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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.

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