Mixed Terminal, Running, and Exit-Time Problems

One Expectation, Several Payoff Mechanisms

A diffusion may accumulate rewards while it runs, receive a terminal payoff if it reaches the end of a fixed horizon, and instead receive a boundary payoff if it exits a domain first. Discounting weights both the accumulated rewards and whichever payoff occurs. The corresponding PDE must account for all four features at once: diffusion and drift in the domain, a potential term from discounting, a source term from running rewards, and data on both the terminal and spatial boundaries.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. For a starting time $t$ and point $x$, consider a strong solution $X_s$, $s\ge t$, of $dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s$, with $X_t=x$. The drift is $\mu$, and the diffusion coefficient is $\sigma$. We assume the coefficients are measurable in time, uniformly Lipschitz in space, and of linear growth, so the usual existence and pathwise uniqueness result for strong solutions applies. All stochastic integrals below are Itô integrals. Their integrands are progressively measurable and belong to $\mathcal{L}^2$ on each stated finite horizon: that is, $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Let $D$ be a bounded open interval or a bounded domain with sufficiently regular boundary, and define the first exit time after the start by $\tau_D=\inf\{s\ge t:X_s\notin D\}$. For a fixed horizon $T>t$, set $\tau=\tau_D\wedge T$. Let $f$ be a running reward, $c$ a discount rate, $h$ a terminal payoff, and $g$ a boundary payoff. The discount factor from the starting time to $s$ is $A_s=\exp(-\int_t^s c(r,X_r)\,dr)$. In the formula below, exit strictly before $T$ triggers $g$; otherwise the terminal payoff is $h$. Assume $h$ and $g$ agree at the terminal-boundary intersection, so the convention at an exit exactly at $T$ causes no conflict.

Definition: The mixed terminal, running, and exit-time value is $$ u(t,x)=\mathbb{E}_{t,x}\!\left[ \int_t^\tau A_s f(s,X_s)\,ds +\mathbf{1}_{\{\tau_D<T\}}A_{\tau_D}g(\tau_D,X_{\tau_D}) +\mathbf{1}_{\{\tau_D\ge T\}}A_T h(X_T) \right], $$ where $\mathbb{E}_{t,x}$ denotes expectation for the diffusion started from $X_t=x$. The running integral stops at the same time as the payoff rule: the boundary payoff is used on exit before $T$, and the terminal payoff is used if the diffusion remains in $D$ until $T$.

On $D$, define the diffusion generator by $\mathcal{L}_s\varphi(x)=\mu(s,x)\varphi'(x)+\tfrac12\sigma(s,x)^2\varphi''(x)$ in one dimension, with the corresponding gradient and Hessian expression in higher dimensions. The factor $\tfrac12$ is the Itô correction; it comes from $dB_s\,dB_s=ds$, while $dB_s\,ds=0$ and $ds\,ds=0$.

The Mixed Feynman–Kac Verification

Theorem: Suppose $c,f$ are bounded on $[t,T]\times\overline D$, the coefficients are bounded there, and $u$ is bounded, continuous up to the parabolic boundary, and of class $C^{1,2}$ in the interior. Suppose also that $u$ satisfies $$ u_t+\mathcal{L}_s u-cu+f=0\quad\text{in }[t,T)\times D,\qquad u(T,x)=h(x),\qquad u(s,y)=g(s,y)\quad(y\in\partial D). $$ If $u_x$ (or $\nabla u$ in higher dimensions) is bounded in the domain, then $u(t,x)$ equals the mixed expectation in the definition.

Proof. Apply Itô’s formula to $A_su(s,X_s)$, stopped at $\tau$. For the function $u(s,z)$, the time derivative is $u_t$, the spatial derivatives are $u_x$ and $u_{xx}$, and the second-order Itô correction is $\tfrac12\sigma(s,X_s)^2u_{xx}(s,X_s)\,ds$. Since $dA_s=-c(s,X_s)A_s\,ds$, the product rule gives $$ d(A_su(s,X_s))=A_s(u_t+\mathcal{L}_s u-cu)(s,X_s)\,ds +A_s\sigma(s,X_s)u_x(s,X_s)\,dB_s. $$ The PDE makes the drift term equal to $-A_sf(s,X_s)\,ds$. Integrating from $t$ to $\tau$ therefore gives $$ A_\tau u(\tau,X_\tau)+\int_t^\tau A_sf(s,X_s)\,ds =u(t,x)+\int_t^T\mathbf{1}_{\{s\le\tau\}}A_s\sigma(s,X_s)u_x(s,X_s)\,dB_s. $$ The stopped integrand is progressively measurable. Boundedness of $A,\sigma,u_x$ and the finite horizon imply that its expected squared integral is finite, so it belongs to $\mathcal{L}^2$. By the Itô isometry the stochastic integral is square-integrable; as an Itô integral with a square-integrable adapted integrand, it is a true martingale and has expectation zero. At exit before $T$, $u(\tau_D,X_{\tau_D})=g(\tau_D,X_{\tau_D})$; if the process reaches $T$ without an earlier exit, $u(T,X_T)=h(X_T)$. Taking expectations proves the claimed representation. $\square$

This is a verification result: a sufficiently regular solution of the PDE gives the expectation. It does not, by itself, establish existence of such a solution. Boundedness and the square-integrability check matter: without them the stochastic integral may only be a local martingale, and taking expectations in the stopped identity would require further justification.

Worked Mixed Payoff: Exit or Terminal Time

Worked Example: A Running Reward with an Elapsed-Time Check

Let $X_s=x+B_s-B_t$ start at $x\in(0,1)$ at time $t$, and let $T=1$. The domain is $D=(0,1)$, there is no discounting, and the running reward is $f=1$. Set the terminal payoff to $h(x)=x$, and set the exit payoff to $g(s,y)=y+1-s$ for $y\in\{0,1\}$. Find and verify the value.

Consider $u(s,y)=y+1-s$. Its derivatives are $u_s=-1$, $u_y=1$, and $u_{yy}=0$. Here $\mu=0$, $\sigma=1$, and $\mathcal{L}u=\tfrac12u_{yy}=0$. Thus $u_s+\mathcal{L}u+f=-1+0+1=0$. At $s=1$, $u(1,y)=y=h(y)$, and at either boundary point $y$, $u(s,y)=g(s,y)$.

For a path that exits before time $1$, the exit payoff plus accumulated reward is $X_{\tau_D}+1-\tau_D+(\tau_D-t)=X_{\tau_D}+1-t$. For a path that survives to time $1$, the terminal payoff plus accumulated reward is $X_1+(1-t)$. In both cases the total equals $X_\tau+1-t$, where $\tau=\tau_D\wedge1$.

To check its expectation directly, $X_{s\wedge\tau}=x+\int_t^s\mathbf{1}_{\{r\le\tau\}}\,dB_r$ is a true martingale: its integrand is bounded and its squared integral has expectation at most $1-t$. Hence $\mathbb{E}_{t,x}[X_\tau]=x$, and the value is $x+1-t$, precisely $u(t,x)$.

Check: for $t=0$ and $x=\tfrac12$, the value is $\tfrac12+1=\tfrac32$. The running reward is the elapsed time $\tau-t$, not $\tau$. In particular, before exit the total payoff is $X_{\tau_D}+1-\tau_D+\tau_D-t=X_{\tau_D}+1-t$, whose expectation is $x+1-t$, not $x$.

Worked Discounted Example Without a Spatial Boundary

Worked Example: Discounted Terminal Data and a Constant Reward

Let $X_s=x+B_s-B_t$ on the whole real line, with horizon $T$, constant discount rate $c=1$, running reward $f=2$, and terminal payoff $h(x)=x^2$. There is no exit boundary. Put $\Delta=T-t$. Compute the expectation.

Since $X_T=x+B_T-B_t$, its mean is $x$ and its variance is $\Delta$, so $\mathbb{E}_{t,x}[X_T^2]=x^2+\Delta$. The discounted terminal contribution is $e^{-\Delta}(x^2+\Delta)$. The discounted running contribution is $\int_t^T2e^{-(s-t)}\,ds=2(1-e^{-\Delta})$, by direct integration. Therefore $$ u(t,x)=e^{-\Delta}(x^2+\Delta)+2(1-e^{-\Delta}). $$

Verify the PDE $u_t+\tfrac12u_{xx}-u+2=0$. Write $u=F(\Delta,x)$, where $\Delta=T-t$. Differentiating the displayed expression gives $u_t=e^{-\Delta}(x^2+\Delta-3)$, $u_x=2xe^{-\Delta}$, and $u_{xx}=2e^{-\Delta}$. Substitution yields $$ u_t+\tfrac12u_{xx}-u+2 =e^{-\Delta}(x^2+\Delta-3)+e^{-\Delta} -e^{-\Delta}(x^2+\Delta)-2(1-e^{-\Delta})+2=0. $$ At $t=T$, $\Delta=0$, so $u(T,x)=x^2$, as required.

Check: for $x=1$ and $T-t=1$, the expectation is $2e^{-1}+2(1-e^{-1})=2$. Independently, the expected terminal payoff is $\mathbb{E}[X_T^2]=2$, but it is discounted; the running reward contributes the complementary amount $2(1-e^{-1})$, giving the stated total.

Worked Example: All Four Ingredients on an Interval

Worked Example: Discounting, Running Reward, Terminal Payoff, and Exit

Let $X_s=x+B_s-B_t$ in $D=(0,1)$, with $c=1$ and running reward $f=1$. Give boundary rewards $g(s,0)=0$ and $g(s,1)=1$, and choose the terminal payoff $h$ to equal the value function found below. Set $k=\sqrt{2}$, and define $u(s,x)=1-\sinh(k(1-x))/\sinh(k)$, independent of $s$.

Its spatial derivatives are $u_x=k\cosh(k(1-x))/\sinh(k)$ and $u_{xx}=-k^2\sinh(k(1-x))/\sinh(k)$. Since $k^2=2$, $$ \tfrac12u_{xx}-u+1 =-\frac{\sinh(k(1-x))}{\sinh(k)} -\left(1-\frac{\sinh(k(1-x))}{\sinh(k)}\right)+1=0. $$ Also $u_t=0$. At $x=0$, $u=1-\sinh(k)/\sinh(k)=0$; at $x=1$, $u=1-\sinh(0)/\sinh(k)=1$. Taking $h(x)=u(x)$ makes the terminal condition hold, including compatibility at the two corners.

The theorem now gives the value as the expectation of the discounted running reward up to $\tau=\tau_D\wedge T$, plus the discounted boundary reward if exit occurs before $T$, or the discounted terminal reward if the process survives to $T$. For example, at $x=\tfrac12$, the value is exactly $1-\sinh(k/2)/\sinh(k)$. No numerical approximation is needed to verify it: the PDE and both boundary values have just been checked by substitution.

Check: if the running reward were removed while all other data were unchanged, the PDE would instead contain $\tfrac12u_{xx}-u=0$. Thus the $+1$ in the verified equation is specifically the source term from the running reward; it is not part of the discount factor.

Why the Stopping Rule and the Conditions Matter

The mixed formula is useful because it separates three economically and probabilistically distinct contributions: the running integral is accumulated only until stopping, the boundary payoff is used only on exit before the horizon, and the terminal payoff is used only if the process survives. The discount factor starts at the process’s starting time $t$, so the discount to a later time $s$ is $e^{-\int_t^s c(r,X_r)\,dr}$, not a factor measured from time zero.

A frequent error is to confuse calendar time with elapsed time. If a process starts at $t$, a unit running reward accumulated until an exit at $\tau_D$ contributes $\tau_D-t$. In the first example the exit payoff is $X_{\tau_D}+1-\tau_D$; adding the running reward gives $X_{\tau_D}+1-t$. Since the stopped Brownian motion has expectation $x$, the expected total is $x+1-t$. Omitting the subtraction of $t$ changes the answer.

Other checks are equally important. The terminal datum must be imposed at $T$, whereas the spatial boundary datum is imposed at exit. If they disagree at the corner, the value’s boundary convention must be stated and regularity at that corner may fail. Finally, a formal Itô calculation is not enough to justify taking expectations: one must establish that the stopped stochastic integral is a true martingale, for example by showing its integrand belongs to $\mathcal{L}^2$. Without that integrability, a local martingale need not have constant expectation.

Check Your Understanding

Use the stopped representation and check both the payoff convention and the corresponding PDE.

For a process started at time $t$, what is the accumulated reward when the running reward is the constant $q$ and exit occurs at $\tau_D$?
Write the PDE and terminal condition for a discounted diffusion with no boundary, running reward $f$, and terminal payoff $h$.
In the first worked example, substitute $u(s,y)=y+1-s$ into the PDE and verify the boundary and terminal data.
Why does boundedness of the stopped Itô integrand imply that the stochastic integral in the verification proof has expectation zero?
In the final example, verify the two boundary values of $u(x)=1-\sinh(\sqrt{2}(1-x))/\sinh(\sqrt{2})$.