Deriving the Risk-Neutral Up Probability in the Binomial Model

Summary

The discussion explains an algebraic step in proving that the one-period binomial model has a unique risk-neutral probability. At each step, the discounted asset price is a martingale, while the next asset value can move to an up or down state. Since the two conditional probabilities sum to one, the expected next value can be rewritten as the current discounted price multiplied by the down factor plus the difference between the up and down factors times the up probability.

Imposing the martingale condition determines that probability uniquely as the risk-free gross return adjusted by the down factor and divided by the spread between the up and down factors. The answer focuses on expanding and collecting terms; it does not discuss the assumptions needed for an arbitrage-free binomial model or broader proofs of market completeness. Its main lesson is the simple algebra behind the probability calculation.

Key ideas

• The conditional probabilities of the up and down states sum to one.
• Substituting one probability as the complement of the other makes the expectation easier to simplify.
• The martingale condition determines the risk-neutral up probability from the model’s return factors.
• A unique risk-neutral probability in the binomial setting supports the completeness conclusion.

Tags

# completeness of the binomial model - proof


# 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.

## Answer by Richi Wa (score 1, accepted)

https://quant.stackexchange.com/a/22826

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.

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