From a Terminal Payoff to a Backward Equation
A terminal payoff depends on the state at a future time, while its conditional expectation depends on the current time and state. The backward equation describes how that expectation changes as the starting time moves toward the terminal time. Its terminal condition is the payoff itself; the differential operator in the equation is the generator of the diffusion.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. For a diffusion \(X_s\) with drift \(\mu\) and diffusion coefficient \(\sigma\), write \(dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s\). Here the drift is the coefficient of \(ds\), and the diffusion coefficient is the coefficient of \(dB_s\). The generator acting on a twice differentiable function \(f\) is \(\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12\sigma(x)^2f''(x)\) in the time-homogeneous case.
The notation conditioning on \(X_t=x\) means expectation under the diffusion started at state \(x\) at time \(t\); it does not mean conditioning on the probability-zero event \(\{X_t=x\}\). For the Brownian and Ornstein–Uhlenbeck processes considered below, the transition law specifies this expectation for every real \(x\). We use the backward equation to find candidate functions, then check both the PDE and the conditional expectation.
The Representation Behind the Equation
The following result explains why a sufficiently regular solution of the terminal-value problem gives the conditional expected payoff. It also supplies a useful uniqueness statement within the specified integrability class.
Proof. Apply Ito's formula to the function \(f(s,x)=u(s,x)\) and the process \(X_s\). Its partial derivatives are \(f_s=u_t\), \(f_x=u_x\), and \(f_{xx}=u_{xx}\). The quadratic-variation correction is \(\tfrac12 f_{xx}\sigma^2\,ds\), since \(dB_s\,dB_s=ds\), \(dB_s\,ds=0\), and \(ds\,ds=0\). Thus
Consequently \(u(t,X_t)=u(0,X_0)+\int_0^t H_s\,dB_s\). The Ito integral is defined for progressively measurable integrands in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Ito isometry gives a square-integrable integral. It is a martingale: for simple adapted integrands, each increment has conditional mean zero given the past, by the independent mean-zero Brownian increment; the general \(\mathcal{L}^2\) integral is their \(L^2\) limit, and conditional expectation is continuous in \(L^1\). Therefore \(\mathbb{E}[u(T,X_T)\mid\mathcal{F}_t]=u(t,X_t)\). Since \(u(T,X_T)=g(X_T)\), the claimed representation follows. If \(u\) and \(\widetilde u\) both satisfy the hypotheses with the same terminal payoff, both equal this conditional expectation along \(X_t\), so they agree almost surely there. \(\square\)
This argument verifies the representation when the stated smoothness and integrability conditions hold; it does not assert that every payoff has a classical \(C^{1,2}\) solution. In the examples below, the candidate functions are explicit, and their derivatives and integrability can be checked directly.
Brownian Motion: Polynomial and Oscillating Payoffs
For Brownian motion started at \(x\) at time \(t\), the terminal state is \(x+B_T-B_t\). The increment \(B_T-B_t\) is independent of \(\mathcal{F}_t\) and has distribution \(N(0,T-t)\). Thus direct Gaussian expectations provide an independent way to check solutions of the backward equation. The Brownian generator is \(\mathcal{L}f=\tfrac12 f''\), so the equation is \(u_t+\tfrac12u_{xx}=0\).
Worked Example: A Quadratic Brownian Payoff
Take \(g(y)=y^2\). Since \(B_T-B_t\) has mean zero and variance \(T-t\), direct expansion gives
Set \(u(t,x)=x^2+(T-t)\). Its derivatives are \(u_t=-1\), \(u_x=2x\), and \(u_{xx}=2\). The backward equation is satisfied because \(u_t+\tfrac12u_{xx}=-1+\tfrac12(2)=0\), and \(u(T,x)=x^2\). For the Ito representation, the stochastic-integral integrand is \(2X_s\). For Brownian motion started at \(x\) at time \(t\), \(\mathbb{E}[X_s^2]=x^2+s-t\), so \(\int_t^T\mathbb{E}[(2X_s)^2]\,ds=4\int_t^T(x^2+s-t)\,ds<\infty\). The representation theorem applies.
Check: at \(T=2,t=\tfrac12,x=1\), direct calculation gives \(1^2+(2-\tfrac12)=\tfrac52\). Substitution in the candidate gives \(u(\tfrac12,1)=1+\tfrac32=\tfrac52\), as required.
Worked Example: A Cosine Payoff for Brownian Motion
Let \(g(y)=\cos(\xi y)\), where \(\xi\) is a real constant. A centered normal variable \(Z\) with variance \(T-t\) has characteristic function \(\mathbb{E}[e^{i\xi Z}]=e^{-\xi^2(T-t)/2}\). Taking the real part after multiplying by \(e^{i\xi x}\) gives
Therefore take \(u(t,x)=e^{-\xi^2(T-t)/2}\cos(\xi x)\). Differentiating explicitly, \(u_t=\tfrac12\xi^2e^{-\xi^2(T-t)/2}\cos(\xi x)\), \(u_x=-\xi e^{-\xi^2(T-t)/2}\sin(\xi x)\), and \(u_{xx}=-\xi^2e^{-\xi^2(T-t)/2}\cos(\xi x)\). Hence \(u_t+\tfrac12u_{xx}=0\), and \(u(T,x)=\cos(\xi x)\). Also, \(|u_x|\le|\xi|\), so the Ito integrand \(u_x(s,B_s)\) is square-integrable on every finite interval: its squared expectation integrated over \([0,T]\) is at most \(\xi^2T\).
Check: choosing \(\xi=1,T=1,t=\tfrac14,x=0\), direct Gaussian expectation is \(e^{-(1/2)(3/4)}\cos(0)=e^{-3/8}\). The PDE solution gives \(e^{-(1/2)(1-1/4)}\cos(0)=e^{-3/8}\). The terminal value is also correct because setting \(t=T\) makes the exponential factor \(1\).
Ornstein–Uhlenbeck Motion: A Quadratic Payoff
Consider the Ornstein–Uhlenbeck process with parameters \(\kappa>0\) and \(\sigma>0\): \(dX_s=-\kappa X_s\,ds+\sigma\,dB_s\). Its drift is \(-\kappa x\), its diffusion coefficient is \(\sigma\), and its generator is \(\mathcal{L}f(x)=-\kappa x f'(x)+\tfrac12\sigma^2 f''(x)\). Conditional on \(X_t=x\), its terminal law is Gaussian with mean \(e^{-\kappa(T-t)}x\) and variance \(\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa(T-t)})\). These mean and variance follow by solving the linear SDE; the stochastic integral in that solution has deterministic integrand and is normal.
Worked Example: The Squared Ornstein–Uhlenbeck State
For \(g(y)=y^2\), if a Gaussian variable has mean \(m\) and variance \(v\), its second moment is \(m^2+v\). Applying that identity to the conditional terminal law gives
To verify the equation, write \(A(t)=e^{-2\kappa(T-t)}\) and \(V(t)=\frac{\sigma^2}{2\kappa}(1-A(t))\), so \(u=Ax^2+V\). Their time derivatives are \(A_t=2\kappa A\) and \(V_t=-\sigma^2 A\). The spatial derivatives are \(u_x=2Ax\) and \(u_{xx}=2A\); therefore
At \(t=T\), \(A(T)=1\) and \(V(T)=0\), so \(u(T,x)=x^2\). For the representation theorem, the Ito integrand is \(2\sigma A(s)X_s\). On a finite interval, \(A(s)\le1\), and the conditional second moment \(\mathbb{E}[X_s^2]=e^{-2\kappa(s-t)}x^2+\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa(s-t)})\) is bounded by \(x^2+\frac{\sigma^2}{2\kappa}\). Thus the expected integral of the squared integrand is finite.
Check: take \(\kappa=1,\sigma=2,T=1,t=\tfrac12,x=1\). The conditional mean is \(e^{-1/2}\), and the conditional variance is \(\frac{4}{2}(1-e^{-1})=2(1-e^{-1})\). The direct second moment is \(e^{-1}+2(1-e^{-1})=2-e^{-1}\). Substituting into \(u\) gives \(e^{-1}+2(1-e^{-1})=2-e^{-1}\) as well.
What the Backward Equation Does—and Does Not—Determine
For these examples, the equation evolves information backward from \(u(T,x)=g(x)\), while the stochastic process itself runs forward from its current state. This viewpoint is useful when a payoff is specified at a fixed future time: one can seek a PDE solution and then interpret it as a conditional expectation. The Gaussian checks are especially valuable because they test both the time direction and the variance contribution without relying on the PDE calculation.
A common error is to use \(t\) instead of the remaining time \(T-t\) in the Brownian quadratic example. The incorrect candidate \(\widetilde u(t,x)=x^2+t\) has \(\widetilde u_t=1\), \(\widetilde u_{xx}=2\), and therefore \(\widetilde u_t+\tfrac12\widetilde u_{xx}=2\), not zero. It also gives \(\widetilde u(T,x)=x^2+T\), not the required terminal payoff \(x^2\). The corrected function \(x^2+(T-t)\) has time derivative \(-1\), satisfies the PDE, and matches the terminal condition.
| Process | Generator | Backward equation | Terminal condition |
|---|---|---|---|
| Brownian motion | \(\tfrac12 f''\) | \(u_t+\tfrac12u_{xx}=0\) | \(u(T,x)=g(x)\) |
| Ornstein–Uhlenbeck | \(-\kappa x f'+\tfrac12\sigma^2f''\) | \(u_t-\kappa xu_x+\tfrac12\sigma^2u_{xx}=0\) | \(u(T,x)=g(x)\) |
Check Your Understanding
For each question, state the terminal condition as well as the proposed value or equation.
- For Brownian motion and payoff \(g(y)=y\), find the backward value and verify its PDE.
- For Brownian motion and payoff \(g(y)=y^2\), explain why replacing \(T-t\) by \(t\) violates both the terminal condition and the PDE.
- For an Ornstein–Uhlenbeck process, write the conditional mean and variance at time \(T\) given \(X_t=x\).
- For the Ornstein–Uhlenbeck payoff \(g(y)=y^2\), identify the Ito-integral integrand and state the square-integrability condition used in the representation.
- Explain why a solution of the backward PDE is not, by itself, enough for the representation theorem unless the stated regularity and integrability conditions are checked.