The Heston Option-Pricing PDE and Boundary Conditions

From Heston Dynamics to a Pricing Equation

In the Heston model, the stock price and its variance are both state variables. Their Brownian shocks are correlated, so the pricing equation contains a mixed derivative that does not appear in a one-factor diffusion. The equation is a backward equation: its terminal condition is the option payoff, and its earlier values describe prices at earlier times.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{Q})$, where $\mathbb{Q}$ is the risk-neutral measure. This is distinct from the physical measure $\mathbb{P}$: the pricing drift below is specified under $\mathbb{Q}$. Let $B^S$ and $B^V$ be Brownian motions under $\mathbb{Q}$ with covariation $d[B^S,B^V]_t=\rho\,dt$, where $-1\le\rho\le1$. For $r$ constant and $\kappa,\theta,\xi>0$, the Heston dynamics are

dS_t=rS_t\,dt+S_t\sqrt{V_t}\,dB^S_t,\qquad dV_t=\kappa(\theta-V_t)\,dt+\xi\sqrt{V_t}\,dB^V_t,\qquad V_t\ge0.

Here $rS_t$ is the stock drift and $\sqrt{V_t}S_t$ is its diffusion coefficient; $\kappa(\theta-V_t)$ and $\xi\sqrt{V_t}$ are the corresponding variance drift and diffusion coefficient. Assume the variance process is nonnegative and the moments needed below are finite on the time interval considered. In particular, the stochastic integrands in these equations are progressively measurable and have finite expected integrated square on each finite horizon. If using only local integrability, the Ito calculations can first be stopped where the state variables are bounded.

Definition: For a European payoff $g(S_T)$, its risk-neutral price at time $t$, state $(s,v)$, is $C(t,s,v)=\mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}g(S_T)\mid S_t=s,V_t=v]$. For bounded $g$, this conditional expectation is integrable. For unbounded payoffs, such as a call, impose the relevant payoff-integrability condition. A classical PDE derivation additionally assumes $C$ has one continuous time derivative and two continuous state derivatives for $t<T,\ s>0,\ v>0$, with derivatives extending appropriately to a boundary when a boundary equation is used.

The covariance terms determine the second-order part of the equation. The quadratic variations and covariation of the state variables are

d[S]_t=S_t^2V_t\,dt,\qquad d[V]_t=\xi^2V_t\,dt,\qquad d[S,V]_t=\rho\xi S_tV_t\,dt.

These follow by multiplying the diffusion coefficients and using $dB^S_t\,dB^S_t=dt$, $dB^V_t\,dB^V_t=dt$, and $dB^S_t\,dB^V_t=\rho\,dt$. As usual, $dB_t\,dt=0$ and $dt\,dt=0$. The last covariance is why the PDE has a $C_{sv}$ term.

The Heston Backward Pricing PDE

Theorem: Under the preceding dynamics and regularity and integrability assumptions, a classical risk-neutral price $C$ of a European payoff $g(S_T)$ satisfies, for $t<T,\ s>0,\ v>0$, $$ C_t+rsC_s+\kappa(\theta-v)C_v +\frac12vs^2C_{ss}+\rho\xi vsC_{sv} +\frac12\xi^2vC_{vv}-rC=0, \qquad C(T,s,v)=g(s). $$

Proof. Fix an initial time $t_0$, and consider the discounted price process $M_t=e^{-r(t-t_0)}C(t,S_t,V_t)$, for $t_0\le t\le T$. For bounded $g$, the pricing definition and the Markov property give $M_t=\mathbb{E}^{\mathbb{Q}}[e^{-r(T-t_0)}g(S_T)\mid\mathcal{F}_t]$. The random variable inside this conditional expectation is integrable, so $M$ is a true martingale: it is integrable at each time, and the tower property gives $\mathbb{E}^{\mathbb{Q}}[M_u\mid\mathcal{F}_t]=M_t$ for $t\le u$. The same argument applies to an unbounded payoff when the stated integrability holds.

Apply Ito's formula to $C(t,S_t,V_t)$. The time derivative is $C_t\,dt$, the first-order terms are $C_s\,dS_t+C_v\,dV_t$, and the second-order correction is one half of the Hessian contracted with the quadratic covariations. Explicitly,

\begin{aligned} dC(t,S_t,V_t) ={}&\left(C_t+rS_tC_s+\kappa(\theta-V_t)C_v +\frac12S_t^2V_tC_{ss} +\rho\xi S_tV_tC_{sv} +\frac12\xi^2V_tC_{vv}\right)dt\\ &+S_t\sqrt{V_t}C_s\,dB^S_t +\xi\sqrt{V_t}C_v\,dB^V_t . \end{aligned}

All derivatives here are evaluated at $(t,S_t,V_t)$. The mixed term has coefficient $\rho\xi S_tV_t$: the two equal Hessian contributions $C_{sv}$ and $C_{vs}$ cancel the factor $1/2$. Multiplying by the discount factor adds the drift term $-rC$, because $d(e^{-r(t-t_0)})=-r e^{-r(t-t_0)}dt$, and the product of a finite-variation discount factor with a diffusion increment has zero quadratic covariation. Thus the drift of $M$ is the displayed PDE operator times $e^{-r(t-t_0)}$.

The stochastic terms, after localization if necessary, form a local martingale. Since $M$ is also a martingale, its finite-variation drift must vanish: a continuous local martingale of finite variation is constant. This standard fact follows, for example, by stopping to make it bounded and using that its quadratic variation is zero. Consequently the PDE operator is zero along the state process. The Markov pricing definition and the assumed interior regularity give the equation at the interior states where the price is defined. At maturity, the discount factor is one and the payoff is received, so $C(T,s,v)=g(s)$. $\square$

The martingale argument requires care. Ito's formula by itself produces a local martingale term, not automatically a true martingale. A local martingale need not have the integrability and conditional-expectation property of a martingale. In the proof, the pricing representation supplies that property for bounded payoffs; for unbounded payoffs it must be justified by integrability rather than inferred merely from the stochastic differential.

Terminal and Variance-Boundary Conditions

For a European call with strike $K>0$, the terminal condition is $C(T,s,v)=(s-K)^+$. The payoff depends on stock price, not on variance, so the terminal value is the same for every $v$. The terminal condition is not a boundary condition at $v=0$: maturity fixes the price at one time, whereas the variance boundary describes the equation's behavior along $v=0$ at earlier times.

At zero variance, the coefficients of $C_{ss}$, $C_{sv}$, and $C_{vv}$ vanish. If the derivatives extend continuously to $v=0$ and $C$ is a classical solution there, taking the limit $v\downarrow0$ in the PDE gives

C_t(t,s,0)+rsC_s(t,s,0)+\kappa\theta C_v(t,s,0)-rC(t,s,0)=0.

This is the reduced, degenerate equation at the variance boundary. It is not a prescribed Dirichlet value for $C(t,s,0)$; one should not assign an arbitrary value there in addition to the terminal payoff. For $\theta>0$, the variance drift at zero is $\kappa\theta>0$, so the boundary dynamics are not those of a variance process held fixed at zero.

The Feller condition $2\kappa\theta\ge\xi^2$ is the standard condition under which a CIR variance process started at $v>0$ does not hit zero. When it holds, there is no independently chosen boundary value at zero for the process started in the interior; the PDE solution is selected by the pricing representation and appropriate regularity and growth conditions. When it fails, zero can be attainable, and the probabilistic behavior of the variance process and suitable regularity still select the relevant solution. In either case, the reduced equation is the formal boundary equation for a classical solution with derivatives extending to zero; it should not be confused with a universal extra payoff condition. Conditions at large $s$ or $v$, when required for a particular PDE problem, are growth or uniqueness conditions and depend on the payoff class.

Proposition: If $C$ and the derivatives in the Heston PDE extend continuously to $v=0$, then the interior PDE implies the reduced equation above at $v=0$.

Proof. Write the interior PDE at $v>0$. As $v\downarrow0$, the terms $\tfrac12vs^2C_{ss}$, $\rho\xi vsC_{sv}$, and $\tfrac12\xi^2vC_{vv}$ converge to zero because the derivatives extend continuously and hence remain bounded near zero. Also $\kappa(\theta-v)C_v$ converges to $\kappa\theta C_v(t,s,0)$. The terms $C_t$, $rsC_s$, and $-rC$ converge to their boundary values by continuity. Taking the limit gives the stated equation. $\square$

Worked Example: Constant Payoff

Consider a claim paying the constant $A$ at time $T$, where $A$ is finite. Its candidate price is $C(t,s,v)=Ae^{-r(T-t)}$, independent of $s$ and $v$. Verify both its pricing representation and the PDE.

The discounted payoff is deterministic, so its conditional expectation is $Ae^{-r(T-t)}$. The derivatives are $C_t=rAe^{-r(T-t)}=rC$, $C_s=0$, $C_v=0$, $C_{ss}=0$, $C_{sv}=0$, and $C_{vv}=0$. Substitution into the PDE leaves $C_t-rC=rC-rC=0$; every state-derivative term is zero. At maturity, $C(T,s,v)=Ae^0=A$.

Boundary check: at $v=0$, the reduced equation becomes $C_t-rC=0$, and the same calculation gives zero. Thus this claim satisfies both the terminal condition and the variance-boundary equation. The result also agrees directly with discounting a deterministic payment; no volatility parameter can affect its price.

Worked Example: The Mixed-Derivative Coefficient

At $s=2$, $v=\tfrac14$, take $\rho=-\tfrac12$ and $\xi=\tfrac12$. Compute the coefficient multiplying $C_{sv}$ in the PDE, and compare it with the variance coefficient multiplying $C_{vv}$.

The mixed-derivative coefficient is $\rho\xi vs$. Substitution gives $(-\tfrac12)(\tfrac12)(\tfrac14)(2)=-\tfrac18$. The $C_{vv}$ coefficient is $\tfrac12\xi^2v$, which equals $\tfrac12(\tfrac12)^2(\tfrac14)=\tfrac1{32}$. The sign of the mixed coefficient is negative because the Brownian shocks are negatively correlated; the pure second-derivative coefficient remains nonnegative.

Check: the covariance of the stock and variance increments per unit time is $\rho\xi sv$, so the same substitution gives $(-\tfrac12)(\tfrac12)(2)(\tfrac14)=-\tfrac18$, exactly the mixed-derivative coefficient. If $\rho=0$, that covariance and the mixed coefficient are both zero.

Worked Example: An Affine PDE Check

Consider $C(t,s,v)=e^{-r(T-t)}s$. Check directly that this function solves the interior PDE and has terminal value $s$. This is an algebraic PDE check; it does not replace the integrability conditions needed to interpret an unbounded stock payoff as a risk-neutral expectation.

The derivatives are $C_t=rC$, $C_s=e^{-r(T-t)}$, $C_v=0$, $C_{ss}=0$, $C_{sv}=0$, and $C_{vv}=0$. Therefore the PDE left side is $rC+rs e^{-r(T-t)}-rC$. Since $C=s e^{-r(T-t)}$, this equals $rs e^{-r(T-t)}$, which is not zero—so this candidate fails.

The failed check identifies a discounting error: a stock with risk-neutral drift $rS$ has price $s$, not $e^{-r(T-t)}s$, when the terminal payoff is $S_T$. Instead set $C(t,s,v)=s$. Then $C_t=0$, $C_s=1$, and all other state derivatives are zero. The PDE left side is $0+rs-rs=0$, and $C(T,s,v)=s$. The boundary equation also gives $0+rs-rs=0$.

Independent check: the drift of the stock under $\mathbb{Q}$ is $rS_t$, so discounting the stock by $e^{-rt}$ removes that drift in Ito's formula. This is consistent with the PDE solution $C=s$ for a stock-delivery claim, whenever the discounted stock is a true martingale and the expectation is well-defined.

Common Pitfalls and Why the Equation Matters

A frequent error is to omit the mixed derivative because each state variable has its own variance term. That is correct only when the Brownian shocks are uncorrelated, or when the particular function has $C_{sv}=0$. The covariance $[S,V]$, not just the two individual quadratic variations, must be included in the two-variable Ito formula.

Another error is to treat $v=0$ as an ordinary boundary at which an arbitrary price can be prescribed. The second-order coefficients vanish there, but the variance drift does not: it equals $\kappa\theta$. The resulting first-order boundary equation expresses the limiting PDE, subject to regularity; it is not a replacement for the terminal payoff or a free choice of boundary data.

Finally, distinguish a pricing equation from a proof of uniqueness. The derivation shows that a sufficiently regular risk-neutral price satisfies the PDE. To identify a PDE solution uniquely with a price, one also needs suitable growth and boundary requirements and conditions ensuring the pricing expectation is valid. A local martingale obtained from Ito's formula alone does not establish those facts.

The Heston PDE matters because it translates correlated stochastic dynamics into a deterministic equation for prices, making the roles of mean reversion, volatility of variance, correlation, discounting, and payoff conditions explicit. Its degeneracy at zero variance is part of the model structure, not an algebraic defect.

Check Your Understanding

Use the risk-neutral dynamics and the stated regularity assumptions when answering.

Which quadratic covariation produces the mixed derivative, and what is its coefficient in the pricing PDE?
Write the terminal condition for a European put with strike $K$.
Take the limit of the Heston PDE as $v\downarrow0$ and state which terms vanish.
Why does a local-martingale term in Ito's formula not, by itself, prove a conditional-expectation pricing identity?
For a constant terminal payoff $A$, verify the boundary equation for $C(t,s,v)=Ae^{-r(T-t)}$.

The Heston Option-Pricing PDE and Boundary Conditions

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