Deriving Risk-Neutral Probabilities in the Binomial Model
Summary
The document connects the fundamental theorem of asset pricing to a one-period binomial tree. Choose the probability of an up move so that the stock price discounted by the bond is a martingale under the resulting measure. Solving this condition gives the risk-neutral up probability as the excess of the gross risk-free return over the down factor, divided by the difference between the up and down factors; the down probability is its complement.
The answer links absence of arbitrage to the risk-free growth factor lying between the tree’s down and up factors. In that case, the risk-neutral probabilities are positive and unique in the two-outcome model. Under this measure, discounted option values can be expressed as expected discounted payoffs. The explanation is conceptual and brief: it does not develop multi-period trees, prove the general theorem, or discuss practical calibration of tree parameters.
Key ideas
- Risk-neutral probabilities are chosen to make the discounted underlying price a martingale.
- In a two-branch binomial model, the martingale condition and probabilities summing to one determine a unique pair of probabilities.
- No-arbitrage requires the gross risk-free return to lie between the down and up factors.
- Option values under the risk-neutral measure are expectations of discounted payoffs.
- The explanation applies directly to a simple binomial model and does not derive the full general theorem.
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# How is the fundamental theorem of asset pricing used?
# How is the fundamental theorem of asset pricing used?
I know that a multi-period market model is complete and arbitrage free if there's a unique equivalent martingale measure. The thing is, I have absolutely no clue how to apply this theorem to a simple binomial tree. I just don't get what the two things even have to do with one another.
For example, consider:
Yes, I know that $u = 1.1$. I know that $d = 0.9$. But what does this have anything to do with the complicated theorem which talks about conditional expectations and equivalent martingale measures? I guess $q = (R - d)/(u-d)$ and $1 - q$ is this equivalent martingale measure but why? And why is it unique?
## Answer by Mark Joshi (score 2)
https://quant.stackexchange.com/a/26261
well solve for the value of $q$ that makes the value of the stock divided by the bond be a martingale. You will find that only one value does so. It is the one you posted.
If you then define the discounted value of an option to be its expectation of the discounted pay-off, its discounted value is a martingale.
So the discounted value of everything is a martingale. Arbitrages cannot be martingales since a martingale of zero initial value has expectation zero. So there are no arbitrages under the q measure. But the set of arbitrages doesn't change with probability so there are no arbitrages under any equivalent measure.
(my book "concepts" goes into this stuff in gory detail.)
## Answer by user16651 (score 1)
https://quant.stackexchange.com/a/26266
We know the market model is arbitrage free if and only if there exists a martingale measure $Q$, also the Binomial Model is free of arbitrage if and only if $d\le 1+R\le u\,\,$ (Arbitrage Theory in Continuous Time).It is easy to calculate the martingale probabilities.This Condition is equivalent to saying that $1 + R$ is a convex combination of $u$ and $d$ , i.e. $$1+R=u\,{{q}_{u}}+d\,{{q}_{d}}$$ On the other hand $\,{{q}_{d}}+{{q}_{u}}=1$ then $$\left\{ \begin{align} & u\,{{q}_{u}}+d\,{{q}_{d}}=1+R \\ & {{q}_{u}}+\,{{q}_{d}}=1 \\ \end{align} \right.$$ $$\Rightarrow \left\{ \begin{align} & {{q}_{u}}=\frac{\left| \begin{matrix} 1+R & d \\ 1 & 1 \\ \end{matrix} \right|}{\left| \begin{matrix} u & d \\ 1 & 1 \\ \end{matrix} \right|}=\frac{1+R-d}{u-d} \\ & {{q}_{d}}=\frac{\left| \begin{matrix} u & 1+R \\ 1 & 1 \\ \end{matrix} \right|}{\left| \begin{matrix} u & d \\ 1 & 1 \\ \end{matrix} \right|}=\frac{u-(1+R)}{u-d} \\ \end{align} \right.$$
(Also you can read the second edition of Joshi's Book , it is so good.)Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.