Ornstein-Uhlenbeck Expectations and Quadratic Potentials

Ornstein–Uhlenbeck Expectations

For an Ornstein–Uhlenbeck diffusion, the transition law is Gaussian, so exponential-linear terminal data can be evaluated directly from a normal moment-generating function. A quadratic running potential changes the calculation: the expectation is still explicit in an exponential-quadratic form, but its coefficients satisfy a Riccati equation. The sign of that equation depends on whether time is measured forward from the payoff or backward as remaining time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $\theta>0$, $\sigma>0$, and $m\in\mathbb{R}$, and consider the Ornstein–Uhlenbeck process $dX_t=\theta(m-X_t)\,dt+\sigma\,dB_t$, with $X_0=x$. In integral form, the Brownian integral has integrand $H_t=\sigma$, which is adapted and progressively measurable and satisfies $\mathbb{E}\int_0^S H_t^2\,dt=\sigma^2S<\infty$ for every finite $S$. The drift $\theta(m-X_t)$ and diffusion coefficient $\sigma$ are globally Lipschitz in the state and satisfy linear growth bounds. The explicit solution below is a strong solution, constructed using the given Brownian motion; pathwise uniqueness also follows because the difference of two solutions driven by that same Brownian motion satisfies $d(X_t-\widetilde X_t)=-\theta(X_t-\widetilde X_t)\,dt$ with zero initial value.

X_t=m+(x-m)e^{-\theta t}+\sigma\int_0^t e^{-\theta(t-v)}\,dB_v.

The integral is an Ito integral: its progressively measurable integrand belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^t \sigma^2e^{-2\theta(t-v)}\,dv<\infty$. In particular, $X_t$ is normal with mean $m+(x-m)e^{-\theta t}$ and variance $\sigma^2(1-e^{-2\theta t})/(2\theta)$. All expectations below are under $\mathbb{P}$; no risk-neutral measure $\mathbb{Q}$ is being used.

Exponential-Linear Terminal Data

Let $\alpha\in\mathbb{R}$, let the constant killing rate $\rho\geq0$, and initially take the running potential to be zero. For remaining time $s\geq0$, define $u(s,x)=\mathbb{E}_x[e^{-\rho s}e^{\alpha X_s}]$, where $\mathbb{E}_x$ means that the process starts at $x$. The normal moment-generating function gives the formula directly.

Formula: For $\theta>0,\sigma>0,\rho\geq0$ and $\alpha,x\in\mathbb{R}$, $$ u(s,x)=\exp\left(-\rho s+\alpha\big[m+(x-m)e^{-\theta s}\big] +\frac{\alpha^2\sigma^2}{4\theta}(1-e^{-2\theta s})\right). $$

Worked Example: A Discounted Exponential Payoff

Take $\theta=1,\sigma=2,m=1,x=0,\alpha=1,\rho=\tfrac12$, and $s=\ln 2$. The conditional mean is $1+(0-1)e^{-\ln2}=1-\tfrac12=\tfrac12$. The variance is $\frac{2^2}{2}(1-e^{-2\ln2})=2(1-\tfrac14)=\tfrac32$. Thus the normal moment-generating function gives $\mathbb{E}_0[e^{X_s}]=\exp(\tfrac12+\tfrac12\cdot\tfrac32)=e^{5/4}$. The discount factor is $e^{-\rho s}=e^{-(\ln2)/2}=1/\sqrt2$, so

u(\ln2,0)=\frac{e^{5/4}}{\sqrt2}.

Check: Substituting the same parameters into the displayed formula gives exponent $-\tfrac12\ln2+\tfrac12+\frac{1\cdot 2^2}{4\cdot1}(1-\tfrac14) =-\tfrac12\ln2+\tfrac12+\tfrac34$. Exponentiating gives $e^{5/4}/\sqrt2$, as calculated from the Gaussian mean and variance.

Adding a Quadratic Potential

Now fix $q\geq0$ and include the running potential $qX_v^2$. The quantity of interest is $$ u(s,x)=\mathbb{E}_x\left[\exp\left(-\rho s-q\int_0^s X_v^2\,dv\right)e^{\alpha X_s}\right]. $$ The time integral is an ordinary integral along the continuous sample path. The exponential factor involving it is at most $1$, while $X_s$ is Gaussian, so this expectation is finite for every finite $s$ and every real $\alpha$.

The generator of $X_t$ is $\mathcal{L}f(x)=\theta(m-x)f'(x)+\tfrac12\sigma^2f''(x)$. Feynman–Kac suggests the initial-value equation $u_s=\mathcal{L}u-(\rho+qx^2)u$, $u(0,x)=e^{\alpha x}$. Use the exponential-quadratic ansatz $u(s,x)=\exp(a(s)x^2+d(s)x+c(s))$. Its derivatives are $u_x=(2ax+d)u$ and $u_{xx}=[2a+(2ax+d)^2]u$. In particular, the $\tfrac12 u_{xx}\sigma^2$ term contributes $\sigma^2a+\tfrac12\sigma^2(2ax+d)^2$; omitting this second-derivative correction would give a different equation.

Theorem: For $\theta>0,\sigma>0,q\geq0,\rho\geq0$ and $m,\alpha,x\in\mathbb{R}$, let $a,d,c$ solve $$ \begin{aligned} a'&=-2\theta a+2\sigma^2a^2-q, & a(0)&=0,\\ d'&=2\theta ma-\theta d+2\sigma^2ad, & d(0)&=\alpha,\\ c'&=\theta md+\sigma^2a+\tfrac12\sigma^2d^2-\rho, & c(0)&=0. \end{aligned} $$ Then the expectation with quadratic potential is $u(s,x)=\exp(a(s)x^2+d(s)x+c(s))$ for every finite $s\geq0$.

Proof. First, the coefficient equations follow by substituting the derivatives of the ansatz into $u_s=\mathcal{L}u-(\rho+qx^2)u$ and matching powers of $x$. In detail, the coefficient of $x^2$ on the right is $-2\theta a+2\sigma^2a^2-q$; the coefficient of $x$ is $2\theta ma-\theta d+2\sigma^2ad$; and the constant term is $\theta md+\sigma^2a+\tfrac12\sigma^2d^2-\rho$. The initial conditions give $u(0,x)=e^{\alpha x}$.

These coefficients exist and remain finite on finite time intervals. If $q=0$, uniqueness for the scalar ordinary differential equation gives $a=0$. If $q>0$, the roots of the right-hand side of the $a$-equation are $r_\pm=(\theta\pm\sqrt{\theta^2+2\sigma^2q})/(2\sigma^2)$, with $r_-<0<r_+$. Starting at $0$, the derivative is $-q<0$; uniqueness prevents the solution from crossing the equilibrium $r_-$, so $r_-<a(s)\leq0$ for finite $s$. Thus $a$ is bounded. The equation for $d$ is linear in $d$ with continuous bounded coefficients on each finite interval, so $d$ remains finite there; integrating its equation for $c$ shows $c$ is finite as well.

It remains to verify the expectation, rather than only the differential equation. Fix $S<\infty$ and set $Y_t=\exp(-\rho t-q\int_0^tX_v^2\,dv)\,u(S-t,X_t)$ for $0\leq t\leq S$. Ito’s formula applied to $u(S-t,X_t)$ has drift $-u_s+\mathcal{L}u$ and quadratic-variation correction $\tfrac12\sigma^2u_{xx}\,dt$, since $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. The product rule with the finite-variation discount factor adds drift $-(\rho+qX_t^2)u$. These drift terms cancel by the equation just verified. Hence $$ dY_t=\exp(-\rho t-q\int_0^tX_v^2\,dv)\,\sigma u_x(S-t,X_t)\,dB_t. $$ The integrand is progressively measurable. Boundedness of $a,d,c$ on $[0,S]$, together with $a\leq0$, gives $|u(s,x)|\leq C e^{D|x|}$ and $|u_x(s,x)|\leq C(1+|x|)e^{D|x|}$ for finite constants $C,D$. Each $X_t$ is Gaussian with mean and variance bounded uniformly over $0\leq t\leq S$, so $\sup_{0\leq t\leq S}\mathbb{E}[(1+|X_t|)^2e^{2D|X_t|}]<\infty$. The discount factor is at most $1$. Therefore the Ito integrand belongs to $\mathcal{L}^2$, and its Ito integral is a true martingale: square integrability gives integrability, and the conditional expectation of each future Ito integral increment given the present filtration is zero. It is not merely a local martingale. Consequently $\mathbb{E}[Y_S]=Y_0$. Since $u(0,X_S)=e^{\alpha X_S}$, this identity is precisely the asserted expectation formula. $\square$

Quadratic-Potential Examples

Worked Example: A Pure Quadratic Killing Potential

Set $\theta=\sigma=q=1$, $m=\alpha=\rho=0$. The coefficient equations reduce to $a'=-1-2a+2a^2$, $a(0)=0$, and $c'=a$, $c(0)=0$, with $d=0$. Write $r_\pm=(1\pm\sqrt3)/2$; then $2a^2-2a-1=2(a-r_+)(a-r_-)$. Let $y_0=r_+/r_-$ and $y(s)=y_0e^{2\sqrt3s}$. Separation of variables, or direct differentiation, gives

a(s)=\frac{r_+-r_-y(s)}{1-y(s)},\qquad c(s)=r_+s-\frac12\log\left(\frac{1-y(s)}{1-y_0}\right).

Indeed, $y(0)=y_0$, so $a(0)=0$ and $c(0)=0$. Since $y_0<0$, the denominators and logarithm arguments are positive for every $s\geq0$. Differentiating the displayed $a$ gives $a'=2(a-r_+)(a-r_-)=-1-2a+2a^2$; differentiating $c$ gives $c'=a$. The expectation is therefore $u(s,x)=e^{a(s)x^2+c(s)}$. Check: for $s>0$, the solution starts with $a'(0)=-1$ and stays below zero, so $c(s)=\int_0^s a(v)\,dv<0$. Thus at $x=0$, $u(s,0)=e^{c(s)}<1$, consistent with the pathwise bound $e^{-\int_0^sX_v^2\,dv}\leq1$.

Worked Example: A Nonzero Linear Terminal Exponent

Keep $\theta=\sigma=q=1$, $m=\rho=0$, and take $\alpha=1$. Use the $a(s)$ and $y(s)$ just computed. Define $c_0(s)=r_+s-\tfrac12\log((1-y(s))/(1-y_0))$, the constant coefficient when $\alpha=0$. The $d$-equation is $d'=(-1+2a)d$, with $d(0)=1$. Since $c_0'=a$, integration gives $d(s)=\exp(-s+2c_0(s))=(1-y_0)e^{\sqrt3s}/(1-y(s))$. The remaining coefficient satisfies $c'=a+\tfrac12d^2$, so $c=c_0+\tfrac12\int_0^s d(v)^2\,dv$.

This last integral can also be evaluated. Put $z=e^{\sqrt3v}$, so $y(v)=y_0z^2$ and $dv=dz/(\sqrt3z)$. Then $$ \int_0^s d(v)^2\,dv =\frac{(1-y_0)^2}{\sqrt3}\int_1^{e^{\sqrt3s}}\frac{z}{(1-y_0z^2)^2}\,dz =\frac{(1-y_0)^2}{2\sqrt3y_0} \left(\frac{1}{1-y(s)}-\frac{1}{1-y_0}\right). $$ For the specific remaining time $s=(\ln2)/\sqrt3$, $e^{\sqrt3s}=2$ and $y(s)=4y_0$. Thus $d(s)=2(1-y_0)/(1-4y_0)$ and $$ c(s)=\frac{r_+\ln2}{\sqrt3} -\frac12\log\left(\frac{1-4y_0}{1-y_0}\right) +\frac{(1-y_0)^2}{4\sqrt3y_0} \left(\frac{1}{1-4y_0}-\frac{1}{1-y_0}\right). $$ The answer at initial state $x$ is $u(s,x)=\exp(a(s)x^2+d(s)x+c(s))$, with $a(s)=(r_+-4r_-y_0)/(1-4y_0)$. Check: the integral term added to $c_0$ is positive: $y_0<0$, and the bracket $1/(1-4y_0)-1/(1-y_0)$ is negative, so their quotient is positive. Differentiating the general expression for $d$ gives $d'/d=-1+2a$, and differentiating $c=c_0+\tfrac12\int_0^s d^2$ gives $c'=a+\tfrac12d^2$, as required.

Remaining-Time Signs and Scope

The equations above use $s$ as time remaining, with initial values at $s=0$. This is why the potential contributes $-q$ to $a'$. In particular, when $\theta=\sigma=q=1$ and $a(0)=0$, the correct initial derivative is $a'(0)=-1$, not $+1$. A quick sign check is that a nonnegative killing potential should suppress the expectation; for zero terminal exponent and zero initial state, the quadratic-killing example indeed gives a value below $1$.

The quadratic-potential formula is not the same as simply inserting a modified variance into the Gaussian formula. The potential weights entire paths through $\int_0^sX_v^2\,dv$, not just the terminal value $X_s$. Its effect appears in the Riccati coefficient $a$, while the exponential-linear terminal payoff determines $d$. For $q=0$, $a=0$ and the equations reduce to the Gaussian moment calculation. The stochastic integral in the verification is Ito, not Stratonovich; no Stratonovich conversion or ordinary chain rule is used. The common shorthand $dB_t\,dB_t=dt$, with $dB_t\,dt=dt\,dt=0$, records the Ito quadratic-variation terms in that calculation.

Check Your Understanding

Use the stated assumptions $\theta,\sigma>0$, $q,\rho\geq0$, and finite remaining time in your answers.

Find the mean and variance of $X_s$ when $X_0=x$, and use them to calculate $\mathbb{E}_x[e^{\alpha X_s}]$.
For $\theta=\sigma=q=1$, what is the initial derivative $a'(0)$, and why does its sign agree with the effect of quadratic killing?
When $q=0$, identify the solution for $a$ and explain how the coefficient equations relate to the Gaussian formula.
In the Ito verification, why is the stochastic integral a true martingale rather than only a local martingale?
For the second worked example, verify directly that $y(0)=y_0$ implies $a(0)=c(0)=0$.

Ornstein-Uhlenbeck Expectations and Quadratic Potentials

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