What are the boundary conditions for the Forward contract PDE?

[Rabelais]

European call

When solving the PDE for the value $V$ of a European call option under the Black-Scholes model using a finite difference scheme, we have that

• Initial/terminal condition. $V(S_T,T) = \text{payoff}(S_T) = \max(S_T-K,0)$ (initial since the scheme is solved backward, terminal since it holds at the final time $T$)
• Left end boundary condition. If $S_0=0$ then $S_t=0$ for all $t$ and the option will surely be out of the money with payoff $\max(0-K,0)=0$ and $V(0,t)=0$ for all $t$
• Right end boundary condition. If $S_0$ is large enough (w.r.t. to $K$) then $S_t>K$ for all $t$ and the option will end up in the money with payoff $\max(S_t-K,0)=S_t-K$ and $V(S_t,t)=S_t-Ke^{-r(T-t)}$ for all $t$

Forward contract

How to deduce the boundary conditions in a similar manner when solving the PDE for the value of a forward contract whose underlying follows the Schwartz mean reverting model? What I understand until now about forwards is

• There is no money exchanged when signing a forward contract so, using the notation below, I think that the value of the forward at time 0 is $F(S_0,0)=0$
• The payoff (of the option equivalent to this forward contract) is $S_T-K$ since the holder is obliged to buy the underlying at expiry

The Schwartz model is $dS = \alpha(\mu-\log S)Sdt + \sigma S dW$ with $\alpha$ speed of mean reversion. From this eq it is possible to derive the PDE for the value of the forward
$$\tag1 \frac{\partial F}{\partial t} + \alpha\Big(\mu-\lambda -\log S\Big)S\frac{\partial F}{\partial S}+\frac12\sigma^2S^2\frac{\partial^2F}{\partial S^2} = 0, \qquad \text{with }\lambda = \sigma\frac{\mu-r}\alpha$$ whose solution is, letting $\tau=T-t$
$$\tag2 F(S_t,\tau) = \mathbb E[S_t] = \exp\bigg(e^{-\alpha\tau}\log S_t +\Big(\mu-\frac{\sigma^2}{2\alpha}-\lambda\Big)(1-e^{-\alpha\tau})+\frac{\sigma^2}{4\alpha}(1-e^{-2\alpha\tau})\bigg)$$

• Initial/terminal condition. Plugging $\tau=0$ (ie $t=T$) in $(2)$ we get $F(S_T,0) = \exp(\log S_T) = S_T$. This value appears also in the original paper by Schwartz (at page 5), so it is correct.
• Left end boundary condition. As in the European call case, if $S_0=0$ then $S_t=0$ for all $t$ and from $(1)$ we get $\dfrac{\partial F}{\partial t}=0$ ie $F$ does not change in time, and since as said before $F(S_0,0)=0$ (not sure though) it follows $F(0,t)=0$ for all $t$. But this means that the payoff of the option would be $0-K$ that is negative, since prices cannot be negative how to deal with this fact?
• Right end boundary condition. If $S_0$ is large enough (w.r.t. to $K$) then $S_t>K$ for all $t$ and the payoff of the option will be $F(S_t,t)-K$ and $V(S_t,t)=(F(S_t,t)-K)e^{-r(T-t)}$ for all $t$, but what can we say about the value of $F(S_t,t)$?

At the end of the finite difference scheme, to obtain the value of the option at time 0 we compute $V(S_0,0) = (F(S_0,0)-K)e^{-rT}$ which in general is not 0, in contrast with what I said above about $F$ being 0 at time 0. What is wrong in the reasoning?

Code

What follows is the Matlab code that computes the exact value of the option at time 0 (V_exact in the code) and the value approximated by the Euler explicit finite difference scheme (V_euler). The initial/terminal condition is applied at line 26 (F = ST), the next two lines are for the left end condition (F(1) = 0) and the right end condition (I don't know what to put here).

Moreover, I'm not sure that V_exact is computed correctly, should be $(F-K)\exp(-rT)$ or $F-K\exp(-rT)$?

 

[Rabelais]

This is the plot with spot prices at time T (vector ST in the code) on x-axis and option prices at time 0 ((F-K)*exp(-r*T) in the code) on y-axis. As you can see the there is a problem when the option price approaches the biggest value of the spot price, since the curve goes from linear to exponential, I guess this is due to the fact that the right end condition is missing

 

[Daniel Duffy]

why post the same question on 2 forums (btw I am Cuchulainn).

 

[Rabelais]

why post the same question on 2 forums (btw I am Cuchulainn).

I did not know that there are the same people on the 2 forums

 

[Daniel Duffy]

I did not know that there are the same people on the 2 forums

Just me!
Anyways, how is it working out?

 

[Rabelais]

Just me!
Anyways, how is it working out?

I confirm you that when using the exact solution at Smax as BC then the plot is correct, ie it is a straight line.
I'm now trying to implement the BC $F_{SS}=0$, I see that you are talking about the tridiagonal matrix, but since I'm using Euler explicit i did not use the matrix form but I just do

for N = 1:n
    F(2:end-1) = a.*F(1:end-2) ...
               + b.*F(2:end-1) ...
               + c.*F(3:end);
end

that is F is a vector, and for example F(1:end-2) is the vector F without the last 2 elements.
Is it possible to implement that BC also in this case? Thank you very much for support

 

[Daniel Duffy]

Taking the exact solution as "fallback" is a good test to see is the full assembled scheme is OK.
I don't use $F_{SS}=0$ myself (I use domain transformation or price a put) but I reckon it should work. In general, the matrix will no longer be tridiagonal.

A possibly good option is Neumann BC at boundary: $F_{S}=1$ since F is essentially S -K there.


Here's a scheme that uses Neumann BC, might be worth investigating (N.B. they assume convection = 0).

http://wwwf.imperial.ac.uk/~ajacquie/IC_Num_Methods/IC_Num_Methods_Docs/Literature/Reports/LocalMart_Report.pdf

 

[JohnLeM]

I agree with Daniel : true boundary conditions are good, but just for testing purposes, because you will end having a payoff-dependent pricer.
F_SS = 0 is better, but you will loose some nice properties doing so.
If you are working with schwartz model, try F_S= 0 (Neumann), it might work.