Completeness in Black–Scholes Models with Multiple Brownian Drivers

Summary

The discussion relates market completeness to the number of traded risky assets and independent sources of randomness. In the standard Black–Scholes setting, one risky asset is driven by one Brownian motion, so the example has matching counts. The answer then considers a model with two risky assets driven by two independent Brownian motions and concludes that it is complete under the same count-based reasoning.

The count rule is presented as a generic heuristic: matching asset and risk-source counts is associated with a market that is both arbitrage-free and complete, while fewer assets than sources suggests incompleteness. Counts alone are not a full test, however. In diffusion models, the relevant condition involves the rank of the assets’ exposure matrix, along with assumptions about admissible trading and no arbitrage. The two sample assets have distinct exposure vectors, but the response does not develop the rank argument or discuss those assumptions, so its broad claims should be read with that limitation.

Key ideas

• Completeness concerns whether traded assets can replicate contingent claims driven by the model’s randomness.
• The standard one-asset Black–Scholes model pairs one risky asset with one Brownian driver.
• The answer uses the number of traded assets and random sources as a generic completeness heuristic.
• In multi-factor diffusion models, the rank of asset exposures matters in addition to the counts.
• The examples do not specify all assumptions needed to establish no arbitrage and completeness.

Tags

# Is Black-Scholes complete?


# Is Black-Scholes complete?












> If we have a Black-Scholes model $B_t = \exp{(rt)}$ and $S_t = S_0\exp{(\sigma W_t + \mu t)}$ then is it complete? What if $W_1$ and $W_2$ are independent Brownian motions. Then the two-stage Black-Scholes model $$B_t = \exp{(rt)}$$ $$S_1(t) = \exp{(W_1(t) + W_2(t) + t)}$$ $$S_2(t) = \exp{(W_1(t) + 2W_2(t) + 2t)}$$ is complete?

I know that we have a completeness if there is a unique martingale measure but I am not sure if this is the case for these two models.

## Answer by user16651 (score 3, accepted)

https://quant.stackexchange.com/a/31519

Meta-theorem : 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 $M \le R$ .

- The model is complete if and only if $M \ge R$.

- The model is complete and arbitrage free if and only if $M = R$.

In the Black–Scholes model, we have one underlying asset $S_t$ plus the risk free asset so $M = 1$. Also we have one driving Wiener process, giving us $R = 1$, so in fact $M = R$.

In the second model, we have two underlying assets $S_1(t)$ and $S_2(t)$ plus the risk free asset so $M = 2$. Also we have two driving Wiener process, giving us $R = 2$, so in fact $M = R$.

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