# CRR model arbitrage free

I'm currently studying this proof

In this proof the author defines a probability measure

$$P^*[\{\omega\}]=(p^*)^{k(\omega)}(1-p^*)^{T-k(\omega)}$$ on $$\Omega=\{\omega=(y_1,\ldots,y_T)|y_i=\pm1\}$$

where $$p^*=(r-a)/(b-a)$$ and $$k(\omega)$$ is the number of ones in $$\omega$$.

$$a.

Unfortunately I can't prove that $$P^*$$ is indeed a probability measure.

$$P^*[\{\omega\}]\ge0$$ is clear.

I don't see $$P^*(\Omega)=1$$

$$P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)}$$, but I don't know how to continue

## 1 Answer

If you consider $$\omega$$ and $$\tilde{\omega}$$ with $$k(\omega)=k(\tilde{\omega})$$ it holds that $$P^*(\omega)=P^*(\tilde{\omega})$$. Now instead of summing up over every $$\omega_i \in \Omega$$ you can sum up from $$n =0 ... T$$ and count the elements with $$k(\omega_i)= n$$. There are $$\dbinom{T}{n}$$ elements in $$\Omega$$ which fullfill $$k(\omega_i)= n$$.

Therefore $$P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)} = \sum_{n=0}^T \dbinom{T}{n} (p^*)^n(1-p^*)^{T-n}=1$$ as a result from the binomial theorem.

• Thanks a lot, that's a clever way. How would you show additivity? – user674879 May 17 '19 at 16:40