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Consider a Black Scholes model with two risky assets that are driven by the same $W$-process, and then 1 risk-free asset.

When is this model arbitrage-free and complete?

We have only 1 driving Wiener process, so it does not fit into the theory I learned from Bjork, chapter 13.

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    $\begingroup$ Homework? What are your thoughts? $\endgroup$ – LocalVolatility Jan 22 '17 at 3:19
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You must have $ r= \mu_1 = \mu_2$ for arbitrage-ness. The completeness follows because $n < k$ (k is the number of random sources, n is the number of risky assets).

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    $\begingroup$ There is no need to become insulting here. I think it is perfectly legitimate to ask for an attempt. Especially, when the question sounds like it wasn't formulated by the one posing it. $\endgroup$ – LocalVolatility Jan 22 '17 at 9:34
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    $\begingroup$ I also disagree with your answer on a technical level: 1) It is not necessary for $\mu_1 = \mu_2 = r$ under $\mathbb{P}$. Of course this has to hold under $\mathbb{Q}$ if such a $\mathbb{Q}$ exists - but this was exactly the question. 2) You say that completeness follows since the number of risky assets is less than the number of random sources. First, this usually indicates that the market is incomplete (except for special cases). Second, we have the exact opposite case here. $\endgroup$ – LocalVolatility Jan 22 '17 at 9:44
  • $\begingroup$ Tone it down Chamin. @LocalVolatility is quite right that we normally like to see an attempt first. $\endgroup$ – Bob Jansen Jan 22 '17 at 10:01

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