0
$\begingroup$

I'm considering an extension of the binomial model where the risky asset can take three values at each node, that is $ S_{t+1}=\left\{ \begin{array}{ll} S_t\cdot u\\\nonumber S_t\cdot c\\ S_t\cdot d \end{array} \right.$

with $0<d<c<u$

If we consider $r\in]d,u[$ the market is arbitrage free but for sure it is not complete. I don't think I can find a unique price for the contingent claim so my question is what is possible ? I tried to solve the system by backward induction to find a hedging strategy but the system has no solution . By the way, the fact that we add a third value invalid all we have about the price of the contingent claim as an expectation since the binomial representation is broken ?

Thank you a lot

$\endgroup$
2
  • 3
    $\begingroup$ Don't worry, nothing is broken. You're describing a trinomial tree. $\endgroup$
    – Kevin
    Commented Jun 7, 2022 at 14:47
  • 1
    $\begingroup$ You can have a look at this answer, maybe it helps you. $\endgroup$ Commented Jun 7, 2022 at 15:01

1 Answer 1

2
$\begingroup$

In all brevety: The model has 6 degrees of freedom:

$$p_u,p_c,p_d, u,c,d$$

We have the following two 'natural' constraints:

$$ p_u+p_c+p_d=1,\quad\quad E^{\mathbb{Q}}(S_{t+\Delta t})=F_{t+\Delta t} $$

leaving four d.o.f. Adding the constraints $u=1/d$ and $c=1$ induces a recombining tree that grows polynomially instead of exponentially. This leaves us with two degrees of freedom. Commonly, we want to moment-match the not only the first but also the second moment of the distribution of $S_{t+\Delta t}$,

$$Var(S_{t+\Delta t})=\Delta_tS_t\sigma^2$$

This leaves the modeler with one degree of freedom. You may close this d.o.f. so that it solves one additional requirement, i.e. the quality of the variance approximation or a higher moment of the distribution. Canonically, $u=e^{\sigma\sqrt{2\Delta t}}$ is chosen.

Note that this choice for $u$ results from a trade-off between convenience and accuracy. At any point, you can simply solve for the four degrees of freedom directly using some numerical multidimensional root solver.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.