Pricing an Up-and-Out Call with the Heat Equation
Summary
The post asks how to solve the Black–Scholes heat-equation formulation for an up-and-out call. It contrasts the option's absorbing barrier condition with the vanilla call problem and the down-and-out case, where a reflection construction is suggested. The central implementation idea in the answer is to define an auxiliary full-line heat problem whose initial payoff is restricted to the option's active region using an indicator function.
That auxiliary solution can be obtained by convolving the restricted initial condition with the heat kernel, after which it can be used to construct the barrier solution. The discussion emphasizes that the initial condition must match the payoff as well as the boundary condition. It outlines a PDE method, but does not carry the construction through to a closed-form price or provide numerical examples; the coordinate and boundary conventions must be handled consistently when applying it.
Key ideas
- A barrier option's heat-equation problem is determined by both its boundary condition and its payoff-based initial condition.
- For an up-and-out call, restrict the initial payoff to the domain where the option remains active.
- Solve an auxiliary full-line heat equation with that restricted initial condition, for example by heat-kernel convolution.
- The post describes a construction method but does not derive a final pricing formula or demonstrate numerical results.
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# How to price up-out-call by solving heat equation like down-out-call
# How to price up-out-call by solving heat equation like down-out-call
We know that by changing the variables we can obtain the `Black-Scholes` formula of `vanilla call` through solving the `heat equation`: $$S = Be^{x},\quad t = T - \tau/\dfrac{1}{2}\sigma^2,\quad C(t,S) = Bu(\tau, x),\quad v(\tau,x)=e^{\alpha x+\beta\tau}u(\tau,x)$$ $$\dfrac{\partial v}{\partial \tau} = \dfrac{\partial^2 v}{\partial x^2}$$ with boundary and initial conditions: $$v(\tau,x)=0, x\rightarrow -\infty;$$ $$v(\tau,x)\sim (e^x-e^{-r\tau}), x\rightarrow \infty;$$ $$v(0,x) = u_0(x).$$ And, for the `down-out-call`, only the boundary conditions change: $$v(\tau,x)=0, x=0;$$ $$v(\tau,x)\sim (e^x-e^{-r\tau}), x\rightarrow \infty;$$ $$x\in(0,\infty)$$ then we can use the `reflection property` of `heat equation`
Let $$v(\tau,x) = V(\tau,x)-V(\tau,-x)$$ here $V(\tau,x)$ is the solution in the first `vanilla call heat equation`, to extend $v(\tau,x)$ to $x\in(-\infty,\infty);$
But, for the `up-out-call`, the boundary conditions become: $$v(\tau,x)=0, x\rightarrow-\infty;$$ $$v(\tau,x)=0,x=0;$$ $$x\in(-\infty,0)$$ I don't how to use $V(\tau,x)$ to construct the above conditions(surely, include the initial condition)？
## Answer by LocalVolatility (score 1)
https://quant.stackexchange.com/a/32756
What you seem to omit is the initial condition for $v(\tau, x)$? Assume you have an up-and-out barrier option for which $v$ satisfies the initial boundary value problem
\begin{eqnarray} \mathcal{H} \{ v \} (\tau, x) & = & 0 \qquad \text{for } (\tau, x) \in \mathbb{R}_+ \times \mathbb{R}_-\\ v(0, x) & = & f(x)\\ v(\tau, 0) & = & 0 \qquad \text{for } \tau \in [0, \infty). \end{eqnarray}
Here, $\mathcal{H}$ is the heat equation operator and $f(x)$ is the payoff specific initial condition - in your case that of a call option. Then you define the auxiliary initial value problem $V(\tau, x)$ which satisfies
\begin{eqnarray} \mathcal{H} \{ V \} (\tau, x) & = & 0 \qquad \text{for } (\tau, x) \in \left( \mathbb{R}_+, \mathbb{R} \right)\\ V(0, x) & = & f(x) \mathrm{1} \{ x < 0 \}. \end{eqnarray}
$V(\tau, x)$ is the corresponding full-range problem where you keep the initial condition but restrict it to the active domain of the barrier option via the indicator. You solve for $V(\tau, x)$ by e.g. a convolution of the initial condition with the heat kernel.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.