\right) g(y)\,dy\]

Example 3: Ornstein–Uhlenbeck

SDE: \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\)

Backward equation:

\[\frac{\partial u}{\partial t} = -\kappa x\frac{\partial u}{\partial x} + \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}\]

Explicit example: For \(g(x) = x\) (expected position):

\[u(t, x) = \mathbb{E}_x[X_t] = x e^{-\kappa t}\]

Verification: \(\partial_t u = -\kappa x e^{-\kappa t}\), \(\partial_x u = e^{-\kappa t}\), \(\partial_{xx} u = 0\)

\(-\kappa x \cdot e^{-\kappa t} + 0 = -\kappa x e^{-\kappa t}\). ✓

Example 4: Geometric Brownian Motion

SDE: \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\)

Generator: \(\mathcal{L} = \mu s\frac{\partial}{\partial s} + \frac{\sigma^2 s^2}{2}\frac{\partial^2}{\partial s^2}\)

Backward equation:

\[\frac{\partial u}{\partial t} = \mu s\frac{\partial u}{\partial s} + \frac{\sigma^2 s^2}{2}\frac{\partial^2 u}{\partial s^2}\]

Connection to Black–Scholes

Under the risk-neutral measure (\(\mu \to r\)), the terminal value form with discounting:

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

is the Black–Scholes PDE.

This follows from Feynman–Kac: the discounted price \(e^{-r(T-t)}V(t,S)\) should be a martingale.

Connection to Feynman–Kac

The backward equation is a special case of the Feynman–Kac formula with zero potential.

Problem PDE Probabilistic Solution
Backward equation \(\partial_t u = \mathcal{L}u\) \(u(t,x) = \mathbb{E}_x[g(X_t)]\)
Feynman–Kac \(\partial_t u = \mathcal{L}u - V(x)u\) \(u(t,x) = \mathbb{E}_x\left[e^{-\int_0^t V(X_s)ds} g(X_t)\right]\)

The potential \(V(x)\) introduces discounting or killing:

  • Finance: \(V = r\) (constant risk-free rate) gives discounted expectations
  • Physics: \(V(x)\) represents a potential field that "kills" particles
  • Probability: \(V(x)\) is the rate of an independent exponential killing time

See Feynman–Kac Formula.

Connection to Dynkin's Formula

Dynkin's formula is the integrated form of the backward equation.

Backward equation (differential):

\[\frac{\partial}{\partial t}\mathbb{E}_x[g(X_t)] = \mathcal{L}_x \mathbb{E}_x[g(X_t)]\]

Dynkin's formula (integral):

\[\mathbb{E}_x[g(X_t)] = g(x) + \mathbb{E}_x\left[\int_0^t (\mathcal{L}g)(X_s)\,ds\right]\]

Differentiating Dynkin at \(t = 0\):

\[\frac{\partial}{\partial t}\mathbb{E}_x[g(X_t)]\Big|_{t=0} = (\mathcal{L}g)(x)\]

recovers the backward equation at \(t = 0\).

Dynkin = E[Itô]

Dynkin's formula is simply Itô's formula with the martingale part averaged out:

\[g(X_t) = g(X_0) + \int_0^t (\mathcal{L}g)(X_s)\,ds + \int_0^t g'(X_s)\sigma(X_s)\,dW_s\]

Taking expectations eliminates the stochastic integral (which is a martingale).

Boundary Conditions

For diffusions on bounded domains \(D\), boundary conditions are needed.

Type Condition Interpretation Example
Dirichlet \(u = g\) on \(\partial D\) Absorbing boundary (process killed) Barrier options
Neumann \(\frac{\partial u}{\partial n} = 0\) on \(\partial D\) Reflecting boundary Particles in a box
Robin \(\alpha u + \beta \frac{\partial u}{\partial n} = g\) Partial absorption/reflection Sticky boundaries

The choice depends on the physical/financial problem.

First Passage Problems

For the first exit time \(\tau = \inf\{t \geq 0 : X_t \notin D\}\):

\[u(x) = \mathbb{E}_x[g(X_\tau)]\]

solves the Dirichlet problem:

\[\mathcal{L}u = 0 \text{ in } D, \quad u = g \text{ on } \partial D\]

This connects to harmonic functions and potential theory.

Well-Posedness

Scope

Full PDE theory is beyond this document. We state key facts.

Under standard conditions (Lipschitz \(\mu\), \(\sigma\); bounded or with growth control):

Property Condition
Existence Smooth \(g\) → classical solution
Uniqueness In appropriate function spaces (e.g., bounded, polynomial growth)
Regularity Solution inherits smoothness from coefficients and data

Maximum Principle

If \(u\) solves the backward equation on \([0,T] \times D\):

\[\max_{[0,T] \times \bar{D}} u = \max\left(\max_{\bar{D}} u(0, \cdot), \max_{[0,T] \times \partial D} u\right)\]

The maximum is attained either at the initial time or on the boundary—never in the interior.

Viscosity Solutions

For irregular data or degenerate diffusions (where \(\sigma\) can vanish), viscosity solutions provide the right framework:

  • Weaker notion that doesn't require differentiability
  • Comparison principle ensures uniqueness
  • Connects to optimal control (Hamilton–Jacobi–Bellman equations)

Numerical Methods

Finite Differences (Explicit Scheme)

For Form 1 (\(\partial_t u = \mathcal{L}u\)), discretize on grid \((x_i, t_n)\):

\[\frac{u_i^{n+1} - u_i^n}{\Delta t} = \mu(x_i)\frac{u_{i+1}^n - u_{i-1}^n}{2\Delta x} + \frac{\sigma^2(x_i)}{2}\frac{u_{i+1}^n - 2u_i^n + u_{i-1}^n}{\Delta x^2}\]

Stability (CFL condition): \(\Delta t \leq \frac{\Delta x^2}{\sigma^2}\)

Finite Differences (Terminal Value Problem)

For Form 2 (\(\partial_t v + \mathcal{L}v = 0\)), march backward from \(v(T, \cdot) = g\):

\[v_i^{n} = v_i^{n+1} + \Delta t \cdot (\mathcal{L}v)_i^{n+1}\]

This is the standard approach for option pricing.

Monte Carlo

The probabilistic representation provides a natural Monte Carlo method:

\[u(t, x) = \mathbb{E}_x[g(X_t)] \approx \frac{1}{N}\sum_{j=1}^N g(X_t^{(j)})\]

where \(X_t^{(j)}\) are simulated paths starting from \(x\).

Advantages: Works in high dimensions, handles complex payoffs

Disadvantages: Slow convergence (\(O(1/\sqrt{N})\)), gives answer at one point only

Summary

Form PDE Condition Solution
IVP \(\partial_t u = \mathcal{L}u\) \(u(0,x) = g(x)\) \(u(t,x) = \mathbb{E}_x[g(X_t)]\)
TVP \(\partial_t v + \mathcal{L}v = 0\) \(v(T,x) = g(x)\) \(v(t,x) = \mathbb{E}[g(X_T) \mid X_t = x]\)
\[\boxed{\frac{\partial u}{\partial t} = \mathcal{L}_x u = \mu(x)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(x)\frac{\partial^2 u}{\partial x^2}}\]

The Kolmogorov backward equation is the PDE for expected values, with the generator acting on the initial condition.

Aspect Description
Input Initial position \(x\), test function \(g\)
Output Expected value \(\mathbb{E}_x[g(X_t)]\)
Operator Generator \(\mathcal{L}\) (not adjoint)
Key insight Expectations satisfy deterministic PDEs
Applications Option pricing, optimal stopping, filtering

See Also

Exercises

Exercise 1. For the Ornstein-Uhlenbeck process \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\), write down the Kolmogorov backward equation. Verify that \(u(t, x) = xe^{-\kappa t}\) solves the backward equation with initial condition \(g(x) = x\). What is the probabilistic interpretation of this solution?

Solution to Exercise 1

For the OU process \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\), the generator is:

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

The backward equation is:

\[ \frac{\partial u}{\partial t} = -\kappa x\frac{\partial u}{\partial x} + \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2} \]

Verification that \(u(t, x) = xe^{-\kappa t}\) solves it with \(g(x) = x\):

\[ \frac{\partial u}{\partial t} = -\kappa x e^{-\kappa t}, \qquad \frac{\partial u}{\partial x} = e^{-\kappa t}, \qquad \frac{\partial^2 u}{\partial x^2} = 0 \]
\[ \mathcal{L}u = -\kappa x \cdot e^{-\kappa t} + \frac{\sigma^2}{2}\cdot 0 = -\kappa x e^{-\kappa t} = \frac{\partial u}{\partial t} \;\checkmark \]

Also \(u(0, x) = x \cdot 1 = x = g(x)\). \(\checkmark\)

Probabilistic interpretation: \(u(t, x) = \mathbb{E}[X_t \mid X_0 = x] = xe^{-\kappa t}\). This says the expected position of the OU process at time \(t\) decays exponentially toward zero (the long-run mean when \(\theta = 0\)). The mean-reversion parameter \(\kappa\) controls how fast the expectation converges.

Exercise 2. For Brownian motion \(dX_t = dW_t\), the backward equation is the heat equation \(\partial_t u = \frac{1}{2}\partial_{xx} u\). Starting from the initial condition \(g(x) = e^{\alpha x}\) for a constant \(\alpha\), find the solution \(u(t, x) = \mathbb{E}_x[e^{\alpha X_t}]\) by guessing \(u(t, x) = e^{\alpha x + \beta t}\) and determining \(\beta\).

Solution to Exercise 2

For Brownian motion \(dX_t = dW_t\), the backward equation is \(\partial_t u = \frac{1}{2}\partial_{xx}u\). We guess \(u(t, x) = e^{\alpha x + \beta t}\). Then:

\[ \frac{\partial u}{\partial t} = \beta e^{\alpha x + \beta t}, \qquad \frac{\partial^2 u}{\partial x^2} = \alpha^2 e^{\alpha x + \beta t} \]

Substituting into \(\partial_t u = \frac{1}{2}\partial_{xx}u\):

\[ \beta e^{\alpha x + \beta t} = \frac{\alpha^2}{2}e^{\alpha x + \beta t} \]

Dividing by \(e^{\alpha x + \beta t}\) gives \(\beta = \frac{\alpha^2}{2}\).

Therefore the solution is:

\[ u(t, x) = e^{\alpha x + \alpha^2 t / 2} \]

Verification: At \(t = 0\), \(u(0, x) = e^{\alpha x} = g(x)\). \(\checkmark\)

Probabilistic check: \(u(t, x) = \mathbb{E}_x[e^{\alpha X_t}] = \mathbb{E}[e^{\alpha(x + W_t)}] = e^{\alpha x}\mathbb{E}[e^{\alpha W_t}]\). Since \(W_t \sim N(0, t)\), the moment-generating function gives \(\mathbb{E}[e^{\alpha W_t}] = e^{\alpha^2 t/2}\). Therefore \(u(t, x) = e^{\alpha x + \alpha^2 t/2}\). \(\checkmark\)

Exercise 3. The backward equation has two forms: the initial value problem \(\partial_t u = \mathcal{L}u\) with \(u(0, x) = g(x)\) and the terminal value problem \(\partial_t v + \mathcal{L}v = 0\) with \(v(T, x) = g(x)\). Show that these are related by the substitution \(v(t, x) = u(T - t, x)\). Why is the terminal value form more natural for option pricing?

Solution to Exercise 3

Given \(v(t, x) = u(T - t, x)\), we compute:

\[ \frac{\partial v}{\partial t}(t, x) = -\frac{\partial u}{\partial s}(s, x)\bigg|_{s = T - t} \]

where \(s = T - t\). Since \(u\) satisfies \(\frac{\partial u}{\partial s} = \mathcal{L}u\):

\[ \frac{\partial v}{\partial t} = -\mathcal{L}u(T - t, x) = -\mathcal{L}v(t, x) \]

Therefore \(\frac{\partial v}{\partial t} + \mathcal{L}v = 0\). \(\checkmark\)

For the condition: \(v(T, x) = u(T - T, x) = u(0, x) = g(x)\). \(\checkmark\)

Why Form 2 is more natural for option pricing: In finance, the payoff \(g(S_T) = (S_T - K)^+\) is known at maturity \(T\). We want to find the present value at time \(t < T\). The terminal value formulation \(\partial_t v + \mathcal{L}v = 0\) with \(v(T, x) = g(x)\) directly models this situation: we know the boundary condition at the future time \(T\) and solve backward to find the value at earlier times. The time variable \(t\) represents calendar time, running from now to expiry, and \(v(t, S)\) gives the option price at each intermediate time.

Exercise 4. For geometric Brownian motion \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\), write the backward equation in the variable \(S\). Show that under the risk-neutral measure (replacing \(\mu\) with \(r\)) and adding discounting \(-rV\), you recover the Black-Scholes PDE:

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]
Solution to Exercise 4

For GBM \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\), the generator is \(\mathcal{L} = \mu S\partial_S + \frac{\sigma^2 S^2}{2}\partial_{SS}\), so the backward equation is:

\[ \frac{\partial u}{\partial t} = \mu S\frac{\partial u}{\partial S} + \frac{\sigma^2 S^2}{2}\frac{\partial^2 u}{\partial S^2} \]

Under the risk-neutral measure, replace \(\mu\) with \(r\). In the terminal value form, the backward equation becomes:

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

Now, the option price involves discounting: \(V(t, S) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[g(S_T) \mid S_t = S]\). To account for the discounting factor, define \(\tilde{V} = e^{r(T-t)}V\), so \(\tilde{V}\) satisfies the backward equation without discounting. From \(V = e^{-r(T-t)}\tilde{V}\):

\[ \frac{\partial V}{\partial t} = re^{-r(T-t)}\tilde{V} + e^{-r(T-t)}\frac{\partial\tilde{V}}{\partial t} = rV + e^{-r(T-t)}\frac{\partial\tilde{V}}{\partial t} \]

Since \(\tilde{V}\) satisfies \(\frac{\partial\tilde{V}}{\partial t} + rS\frac{\partial\tilde{V}}{\partial S} + \frac{\sigma^2 S^2}{2}\frac{\partial^2\tilde{V}}{\partial S^2} = 0\), and noting \(\frac{\partial V}{\partial S} = e^{-r(T-t)}\frac{\partial\tilde{V}}{\partial S}\) (similarly for the second derivative), substituting gives:

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

This is the Black-Scholes PDE. \(\checkmark\)

Exercise 5. Verify the backward equation using Ito's lemma: if \(v(t, x)\) solves \(\partial_t v + \mathcal{L}v = 0\), apply Ito's lemma to \(v(t, X_t)\) to show that \(v(t, X_t)\) is a local martingale. Taking expectations, conclude that \(v(0, x) = \mathbb{E}_x[g(X_T)]\).

Solution to Exercise 5

Suppose \(v(t, x)\) solves \(\frac{\partial v}{\partial t} + \mathcal{L}v = 0\) with \(v(T, x) = g(x)\). Apply Ito's lemma to \(v(t, X_t)\):

\[ dv(t, X_t) = \frac{\partial v}{\partial t}\,dt + \frac{\partial v}{\partial x}\,dX_t + \frac{1}{2}\frac{\partial^2 v}{\partial x^2}\,(dX_t)^2 \]

Substituting \(dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t\) and \((dX_t)^2 = \sigma^2(X_t)\,dt\):

\[ dv = \left(\frac{\partial v}{\partial t} + \mu\frac{\partial v}{\partial x} + \frac{\sigma^2}{2}\frac{\partial^2 v}{\partial x^2}\right)dt + \sigma\frac{\partial v}{\partial x}\,dW_t \]
\[ = \left(\frac{\partial v}{\partial t} + \mathcal{L}v\right)dt + \sigma\frac{\partial v}{\partial x}\,dW_t = 0 \cdot dt + \sigma\frac{\partial v}{\partial x}\,dW_t \]

So \(v(t, X_t)\) is a local martingale (the \(dt\) term vanishes because \(v\) solves the PDE).

Under suitable integrability conditions (e.g., \(v\) and \(\partial_x v \cdot \sigma\) are bounded), \(v(t, X_t)\) is a true martingale. Therefore:

\[ v(0, x) = \mathbb{E}_x[v(0, X_0)] = \mathbb{E}_x[v(T, X_T)] = \mathbb{E}_x[g(X_T)] \]

This confirms that the solution to the backward PDE gives the expected value of the terminal payoff. \(\checkmark\)

Exercise 6. Dynkin's formula states \(\mathbb{E}_x[g(X_t)] = g(x) + \mathbb{E}_x\left[\int_0^t (\mathcal{L}g)(X_s)\,ds\right]\). Differentiate both sides with respect to \(t\) at \(t = 0\) to recover the backward equation at \(t = 0\). Why does the Markov property allow you to extend this to all \(t > 0\)?

Solution to Exercise 6

Dynkin's formula states:

\[ \mathbb{E}_x[g(X_t)] = g(x) + \mathbb{E}_x\left[\int_0^t(\mathcal{L}g)(X_s)\,ds\right] \]

Differentiating both sides with respect to \(t\):

\[ \frac{\partial}{\partial t}\mathbb{E}_x[g(X_t)] = \frac{\partial}{\partial t}\mathbb{E}_x\left[\int_0^t(\mathcal{L}g)(X_s)\,ds\right] = \mathbb{E}_x[(\mathcal{L}g)(X_t)] \]

At \(t = 0\):

\[ \frac{\partial}{\partial t}\mathbb{E}_x[g(X_t)]\bigg|_{t=0} = \mathbb{E}_x[(\mathcal{L}g)(X_0)] = (\mathcal{L}g)(x) \]

Writing \(u(t, x) = \mathbb{E}_x[g(X_t)]\), this gives \(\partial_t u(0, x) = (\mathcal{L}u)(0, x)\), which is the backward equation at \(t = 0\).

Extension to all \(t > 0\) via the Markov property: By the Markov property, \(\mathbb{E}_x[g(X_{t+h}) \mid \mathcal{F}_t] = \mathbb{E}_{X_t}[g(X_h)]\). Therefore:

\[ u(t + h, x) = \mathbb{E}_x[g(X_{t+h})] = \mathbb{E}_x[\mathbb{E}_{X_t}[g(X_h)]] = \mathbb{E}_x[u(h, X_t)] \]

This shows that \(u(t + h, x)\) can be viewed as computing the expectation \(\mathbb{E}_x[\tilde{g}(X_t)]\) where \(\tilde{g}(\cdot) = u(h, \cdot)\). Applying the same Dynkin argument to \(\tilde{g}\) and differentiating with respect to \(h\) at \(h = 0\) gives \(\partial_t u(t, x) = (\mathcal{L}u)(t, x)\) for all \(t > 0\).

Exercise 7. Consider the first exit time \(\tau = \inf\{t \geq 0 : X_t \notin (a, b)\}\) for a Brownian motion with drift \(dX_t = \mu\,dt + \sigma\,dW_t\). The expected exit time \(u(x) = \mathbb{E}_x[\tau]\) satisfies the boundary value problem \(\mathcal{L}u = -1\) in \((a, b)\) with \(u(a) = u(b) = 0\). Solve this ODE explicitly and verify that \(u(x) > 0\) for \(x \in (a, b)\).

Solution to Exercise 7

The ODE is \(\mathcal{L}u = -1\) with \(\mathcal{L} = \mu\frac{d}{dx} + \frac{\sigma^2}{2}\frac{d^2}{dx^2}\):

\[ \mu u' + \frac{\sigma^2}{2}u'' = -1, \qquad u(a) = u(b) = 0 \]

Case 1: \(\mu = 0\). The ODE becomes \(\frac{\sigma^2}{2}u'' = -1\), so \(u'' = -2/\sigma^2\). Integrating twice:

\[ u(x) = -\frac{x^2}{\sigma^2} + C_1 x + C_2 \]

Boundary conditions: \(u(a) = 0\) and \(u(b) = 0\) give \(C_1 = \frac{a + b}{\sigma^2}\) and \(C_2 = -\frac{ab}{\sigma^2}\). Thus:

\[ u(x) = \frac{(x - a)(b - x)}{\sigma^2} \]

Since \(a < x < b\), both factors are positive, so \(u(x) > 0\). \(\checkmark\)

Case 2: \(\mu \neq 0\). Let \(\lambda = 2\mu/\sigma^2\). The ODE is \(u'' + \lambda u' = -2/\sigma^2\). The homogeneous solution is \(u_h = A + Be^{-\lambda x}\). A particular solution is \(u_p = -x/\mu\) (verify: \(u_p' = -1/\mu\), \(u_p'' = 0\), so \(\mu(-1/\mu) + 0 = -1\)). The general solution:

\[ u(x) = A + Be^{-\lambda x} - \frac{x}{\mu} \]

Applying \(u(a) = 0\) and \(u(b) = 0\):

\[ A + Be^{-\lambda a} = \frac{a}{\mu}, \qquad A + Be^{-\lambda b} = \frac{b}{\mu} \]

Subtracting: \(B(e^{-\lambda a} - e^{-\lambda b}) = \frac{a - b}{\mu}\), so:

\[ B = \frac{a - b}{\mu(e^{-\lambda a} - e^{-\lambda b})} \]

and \(A = \frac{a}{\mu} - Be^{-\lambda a}\).

Positivity: By the maximum principle for elliptic equations, since \(\mathcal{L}u = -1 < 0\) in \((a, b)\) and \(u = 0\) on the boundary, the minimum of \(u\) must be attained on the boundary. Since \(u(a) = u(b) = 0\) and \(u\) cannot have an interior minimum below zero (that would contradict \(\mathcal{L}u = -1 < 0\)), we conclude \(u(x) \geq 0\) in \([a, b]\). The strong maximum principle sharpens this to \(u(x) > 0\) for \(x \in (a, b)\).