# completeness of the binomial model - proof

I am reviewing the steps of proof that the binomial model is complete and don't understand the marked in red transition. Could anybody explain this step?

If $P^{**}$ is a risk-neutral measure, so that

$E^{**}[\bar{S}_{n+1} | \ F_n] = \bar{S}_n \ \$ for all n

So given the structure of the model,

$\frac{1}{(1 + r)^{n+1}}E^{**}[S_{n+1} | \ F_n]=\frac{1}{(1 + r)^{n+1}}(uS_nP^{**}[R_{n+1}=u \ | \ F_n]+dS_nP^{**}[R_{n+1}=d \ | \ F_n])=$

$=\color{red}{\bar{S}_{n} \{ d+(u-d)\ P^{**}[R_{n+1}=u \ | \ F_n] \} }$

by martingale condition
$\frac{1}{1+r} \{ d+(u-d)\ P^{**}[R_{n+1}=u \ | \ F_n] \}=1$

so $P^{**}[R_{n+1}=u \ | \ F_n]=\frac{1+r-d}{u-d}=p^*$

$P^{**}=P^{*}$

Therefore the binomial model is complete.

• Please explain your notation. What is $P*$? $\bar{S}$ is the discounted stock price? Where is a reference to the above notation? In the red part, where does the curly bracket end? – Richard Jan 19 '16 at 11:40

What if you write $$P[R_{n+1} = d|F_n] = 1 - P[R_{n+1} = u|F_n] ?$$ Let us write $P(u) = P[R_{n+1} = u|F_n]$ Then the part to show is $$u \bar{S}_n P(u) + d \bar{S}_n (1-P(u))$$ and this $$\bar{S}_n \left(d +(u-d)P(u) \right),$$ where we just expanded terms and then extracted the coefficients.