# About the definition of a complete market

In Steven Shreve's book "Stochastic Calculus for Finance 2", Definition 5.4.8 says a market is complete if every derivative security can be hedged. What exactly does every derivative security mean? The book up to Ch.5 has only considered European options. But in the proof of Theorem 5.4.9, it constructs a derivative security whose payoff $V(T)$ is path dependent. It looks like this is a case for American option. Is it that every payoff $V(T)$ which is a measurable function in $\mathcal{F}(T)$ is a derivative security, and every derivative security can be defined in such a way?

The best way to understand is to go back to one-period binomial option pricing. This is also discussed in Shreve's: "Stochastic Calculus for Finance 1" book.

There's a nice article on one-period binomial option pricing here. You will need to understand:

• Although there are infinite ways to assign a one-period probability, there is a unique solution to the risk-neutral probability. Anything else is not a risk-neutral probability.
• If we have the unique risk-neutral probability, we know there won't be arbitrage opportunity.
• If we know there won't be arbitrage opportunity, we know we can hedge (or replicate) any payoff.

As you can see, they are linked. If you can't hedge something, there is no risk-neutral probability, and there is arbitrage. Furthermore, if there is arbitrage you can't really price the option.

Now, let's go back to your question. Recall continuous option pricing is really just infinite one-period binomial trees. We don't talk about discrete probability anymore, we use now use the term probability measure (e.g. risk neutral probability measure).

Complete model basically means everything we just said:

• It is a complete model if we can hedge (replicate) everything
• If we can hedge everything, there is an unique risk-neutral probability (Second fundamental theorem - Theorem 5.4.9 in the book)
• If we have a unique risk-neutral probability, there is no arbitrage (First fundamental theorem of asset pricing - Theorem 5.4.7 in the book)

Shreve then talks about Radon-Nikodym derivative, which is just a way to relate the real and risk-neutral measure.

So, when the book says "complete market", it means a perfect market where all risk factors can be hedged perfectly, no transaction costs, no surprise, we can price an option (Black-Scholes), and it's fake (our market is not complete).

• The OP asked for the definition of a derivative security. This is not addressed by your answer. – g g Sep 22 '16 at 19:32
• @gg no he didn't – SmallChess Sep 22 '16 at 21:43

The precise definition of a derivative security, in the framework used by Shreve is the following: let $S(t)$ denote a security or a family of securities (in the multidimensional market model, that is $S(t) = (S_1(t), \cdots, S_n(t))$. Then a derivative security is determined at time $T$ by the condition of having a payoff $V(T)$, which can be expressed as $V(T)=g(S(T))$, with $g$ an $\mathcal{F}(T)$-measurable function.

This is the mathematically correct definition, so that in particular for a European call option with strike $K$ you have $g(x)=(x-K)^+$, and so on. Now, look at the proof of Theorem 5.4.9. in this case you have a payoff defined as $V(T)=\mathbf{1}_A\cdot (D(T))^{-1}$. In this case the security $S(T)$ is just $D(T)$, and $g(x)$ is measurable because $A$ is clearly an $\mathcal{F}(T)$-measurable set.

However, you're right that Shreve never explictly defines in the book what he precisely means by "derivative security".

• I disagree. Shreve has made crystal clear what he means by "derivative security". – SmallChess Jul 29 '16 at 14:12
• In what page of the book, precisely? – RandomGuy Jul 29 '16 at 14:20
• @RandomGuy A $\mathcal{F}(T)$ measurable set $A$ does not only depend on the eventual value of the stock price $S(T)$, but also depend on the path of $S$, right? In your definition, the function $g$ only depends on the eventual price $S(T)$. This is why I thought that the way he constructed $A$ in his proof of 5.4.9 has very different meaning than the european option case. Can you help me to understand this? Thank you. – Resorter Aug 21 '16 at 3:35
• @Resorter: You are right and RandomGuy's $V(T)=g(S(T))$ is wrong. The best (or maybe better: most general) way to define a derivative is indeed to say $f$ is $\mathcal{F}(T)$ measurable. That said, you or the author might want to impose more constraints beyond mere measurability such as $f$ being in $L^1$ or $L^2$ or such. – g g Sep 22 '16 at 19:31