The Ornstein-Uhlenbeck Backward Equation

The backward equation for a mean-reverting diffusion

The Ornstein-Uhlenbeck process is a diffusion that pulls its state toward a fixed level while adding Brownian noise. Its backward equation describes how an expected terminal payoff depends on the current time and state. For exponential-affine payoffs, this equation can be solved explicitly, and the answer can be checked directly from the conditional distribution of the process.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a finite horizon $T>0$, a mean-reversion rate $\kappa>0$, a level $m\in\mathbb{R}$, and a volatility $\sigma\geq0$. Starting from $X_t=x$, the Ornstein-Uhlenbeck process satisfies

dX_s=\kappa(m-X_s)\,ds+\sigma\,dB_s,\qquad s\geq t.

The drift coefficient is $\kappa(m-x)$, and the diffusion coefficient is the constant $\sigma$. For a twice differentiable function $f$, the generator is $\mathcal{L}f(x)=\kappa(m-x)f'(x)+\tfrac12\sigma^2f''(x)$. The backward equation with terminal payoff $g$ is $u_t+\mathcal{L}u=0$, with $u(T,x)=g(x)$. Here we take $g(x)=e^{a+qx}$, where $a,q\in\mathbb{R}$. The case $q=0$ is included and gives a constant payoff.

Definition: The backward value for this terminal payoff is $u(t,x)=\mathbb{E}[e^{a+qX_T}\mid X_t=x]$, for $0\leq t\leq T$. In the expression $X_t=x$, the process is understood to start at state $x$ at time $t$; equivalently, $u(t,x)$ is the expectation under that initial condition.

Solving the equation by an exponential-affine ansatz

Seek a solution in the form $u(t,x)=\exp(A(t)x+C(t))$. For this form, $u_t=(A'(t)x+C'(t))u$, $u_x=A(t)u$, and $u_{xx}=A(t)^2u$. Substitution into $u_t+\kappa(m-x)u_x+\tfrac12\sigma^2u_{xx}=0$, followed by division by the strictly positive $u$, gives

\big(A'(t)-\kappa A(t)\big)x+C'(t)+\kappa mA(t)+\tfrac12\sigma^2A(t)^2=0.

For this identity to hold for every $x$, its coefficient of $x$ and its constant term must both vanish. The terminal condition $u(T,x)=e^{a+qx}$ therefore requires $A(T)=q$, $C(T)=a$, and the coefficient equations are $A'=\kappa A$ and $C'=-\kappa mA-\tfrac12\sigma^2A^2$.

Theorem: For $\kappa>0$, $\sigma\geq0$, and $a,q\in\mathbb{R}$, the backward equation with terminal data $u(T,x)=e^{a+qx}$ has the following solution in the exponential-affine class: $$ u(t,x)=\exp\!\left(a+q\big[m+(x-m)e^{-\kappa(T-t)}\big] +\frac{q^2\sigma^2}{4\kappa}\big(1-e^{-2\kappa(T-t)}\big)\right). $$

Proof. Solving $A'=\kappa A$ with $A(T)=q$ gives $A(t)=qe^{-\kappa(T-t)}$. Integrating the equation for $C$ backward from $T$ gives $C(t)=a+\int_t^T[\kappa mA(s)+\tfrac12\sigma^2A(s)^2]\,ds$. In the first integral, $\int_t^T\kappa m q e^{-\kappa(T-s)}\,ds=mq(1-e^{-\kappa(T-t)})$. In the second, $\int_t^T\tfrac12\sigma^2q^2 e^{-2\kappa(T-s)}\,ds=\tfrac{q^2\sigma^2}{4\kappa}(1-e^{-2\kappa(T-t)})$. Thus $A(t)x+C(t)$ is the exponent displayed in the theorem. Substituting these coefficient equations into the PDE makes both the coefficient of $x$ and the constant term zero, and $A(T)x+C(T)=qx+a$. Finally, within the exponential-affine class, any solution must satisfy these same two terminal-value ODEs. Their explicit solutions are unique: multiplying the difference of two solutions of $A'=\kappa A$ by $e^{-\kappa t}$ gives a constant difference, which is zero at $T$; then the difference of the corresponding $C$'s has zero derivative and is zero at $T$. This proves the formula and uniqueness within that class. $\square$

The qualification “within the exponential-affine class” is deliberate: the coefficient calculation establishes uniqueness for this ansatz, not a general uniqueness theorem for every possible growth class of PDE solutions. The formula remains valid when $q=0$ or $\sigma=0$. For $q=0$, it reduces to $e^a$; for $\sigma=0$, the variance contribution disappears.

Why the conditional Gaussian law gives the same answer

The same formula follows from the process itself. Multiplying the SDE by the integrating factor $e^{\kappa s}$ and applying the product rule gives the explicit solution over $[t,T]$:

X_T=m+(x-m)e^{-\kappa(T-t)} +\sigma\int_t^T e^{-\kappa(T-s)}\,dB_s.

Each stochastic integral here is an Itô integral. More generally, $\int_0^v H_s\,dB_s$ is taken for an adapted, progressively measurable integrand $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^v H_s^2\,ds<\infty$. The deterministic integrand $e^{-\kappa(T-s)}$ is square-integrable on the finite interval. The integral from $t$ to $T$ is independent of $\mathcal{F}_t$ and is Gaussian with mean zero and variance $\int_t^T e^{-2\kappa(T-s)}\,ds$. This follows first for deterministic step integrands from the independent Gaussian Brownian increments, and then for this integrand by square-integrable approximation. Consequently, conditional on $\mathcal{F}_t$, $X_T$ is Gaussian with mean and variance

\mathbb{E}[X_T\mid\mathcal{F}_t] =m+(X_t-m)e^{-\kappa(T-t)},\qquad \operatorname{Var}(X_T\mid\mathcal{F}_t) =\frac{\sigma^2}{2\kappa}\big(1-e^{-2\kappa(T-t)}\big).

If $Y$ is Gaussian with mean $\mu$ and variance $v$, completing the square in its density gives $\mathbb{E}[e^{qY}]=e^{q\mu+q^2v/2}$. When $v=0$, the same identity holds because $Y=\mu$ almost surely. Applying this identity conditionally yields

\mathbb{E}[e^{a+qX_T}\mid\mathcal{F}_t] =\exp\!\left(a+q\big[m+(X_t-m)e^{-\kappa(T-t)}\big] +\frac{q^2\sigma^2}{4\kappa}\big(1-e^{-2\kappa(T-t)}\big)\right).

This is exactly $u(t,X_t)$. In particular, the conditional expectation exists for every real $q$, because a Gaussian random variable has finite exponential moments of every real order. This verifies the PDE solution against the conditional Gaussian distribution, rather than relying only on coefficient matching.

Worked calculations

Worked Example: A Positive Exponential Payoff

Take $\kappa=1$, $m=2$, $\sigma=1$, $T-t=\ln 2$, $x=0$, $a=0$, and $q=1$. Compute the backward value.

First, $e^{-\kappa(T-t)}=e^{-\ln2}=\tfrac12$. The conditional mean is $2+(0-2)\tfrac12=1$. The conditional variance is $\tfrac{1}{2}(1-e^{-2\ln2})=\tfrac12(1-\tfrac14)=\tfrac38$. Therefore the Gaussian exponential-moment formula gives

u(t,0)=\exp\!\left(1+\frac{1^2}{2}\cdot\frac38\right) =\exp\!\left(\frac{19}{16}\right).

Check: substituting into the general formula gives the mean term $q[m+(x-m)e^{-\kappa(T-t)}]=1$ and variance term $\tfrac{q^2\sigma^2}{4\kappa}(1-e^{-2\kappa(T-t)})=\tfrac14(1-\tfrac14)=\tfrac{3}{16}$. The exponent is $1+\tfrac{3}{16}=\tfrac{19}{16}$, as above.

Worked Example: A Negative Exponential Parameter

Take $\kappa=2$, $m=-1$, $\sigma=2$, $T-t=\ln2$, $x=3$, $a=0$, and $q=-2$. Compute $\mathbb{E}[e^{-2X_T}\mid X_t=3]$.

Here $e^{-\kappa(T-t)}=e^{-2\ln2}=\tfrac14$. The conditional mean is $-1+(3-(-1))\tfrac14=-1+1=0$. The conditional variance is $\tfrac{2^2}{2\cdot2}(1-e^{-4\ln2})=1(1-\tfrac1{16})=\tfrac{15}{16}$. Thus

\mathbb{E}[e^{-2X_T}\mid X_t=3] =\exp\!\left((-2)\cdot0+\frac{(-2)^2}{2}\cdot\frac{15}{16}\right) =e^{15/8}.

Check: the PDE formula has mean contribution $(-2)\cdot0=0$ and variance contribution $\tfrac{4\cdot4}{4\cdot2}(1-\tfrac1{16})=2\cdot\tfrac{15}{16}=\tfrac{15}{8}$. It gives the same value $e^{15/8}$. The negative value of $q$ causes no integrability problem because the conditional law is Gaussian.

Worked Example: A Longer Mean-Reversion Interval

Take $\kappa=\tfrac12$, $m=0$, $\sigma=2$, $T-t=2\ln2$, $x=2$, $a=0$, and $q=1$. Find the conditional expectation of $e^{X_T}$.

Since $\kappa(T-t)=\ln2$, the decay factor is $\tfrac12$. The conditional mean is $0+(2-0)\tfrac12=1$. The variance is $\tfrac{2^2}{2(1/2)}(1-e^{-2\ln2})=4(1-\tfrac14)=3$. Hence

\mathbb{E}[e^{X_T}\mid X_t=2] =\exp\!\left(1+\frac12\cdot3\right)=e^{5/2}.

Check: in the PDE formula, the first term in the exponent is $q[m+(x-m)e^{-\kappa(T-t)}]=1$. The second term is $\tfrac{1^2\cdot 2^2}{4(1/2)}(1-e^{-2\ln2})=2\cdot\tfrac34=\tfrac32$. Their sum is $\tfrac52$, confirming the conditional calculation.

Itô verification and martingale status

There is also a useful differential check. For fixed terminal data, the displayed $u$ is smooth, and direct differentiation gives $u_x=A(t)u$, $u_{xx}=A(t)^2u$, and $u_t=(A'(t)x+C'(t))u$, where $A(t)=qe^{-\kappa(T-t)}$ and $C(t)$ is as in the proof. Applying Itô's formula to $u(s,X_s)$, with the time derivative and the second-derivative correction shown explicitly, gives

du(s,X_s)= \left[u_s+\kappa(m-X_s)u_x+\tfrac12\sigma^2u_{xx}\right]ds +\sigma u_x\,dB_s =\sigma A(s)u(s,X_s)\,dB_s.

The drift vanishes by the backward equation. A vanishing drift alone identifies a local-martingale calculation; it does not, in general, prove that the process is a true martingale. In this case the conditional Gaussian calculation above gives $u(s,X_s)=\mathbb{E}[e^{a+qX_T}\mid\mathcal{F}_s]$. The terminal random variable is integrable, since its absolute value has finite expectation by the Gaussian exponential-moment formula. The defining properties of conditional expectation therefore show that $u(s,X_s)$ is an integrable martingale: for $r\leq s$, the tower property gives $\mathbb{E}[u(s,X_s)\mid\mathcal{F}_r]=u(r,X_r)$.

The stochastic integrand in the Itô differential also meets the square-integrability condition on the finite time interval. Indeed, $u(s,X_s)$ is an exponential of a Gaussian affine function, so its second moment is finite at every $s$; the explicit Gaussian moment formula makes that second moment a continuous function of $s\in[t,T]$, hence bounded on this compact interval. Since $A(s)$ is bounded there, $\mathbb{E}\int_t^T \sigma^2A(s)^2u(s,X_s)^2\,ds<\infty$. Thus the stochastic integral is an Itô integral with an $\mathcal{L}^2$ integrand, consistent with the martingale verification.

A common error and why the variance term matters

A frequent mistake is to transport the conditional mean correctly but then evaluate the exponential payoff as the exponential of that mean alone. In the first worked example this incorrect calculation gives $e^1=e$, because the conditional mean is $1$. But $X_T$ is not deterministic: its conditional variance is $\tfrac38$. The Gaussian exponential moment includes the correction $\tfrac12q^2\operatorname{Var}(X_T\mid\mathcal{F}_t)=\tfrac{3}{16}$, so the correct value is $e^{19/16}$, not $e$.

The backward equation encodes the same correction through its second derivative: the term $\tfrac12\sigma^2u_{xx}$ contributes $\tfrac12\sigma^2A(t)^2$ to the coefficient equation for $C$. Omitting that term would remove the variance contribution and would fail both the PDE and the conditional Gaussian check. The mean-reversion term determines how strongly the initial state is remembered; the diffusion term determines the additional exponential-moment contribution.

Check Your Understanding

Use the coefficient equations and the conditional Gaussian formula to answer these questions. Assume $\kappa>0$, $T>t$, and the stated parameters are real.

For terminal payoff $e^{a+qx}$, what are the terminal conditions for $A$ and $C$ in the exponential-affine ansatz?
With $\kappa=1$, $\sigma=0$, $m=0$, $x=2$, and $T-t=\ln2$, what is the conditional mean of $X_T$, and what happens to the variance correction?
Why does the backward formula remain finite for every real $q$, including negative $q$?
In the Itô differential for $u(s,X_s)$, identify the second-derivative correction and state why a zero drift by itself would not establish a true martingale.
For the first worked example, compute the incorrect mean-only answer and the missing exponent correction that produces the correct value.