Finite-Difference PDE Boundaries for Fixed-Strike Asian Calls

# Finite difference methods for an Asian call with boundary conditions


# Finite difference methods for an Asian call with boundary conditions












I have a question please.

I have to find the price of a Asian call using a finite diffenrece method.

Here the article, if u want to look it up, it's page 2-4: "https://www.researchgate.net/publication/305974482_The_value_of_an_Asian_option"

The idea of this article is : we define $$ \phi(t, x) \equiv \mathbb{E}\left[\left(\int_{t}^{T} S_{u} \mu(d u)-x\right)^{+} \mid S_{t}=1\right] $$ with $\mu(d u)=T^{-1} I_{[0, T]}(u) d u$

and $$M_{t} = \mathbb{E}\left[\left(\int_{0}^{T} S_{u} \mu(d u)-K\right)^{+} \mid \mathcal{F}_{t}\right]$$ is a martingale and we can prove easily that $$M_{t}=S_{t} \phi(t, \xi_{t})$$ with $$\xi_{t} \equiv \frac{K-\int_{0}^{t} S_{u} \mu(d u)}{S_{t}}$$. Moreover, we have : $$dM=S[r \phi+\dot{\phi}-\left(\rho_{t}+r \xi\right) \phi^{\prime}+\frac{1}{2} \sigma^{2} \xi^{2} \phi^{\prime \prime}] d t$$. Thus If we now write $f(t, x) \equiv e^{-r(T-t)} \phi(t, x)$, we find that $f$ solves

$$ \dot{f}+\mathcal{G} f=0 $$

where $\mathcal{G}$ is the operator

$$ \mathcal{G} \equiv \frac{1}{2} \sigma^{2} x^{2} \frac{\partial^{2}}{\partial x^{2}}-\left(\rho_{t}+r x\right) \frac{\partial}{\partial x} $$

The boundary conditions depend on the problem; in the case of the fixed strike Asian option,

$$ f(T, x)=x^{-}=-max(x,0) $$

I have to resolve this PDEs because using f, i can find the value of a Asian call.

However, here my probleme, i don't undestand how to find the Dirichlet conditions, i only have the one for the first variable of f. I don't even know from where to where the variable x varies.

I must be missing something, if someone could help me, please.

Thanks in advance.

## Answer by Andrea (score 0)

https://quant.stackexchange.com/a/81112

First of all $x^-=-\min(x, 0)$ and it is $\ge0$.

Then, you need an initial condition which is exactly that $f(T,x)=x^-$ for each $x$.

And now you need 2 boundary conditions for the lower and highest $x$ for each $t$.

Towards the bottom of page 4, you can see

$\phi(t,x)=r^{-1}(e^{r(T-t)}-1)-x$, for large negative $x$, which you can use for the lowest bound.

And for the highest bound, I think 0 is good enough.