When Matching Assets to Risk Sources Completes a Market

# Extending an incomplete market to generate a complete one


# Extending an incomplete market to generate a complete one












I am asking a question related to some comments and answers I have seen in the site while investigating characteristics of incomplete markets $-$ see for example @AFK 's answer in How to choose a risk-neutral measure when the market is incomplete?.

Let's assume we have a market $\mathcal{M}$ with $n$ tradeable assets $S \equiv (S^{(1)}, \cdots, S^{(n)})$ and $n+1$ sources of risk, e.g. Brownian motions $W \equiv (W^{(1)}, \cdots, W^{(n+1)})$ $-$ for example assets up to $n-1$ might follow a straightforward Geometric Brownian Motion (GBM), while the $n^{th}$ asset has Heston-like stochastic volatility dynamics. This market is incomplete because the number of (tradeable) assets is lower than the number of (sources of) risks.

Now, WLOG let's assimilate risk sources to Brownian motions and let $W^{(n+1)}$ be the "extra" source of risk. Assume we have built the market model to price a derivative security $f \equiv f(W^{(n+1)})$, meaning that the derivative payoff depends on the risk source $W^{(n+1)}$:

- If we now extend our market, let's call it $\mathcal{M}_{+f}$, by including $f$ as a tradeable asset, then is $\mathcal{M}_{+f}$ complete?

- Is it as simple as saying: "by including $f$, the number of assets is equal to the number of risk factors hence the market $\mathcal{M}_{+f}$ is complete"?

For example, as mentioned above, AFK says in his answer:

> A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete.

Another example is this post, How to prove that markets are incomplete under the Stochastic Volatility model?, see @pbr142 's answer $-$ link to the paper mentioned:

> The paper by Marc Romano and Nizar Touzi, Section 3, contains a general proof that a stochastic volatility model cannot be complete in the sense that the addition of the option completes the market (in the sense of Harrison and Pliska) generated by the underlying and risk-free borrowing/lending.

## Answer by Daneel Olivaw (score 0, accepted)

https://quant.stackexchange.com/a/33495

I believe I have found an exact answer to my question in Thomas Bjork's book, Arbitrage Theory in Continuous Time, on page 122 (third edition):

> Meta-theorem 8.3.1 Let $M$ denote the number of underlying traded assets in the model excluding the risk free asset, and let $R$ denote the number of random sources. Generically we then have the following relations: The model is arbitrage free if and only if $M \leq R$. The model is complete if and only if $M \geq R$. The model is complete and arbitrage free if and only if $M=R$.