In Tomas Björk's Arbitrage Theory in Continuous Time, there exists this proposition

enter image description here

It seems that to show that the model is complete, we must show that the claims are reachable, i.e. we must find replicating portfolios for each claim.

Which part of finding the replicating portfolio makes use of the assumption, where I understand the assumption is equivalent to $d < 1+R < u$ (or $d \le 1+R \le u$, but $d<u$)? All I see so far is the $d<u$, but that's nothing really special: of course the stock price $u$ if it would go up from its current price should be higher than the stock price $d$ if it would go down from the its current price.


Here is a simple fact: In your notation, the model under consideration is complete if and only if the matrix

\begin{bmatrix} 1+R & 1+R \\ d & u \end{bmatrix}

is one-to-one, i.e. $d \neq u$. (Equivalently, its transpose is onto, which is what is shown in your quoted text.)

No-arbitrage holds if and only if $d \leq 1+R \leq u$ and $d < u$. (By assumption, $d \neq u$. Otherwise you have two risk-free assets, and model either trivially collapes or trivially admits arbitrage.) Therefore NA implies market completeness, because $d < u$ implies $d \neq u$. Note that the no-arbitrage condition---that the risk-free return lies in the interval of risky return---is not really used.

In general, market completeness under no-arbitrage (uniqueness of risk-neutral measure) is a property in addition to no-arbitrage (existence of a risk-neutral measure). In the binomial case, however, there is really one "degree of freedom" and completeness holds trivially as long as the model does not collapse.

Leaving the binomial setting, this need not be true. For example, consider a model with $n$ states and $n$ assets, with $n > 2$. The model could have no-arbitrage but not be complete. Similarly, consider a model with $n$ states and $m$ assets, with $m > n > 2$. The model could be complete (in fact, with redundant assets) but admits arbitrage.


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.