The Ito Verification Process for a Candidate PDE Solution

From a Candidate PDE to a Stochastic Process

For a diffusion, Ito’s formula converts derivatives of a smooth function into the drift and diffusion terms of a stochastic process. If the function is intended to solve a backward equation, its PDE residual becomes exactly the drift of an exponentially discounted process. This gives a practical verification method: calculate the residual, check whether it vanishes, and then determine whether the resulting local martingale is a true martingale.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. On a fixed finite horizon $[0,T]$, let $X_t$ be a continuous adapted solution, with deterministic initial value $X_0=x$, of $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$. Here $\mu$ is the drift coefficient and $\sigma$ is the diffusion coefficient. Assume the coefficient processes along the solution are progressively measurable and satisfy $\int_0^T (|\mu(s,X_s)|+\sigma(s,X_s)^2)\,ds<\infty$ almost surely. Integrals with respect to $B$ below are Ito integrals: their integrands are progressively measurable, and when an integrand is asserted to be in $\mathcal{L}^2$, it satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: For $u\in C^{1,2}([0,T]\times\mathbb{R})$, define its discounted backward-equation residual by $$ R_u(t,x)=u_t(t,x)+\mu(t,x)u_x(t,x)+\tfrac12\sigma(t,x)^2u_{xx}(t,x)-r u(t,x), $$ where $r$ is a constant discount rate. The associated verification process is $Y_t=e^{-rt}u(t,X_t)$.

The derivatives in this definition are evaluated at $(t,x)$. The term $\mu u_x+\tfrac12\sigma^2u_{xx}$ is the diffusion generator applied to $u$; subtracting $ru$ accounts for discounting. The candidate backward PDE is $R_u=0$, with terminal condition $u(T,x)=g(x)$ when the desired terminal payoff is $g(X_T)$.

The Ito Residual Identity

Proposition: Under the assumptions above, the discounted process has the decomposition $$ dY_t=e^{-rt}R_u(t,X_t)\,dt+e^{-rt}u_x(t,X_t)\sigma(t,X_t)\,dB_t. $$ The identity holds locally, with localization when the drift and stochastic integral are not globally integrable. If $e^{-rs}u_x(s,X_s)\sigma(s,X_s)$ belongs to $\mathcal{L}^2$ on $[0,T]$ and $\mathbb{E}\int_0^T e^{-rs}|R_u(s,X_s)|\,ds<\infty$, then for every $t\in[0,T]$, $$ \mathbb{E}[Y_t]=u(0,x)+\mathbb{E}\int_0^t e^{-rs}R_u(s,X_s)\,ds. $$

Proof. First apply Ito’s formula to $u(t,X_t)$. Its time derivative is $u_t(t,X_t)$, its first spatial derivative is $u_x(t,X_t)$, and its second spatial derivative is $u_{xx}(t,X_t)$. The quadratic-variation correction is $\tfrac12u_{xx}(t,X_t)\sigma(t,X_t)^2\,dt$, since $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. Thus $$ du(t,X_t)=\left(u_t+\mu u_x+\tfrac12\sigma^2u_{xx}\right)(t,X_t)\,dt+u_x(t,X_t)\sigma(t,X_t)\,dB_t. $$ For the product $e^{-rt}u(t,X_t)$, the time derivative of the discount factor is $-r e^{-rt}$. The discount factor has finite variation, so its covariation with the stochastic part is zero. The product rule therefore subtracts $r e^{-rt}u(t,X_t)\,dt$, giving the stated residual identity.

To justify the calculation without assuming that the coefficients are bounded on compact state intervals, set $\tau_n=\inf\{t\le T: |X_t|\ge n\text{ or }\int_0^t(|\mu(s,X_s)|+\sigma(s,X_s)^2)\,ds\ge n\}\wedge T$. The stopping times increase to $T$ almost surely: each continuous path of $X$ is bounded on $[0,T]$, and the accumulated coefficient integral is finite almost surely. On $[0,\tau_n]$, the derivatives of $u$ are bounded because $(t,X_t)$ stays in the compact set $[0,T]\times[-n,n]$. Also, the integrals of $|\mu|$ and $\sigma^2$ up to $\tau_n$ are bounded by $n$. Consequently, the stopped drift terms are integrable and the stopped stochastic integrand is square-integrable. Ito’s formula applies to the stopped process. If the unstopped integrand belongs to $\mathcal{L}^2$, the Ito integral is a martingale: its increments have conditional expectation zero, by the Ito integral construction, and are integrable by the Ito isometry. Taking expectations in the stopped identity and then passing to the unstopped identity under the stated integrability assumptions proves the expectation formula. $\square$

The local identity and the expectation identity have different hypotheses. Without global integrability, the stochastic integral defines only a local martingale in general; its expectation need not be zero. In particular, an unbounded local martingale need not be a true martingale. Localization justifies the differential calculation, but does not by itself justify removing the stopping in an expectation.

Worked Applications

Worked Example: A Discounted Quadratic Payoff

Let $X_t=x+B_t$, let $r\ge0$, and take terminal payoff $g(x)=x^2$. Verify the candidate $u(t,x)=e^{-r(T-t)}(x^2+T-t)$.

Write $\tau=T-t$. The derivatives are $u_t=e^{-r\tau}\big(r(x^2+\tau)-1\big)$, $u_x=2xe^{-r\tau}$, and $u_{xx}=2e^{-r\tau}$. Here $\mu=0$ and $\sigma=1$, so $$ R_u=e^{-r\tau}\big(r(x^2+\tau)-1\big)+e^{-r\tau}-r e^{-r\tau}(x^2+\tau)=0. $$ Also, $u(T,x)=x^2$. Thus the PDE and terminal condition both hold.

For a direct Ito check, set $F(t,x)=e^{-rt}u(t,x)=e^{-rT}(x^2+T-t)$. Its derivatives are $F_t=-e^{-rT}$, $F_x=2xe^{-rT}$, and $F_{xx}=2e^{-rT}$. Applying Ito’s formula, including the correction $\tfrac12F_{xx}\,dt=e^{-rT}\,dt$, gives $$ dF(t,X_t)=\big(-e^{-rT}+e^{-rT}\big)\,dt+2e^{-rT}X_t\,dB_t=2e^{-rT}X_t\,dB_t. $$ The stochastic integrand is in $\mathcal{L}^2$, because $\mathbb{E}\int_0^T (2e^{-rT}X_s)^2ds=4e^{-2rT}\int_0^T(x^2+s)\,ds<\infty$. Hence $Y$ is a true martingale. Its expectation at time $t$ is $e^{-rT}(x^2+t+T-t)=e^{-rT}(x^2+T)$. At $T$, this is $e^{-rT}\mathbb{E}[X_T^2]$, since $\mathbb{E}[X_T^2]=x^2+T$. The direct moment calculation agrees with the Ito verification.

Worked Example: Detecting a Nonzero Discount Residual

Let $r=\tfrac12$, $X_t=B_t$, and use the candidate $u(t,x)=1$ on $[0,2]$. This candidate has terminal value $1$, but does it solve the discounted PDE?

The derivatives are $u_t=0$, $u_x=0$, and $u_{xx}=0$. Therefore $R_u=0+0+0-\tfrac12=-\tfrac12$, so the candidate does not solve the PDE. Its verification process is $Y_t=e^{-t/2}$. Applying Ito’s formula to $F(t,x)=e^{-t/2}u(t,x)=e^{-t/2}$, we have $F_t=-\tfrac12e^{-t/2}$, $F_x=0$, and $F_{xx}=0$. The correction term is $\tfrac12F_{xx}\,dt=0$, and $$ dY_t=-\tfrac12 e^{-t/2}\,dt=e^{-t/2}R_u(t,B_t)\,dt. $$ Integrating to time $2$ gives $Y_2-Y_0=e^{-1}-1$, which agrees with $e^{-1}-1$ from the explicit values $Y_2=e^{-1}$ and $Y_0=1$. The nonzero residual exactly accounts for the change in the discounted process.

Worked Example: Checking Drift and Diffusion Contributions

Let $X_t=1+2t+3B_t$, so $dX_t=2\,dt+3\,dB_t$, and take $u(t,x)=x$, $r=0$, and $T=1$. Compute the residual and verify the resulting expectation.

The derivatives are $u_t=0$, $u_x=1$, and $u_{xx}=0$. With $\mu=2$ and $\sigma=3$, the residual is $R_u=0+2\cdot1+\tfrac12(3^2)\cdot0=2$. For the direct Ito calculation, $F(t,x)=u(t,x)=x$, with $F_t=0$, $F_x=1$, $F_{xx}=0$; the correction $\tfrac12F_{xx}\sigma^2dt$ is zero. Hence $du(t,X_t)=2\,dt+3\,dB_t$. The stochastic integrand $3$ is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^1 3^2ds=9$, so taking expectations gives $\mathbb{E}[u(1,X_1)]=u(0,1)+\int_0^1 2\,ds=3$. Directly, $X_1=3+3B_1$, and $\mathbb{E}[X_1]=3$. Both calculations agree.

When the Verification Process Is a Martingale

Proposition: If $R_u(t,X_t)=0$ for almost every $t$, almost surely, then $Y_t=e^{-rt}u(t,X_t)$ is a local martingale. If additionally $e^{-rs}u_x(s,X_s)\sigma(s,X_s)\in\mathcal{L}^2$, then $Y$ is a true martingale. Alternatively, a bounded local martingale $Y$ on $[0,T]$ is a true martingale.

Proof. When the residual vanishes, the Ito residual identity reduces to $Y_t=Y_0+\int_0^t e^{-rs}u_x(s,X_s)\sigma(s,X_s)\,dB_s$, locally. This is a local martingale by the definition of the Ito integral. If the integrand is in $\mathcal{L}^2$, the Ito isometry gives a square-integrable martingale, so for $s\le t$, $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s$. For the bounded alternative, take a localizing sequence $\tau_n$ such that each $Y_{t\wedge\tau_n}$ is a martingale. For $s\le t$, its conditional expectation identity is $\mathbb{E}[Y_{t\wedge\tau_n}\mid\mathcal{F}_s]=Y_{s\wedge\tau_n}$. If $|Y_q|\le C$ for all $q\in[0,T]$, both stopped variables are bounded by $C$. Since $\tau_n$ increases to $T$, they converge almost surely to $Y_t$ and $Y_s$, respectively. Bounded convergence, also in conditional expectation, gives $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s$. Thus $Y$ is a true martingale in either case. $\square$

Consequently, if $u(T,x)=g(x)$, $R_u=0$, and one of the true-martingale conditions holds, then $u(0,x)=\mathbb{E}[e^{-rT}g(X_T)]$. This is a verification statement for a candidate: it shows that a sufficiently regular solution with the required integrability has the stochastic representation. It does not, by itself, prove that a classical solution exists.

Common Pitfall: Reversing the Discount Sign

The discount factor has derivative $\frac{d}{dt}e^{-rt}=-re^{-rt}$, so its contribution to the residual is $-ru$, not $+ru$. For the constant candidate $u=1$ with $r=\tfrac12$, using the wrong sign would give residual $+\tfrac12$. That predicts $dY_t=+\tfrac12e^{-t/2}dt$, even though the direct derivative of $Y_t=e^{-t/2}$ is $-\tfrac12e^{-t/2}dt$. The corrected residual is $-\tfrac12$, matching both the direct derivative and Ito’s formula.

A second distinction is equally important: zero residual gives a local martingale first. A true-martingale conclusion requires an additional justification, such as square-integrability of the Ito integrand or boundedness of the process. Only then can expectations be passed to the terminal time without an unjustified localization step.

Check Your Understanding

For each question, keep the residual calculation separate from the martingale-integrability check.

For $dX_t=\mu\,dt+\sigma\,dB_t$ and $u(t,x)=x$, compute $R_u$ for a constant discount rate $r$.
For $u(t,x)=x^2$, $r=0$, and $dX_t=dB_t$, compute the residual and the Ito correction term.
State a sufficient condition that turns the local martingale in the zero-residual case into a true martingale.
Why does stopping when $|X_t|$ reaches $n$ alone not ensure that $\mu(t,X_t)$ and $\sigma(t,X_t)$ are bounded under the assumptions used here?
If $u(T,x)=g(x)$ and the verification process is a true martingale, write the resulting formula for $u(0,x)$.