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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)?

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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.

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  • $\begingroup$ so, how to solve my question? $\endgroup$ – A.Oreo Mar 2 '17 at 7:07
  • $\begingroup$ So your problem is that you don't know what $f(x)$ in my above notation is for a call option? $\endgroup$ – LocalVolatility Mar 2 '17 at 8:57
  • $\begingroup$ sorry, you may misunderstand my point, I mean how to use $V(\tau, x)$ to construct the boundary condition $v(\tau,x)=0,\ x\rightarrow -\infty.$, we know like down-out $v(\tau,x)=V(\tau,x)-V(\tau,-x)$ is not true for up-out case. $\endgroup$ – A.Oreo Mar 2 '17 at 9:23

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