# 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.

• Why did you delete your other question ( $X=S_t^3)$ – user16651 Dec 15 '16 at 19:18
• Sorry must have been an error I'll put it back up – Wolfy Dec 15 '16 at 19:20
• My Answer was true :) – user16651 Dec 15 '16 at 19:21

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 $M = R$.