What are the boundary conditions for the Forward contract PDE?

Rabelais

Member
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 \(F-K\exp(-rT)\)?
 

Attachments

  • boundary_cond.zip
    902 bytes · Views: 5

Rabelais

Member
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

 

Rabelais

Member
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
Code:
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

C++ author, trainer
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).

 
Last edited:

JohnLeM

New Member
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.
 
Top