Binomial Model Probabilities

Joachim Kuczynski, 24 October 2023

Introduction

In this article I want to derive the explicit relationship between an option value and the probability of occurrence of its event states in the binomial model of Cox, Ross and Rubinstein. In many cases I read that the risk neutral probabilities and therefore the option value do not depend on the probabilities of the real state values. But the options values depend on them implicitly. That is what I will derive in this post.

Binomial Model by Cox, Ross and Rubinstein

Options can be valued with the binomial model from Ross, Cox and Rubinstein. The value C_0 of an option at time t_0 is given by:

    \[C_0=\frac{\alpha C_{u,t_1}+(1-\alpha )C_{d,t_1}}{(1+r)^T }\]

C_{u,t_1} and C_{d,t_1} are the option values of the up and down development at time t=1. r is the risk free rate and T is the time between t_0 and t_1, T=t_1-t_0. \alpha is the risk neutral probability of the up movement in t_1, 1-\alpha is the risk neutral probability of the down movement in t_1. The binomial model provides the following relationship:

    \[\alpha=\frac{(1+r)^T-d}{u-d}\]

Including \alpha provides this expression for C_0

    \[C_0=\frac{\frac{(1+r)^T-d}{u-d} C_{u,t_1}+(1-\frac{(1+r)^T-d}{u-d} )C_{d,t_1}}{(1+r)^T }\]

Hence we obtain:

    \[C_0=\frac{( (1+r)^T-d ) C_{u,t_1}+(u-(1+r)^T )C_{d,t_1}}{(1+r)^T (u-d)}\]

u and d are defined as ratio of up and down movement in relation to the expected value in t_0, EV(S_{t_0}):

    \[u= \frac{EV(S_{t_0})}{S_{u,t_1}}\]

    \[d= \frac{EV(S_{t_0})}{S_{d,t_1}}\]

Up to now the probabilities of up state S_{u,{t_1}} and down state S_{d,{t_1}} have not occured. Many times that leads to the argument that these probabilities do not influence the option value. But that is not true. The expected value of the state S_{t1} and therefore S_{t0} depends on the probabilities. The expected value of the event state in t_0 is the discounted value of event state in t_1. With D as yearly constant discount rate we get:

    \[EV(S_{t_0})=\frac{EV(S_{t_1})}{(1+D)^T}=\frac{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}{(1+D)^T}\]

For u and d we get the following:

    \[u=\frac{EV(S_{t_0})}{S_{u,t_1}}=\frac{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}{S_{u,t_1}(1+D)^{T}}\]

    \[d=\frac{EV(S_{t_0})}{S_{u,t_1}}=\frac{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}{S_{d,t_1}(1+D)^{T}}\]

As final result we obtain:

    \[C_0=\frac{( (1+r)^T-\frac{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}{S_{d,t_1}(1+D)^{T}}) C_{u,t_1}}{(1+r)^T ( \frac{{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}}{{S_{u,t_1}(1+D)^{T}}}-\frac{{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}}{{S_{d,t_1}(1+D)^{T}}})}+\]

    \[+\frac{(\frac{{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}}{{S_{u,t_1}(1+D)^{T}}}-(1+r)^T )C_{d,t_1}}{(1+r)^T ( \frac{{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}}{{S_{u,t_1}(1+D)^{T}}}-\frac{{pS_{u,{t_1}}+(1-p)S_{d,{t_1}}}}{{S_{d,t_1}(1+D)^{T}}})}\]

This is the basic relationship between the value of an option at a time t_0 and explicit problem specific variables.

Conclusion

We realize that the option value C_0 expicitely depends on the probability p of the up state S_{u,t_1}, and 1-p of the down state S_{d,t_1} respectively. That is what we wanted to prove. The argument that this dependency does not exist, does not take into account that the value of state S_0 depends on the state probabilities in t_1. Hence there is no disappearance mystery of real life or real states probabilities in options valuation. q.e.d.

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