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

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

1 Answer 1


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.


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