1
$\begingroup$

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:

3

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?

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
0
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.